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Calculus III: Multivariable Calculus Study Notes & Guides

39 AI-generated study notes covering the full Calculus III: Multivariable Calculus curriculum. Showing 10 complete guides below.

Study Guide842 words

Applications of Vectors: Force, Volume, and Geometric Analysis

Applications

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Applications of Vectors: Force, Volume, and Geometric Analysis

This study guide covers the practical utility of vector operations in physical and geometric contexts, specifically focusing on resultant forces and the application of the triple scalar product.

Learning Objectives

  • Calculate Resultant Forces: Combine multiple force vectors into a single resultant vector and determine its magnitude and direction.
  • Evaluate Volumes: Use the triple scalar product to calculate the volume of a parallelepiped.
  • Verify Coplanarity: Apply the triple scalar product to determine if three vectors lie within the same plane.
  • Translate Polar to Cartesian: Express vectors defined by magnitude and angle in terms of standard unit vectors i\mathbf{i}i and j\mathbf{j}j.

Key Terms & Glossary

  • Resultant Force: The vector sum of two or more individual forces acting on an object. Example: If two people push a box from different angles, the resultant force determines where the box actually moves.
  • Standard Unit Vectors: The vectors i=⟨1,0,0⟩\mathbf{i} = \langle 1, 0, 0 \ranglei=⟨1,0,0⟩, j=⟨0,1,0⟩\mathbf{j} = \langle 0, 1, 0 \ranglej=⟨0,1,0⟩, and k=⟨0,0,1⟩\mathbf{k} = \langle 0, 0, 1 \ranglek=⟨0,0,1⟩ that define the axes. Example: Describing a 3D force as 5i+2j5\mathbf{i} + 2\mathbf{j}5i+2j.
  • Triple Scalar Product: An operation involving both a dot product and a cross product, written as u⋅(v×w)\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})u⋅(v×w). Example: Calculating the internal capacity of a slanted storage container.
  • Parallelepiped: A three-dimensional figure formed by six parallelograms. Example: A leaning box or a crystal structure lattice.
  • Coplanar: A set of vectors that lie in the same two-dimensional plane. Example: Three arrows lying flat on a tabletop.

The "Big Idea"

Vectors serve as the bridge between abstract mathematics and the physical world. While a single vector represents a localized quantity (like a pull or a velocity), their combinations reveal the total impact on a system. In physics, we sum vectors to find equilibrium or net motion. In geometry, we combine products to find spatial properties like volume and planar orientation. Understanding these applications is essential for engineering, structural design, and fluid dynamics.

Formula / Concept Box

ConceptFormulaApplication
Component Formv=∥v∥cos⁡θi+∥v∥sin⁡θj\mathbf{v} = \|\mathbf{v}\| \cos \theta \mathbf{i} + \|\mathbf{v}\| \sin \theta \mathbf{j}v=∥v∥cosθi+∥v∥sinθjConverting magnitude/angle to vectors
Resultant VectorFnet=F1+F2+⋯+Fn\mathbf{F}_{net} = \mathbf{F}_1 + \mathbf{F}_2 + \dots + \mathbf{F}_nFnet​=F1​+F2​+⋯+Fn​Finding total force on an object
Volume of Parallelepiped$V =\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})
Coplanarity Checku⋅(v×w)=0\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = 0u⋅(v×w)=0Determining if volume is zero (flat)

Hierarchical Outline

  1. Vector Addition in Physics
    • Resultant Force: Summation of individual vectors.
    • Directional Angles: Using arctan⁡(y/x)\arctan(y/x)arctan(y/x) to find the final heading.
  2. The Triple Scalar Product
    • Definition: Combining dot and cross products.
    • Geometric Interpretation: The absolute value represents volume.
  3. Advanced Spatial Geometry
    • Parallelepipeds: Shapes defined by three non-parallel vectors.
    • Coplanar Analysis: Identifying when three vectors fail to form a 3D volume.

Visual Anchors

Force Addition Logic

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Parallelepiped Geometry

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Definition-Example Pairs

  • Term: Resultant Force
    • Definition: The vector sum of all forces acting on a body.
    • Real-World Example: Two tugboats pulling a large ship at different angles; the ship moves in the direction of the resultant force.
  • Term: Triple Scalar Product
    • Definition: The dot product of one vector with the cross product of two others.
    • Real-World Example: Determining the stability of a tripod; if the vectors representing the legs are coplanar, the tripod will collapse.

Worked Examples

Example 1: Finding Resultant Force

Problem: A truck pulls a car horizontally with 100 lb of force. Two people push the car at an upward angle of 30∘30^\circ30∘ with 50 lb of force. Find the resultant magnitude.

Solution:

  1. Align vectors: Let the truck force be F1=100i\mathbf{F}_1 = 100\mathbf{i}F1​=100i.
  2. Component form for pushing: F2=50cos⁡(30∘)i+50sin⁡(30∘)j=253i+25j\mathbf{F}_2 = 50\cos(30^\circ)\mathbf{i} + 50\sin(30^\circ)\mathbf{j} = 25\sqrt{3}\mathbf{i} + 25\mathbf{j}F2​=50cos(30∘)i+50sin(30∘)j=253​i+25j.
  3. Sum vectors: R=(100+253)i+25j≈143.3i+25j\mathbf{R} = (100 + 25\sqrt{3})\mathbf{i} + 25\mathbf{j} \approx 143.3\mathbf{i} + 25\mathbf{j}R=(100+253​)i+25j≈143.3i+25j.
  4. Magnitude: ∥R∥=143.32+252≈145.5\|\mathbf{R}\| = \sqrt{143.3^2 + 25^2} \approx 145.5∥R∥=143.32+252​≈145.5 lb.

Example 2: Volume of a Parallelepiped

Problem: Find the volume defined by u=⟨1,1,1⟩\mathbf{u} = \langle 1, 1, 1 \rangleu=⟨1,1,1⟩, v=⟨2,0,1⟩\mathbf{v} = \langle 2, 0, 1 \ranglev=⟨2,0,1⟩, and w=⟨0,2,3⟩\mathbf{w} = \langle 0, 2, 3 \ranglew=⟨0,2,3⟩.

Solution:

  1. Cross Product (v×w)(\mathbf{v} \times \mathbf{w})(v×w): v×w=∣ijk201023∣=i(0−2)−j(6−0)+k(4−0)=⟨−2,−6,4⟩\mathbf{v} \times \mathbf{w} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 0 & 1 \\ 0 & 2 & 3 \end{vmatrix} = \mathbf{i}(0-2) - \mathbf{j}(6-0) + \mathbf{k}(4-0) = \langle -2, -6, 4 \ranglev×w=​i20​j02​k13​​=i(0−2)−j(6−0)+k(4−0)=⟨−2,−6,4⟩
  2. Dot Product with u\mathbf{u}u: u⋅⟨−2,−6,4⟩=(1)(−2)+(1)(−6)+(1)(4)=−2−6+4=−4\mathbf{u} \cdot \langle -2, -6, 4 \rangle = (1)(-2) + (1)(-6) + (1)(4) = -2 - 6 + 4 = -4u⋅⟨−2,−6,4⟩=(1)(−2)+(1)(−6)+(1)(4)=−2−6+4=−4
  3. Volume: ∣−4∣=4|-4| = 4∣−4∣=4 units3^33.

Checkpoint Questions

  1. If the triple scalar product of three vectors is exactly 0, what can you conclude about the geometric relationship between those vectors?
  2. How does the resultant force change if one of the individual forces is doubled in magnitude but keeps the same direction?
  3. True or False: The order of vectors in a triple scalar product can change the sign of the result but not the absolute volume.

[!TIP] When solving resultant force problems, always draw a coordinate system first. Aligning one major force with the x-axis simplifies the trigonometry significantly!

▶Click to see Checkpoint Answers
  1. The vectors are coplanar (they lie in the same plane and form no 3D volume).
  2. The resultant vector will shift its direction toward the doubled force and increase in magnitude.
  3. True. Permutations of the triple scalar product may result in a negative value, but the volume is always the absolute value.
Study Guide780 words

Arc Length and Curvature: Multivariable Calculus Study Guide

Arc Length and Curvature

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Arc Length and Curvature

This guide explores how to quantify the distance along a path in space and measure the "sharpness" of its turns using the concepts of arc length and curvature.

Learning Objectives

After studying this material, you should be able to:

  • Calculate the arc length of both plane and space curves using vector-valued functions.
  • Derive the arc-length function s(t)s(t)s(t) and use it to find an arc-length parameterization.
  • Define and calculate the curvature κ\kappaκ of a smooth curve.
  • Understand the geometric significance of the Unit Tangent, Normal, and Binormal vectors.

Key Terms & Glossary

  • Smooth Curve: A curve parameterized by r(t)\mathbf{r}(t)r(t) where r′(t)\mathbf{r}'(t)r′(t) is continuous and r′(t)≠0\mathbf{r}'(t) \neq \mathbf{0}r′(t)=0 for all ttt.
    • Example: A circle is smooth; a curve with a "cusp" (like y2=x3y^2 = x^3y2=x3) is not smooth at the origin.
  • Arc-Length Parameter (sss): A parameter that represents the actual distance traveled along a curve from a starting point.
  • Curvature (κ\kappaκ): A scalar measure of how quickly a curve changes direction at a point.
    • Example: A circle of radius RRR has a constant curvature of $1/R$. A straight line has 0 curvature.
  • Osculating Circle: The "kissing circle" that best approximates a curve at a specific point.

The "Big Idea"

In basic physics, we often describe motion in terms of time (ttt). However, the geometry of a path doesn't care how fast you move; it only cares about the shape. By switching from a time parameter (ttt) to an arc-length parameter (sss), we describe the curve based on its internal geometry. This allows us to define curvature, which tells us how much the unit tangent vector is "twisting" per unit of distance traveled.

Formula / Concept Box

ConceptFormula (Vector Form)Notes
Arc Length$$L = \int_{a}^{b} \\mathbf{r}′(t)∥dt'(t)\| dt′(t)∥dt
Arc-Length Functions(t)s(t) s(t)= \int_{a}^{t} \\mathbf{r}′(u)∥du'(u)\| du′(u)∥du
Curvature (General)κ=∥T′(t)∥∥r′(t)∥\kappa = \frac{\|\mathbf{T}'(t)\|}{\|\mathbf{r}'(t)\|}κ=∥r′(t)∥∥T′(t)∥​Rate of change of Tangent
Curvature (Space)κ=∥r′(t)×r′′(t)∥∥r′(t)∥3\kappa = \frac{\|\mathbf{r}'(t) \times \mathbf{r}''(t)\|}{\|\mathbf{r}'(t)\|^3}κ=∥r′(t)∥3∥r′(t)×r′′(t)∥​Easier for computation

Hierarchical Outline

  1. Arc Length Calculation
    • Plane Curves: $$\int \sqrt{[x'(t)]^2 + [y'(t)]^2}dt dtdt
    • Space Curves: $$\int \sqrt{[x'(t)]^2 + [y'(t)]^2 + [z'(t)]^2}dt dtdt
  2. Arc-Length Parameterization
    • Step 1: Find s(t)s(t)s(t) by integrating speed.
    • Step 2: Solve the resulting equation for ttt in terms of sss.
    • Step 3: Substitute t(s)t(s)t(s) back into r(t)\mathbf{r}(t)r(t).
  3. Curvature Analysis
    • Relation to radius: κ=1/R\kappa = 1/Rκ=1/R.
    • Smoothness requirement: $$\mathbf{r}′(t)'(t)′(t) must not be zero to avoid division by zero.

Visual Anchors

Parameterization Logic

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The TNB Frame

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Definition-Example Pairs

[!IMPORTANT] Definition: Arc-Length Parameterization A curve r(s)\mathbf{r}(s)r(s) is parameterized by arc length if the speed \|\mathbf{r}′(s)∥=1'(s)\| = 1′(s)∥=1 for all sss.

  • Example: Consider \mathbf{r}(t) = \langle 3t$, $4t \rangle.
    • \mathbf{r}$'(t) $= \langle 3$, $4 \rangle, so speed is 32+42=5\sqrt{3^2+4^2} = 532+42​=5.
    • s(t)s(t) s(t)= \int_0^t 5du=5t du = 5tdu=5t.
    • Solving for ttt: t=s/5t = s/5t=s/5.
    • Arc-length version: \mathbf{r}(s) = \langle 3(s/5)$, 4(s/5) $\rangle = \langle \frac{3}{5}s$, $\frac{4}{5}s \rangle.
    • Check: Velocity is \langle 3/5$, $4/5 \rangle, magnitude is 1. Success!

Worked Examples

Example: Arc Length of a Helix

Problem: Find the arc length of the helix \mathbf{r}(t) = \langle \cos t$, $\sin t$, $t \rangle from t=0t=0t=0 to t=2πt=2\pit=2π.

Solution:

  1. Find Derivative: \mathbf{r}$'(t) $= \langle -\sin t$, $\cos t$, $1 \rangle.
  2. Calculate Magnitude (Speed): ∥r′(t)∥=(−sin⁡t)2+(cos⁡t)2+12=sin⁡2t+cos⁡2t+1=1+1=2\|\mathbf{r}'(t)\| = \sqrt{(-\sin t)^2 + (\cos t)^2 + 1^2} = \sqrt{\sin^2 t + \cos^2 t + 1} = \sqrt{1 + 1} = \sqrt{2}∥r′(t)∥=(−sint)2+(cost)2+12​=sin2t+cos2t+1​=1+1​=2​
  3. Integrate: L=∫02π2dt=[2t]02π=2π2L = \int_0^{2\pi} \sqrt{2} dt = \left[ \sqrt{2}t \right]_0^{2\pi} = 2\pi\sqrt{2}L=∫02π​2​dt=[2​t]02π​=2π2​

[!TIP] For a helix, the speed is constant, making the integration very straightforward!

Checkpoint Questions

  1. If a curve has a speed of constant 1, what does the parameter ttt represent?
  2. Why must a curve be "smooth" to calculate curvature using the standard formulas?
  3. What is the curvature of a circle with a diameter of 10 units?
  4. Solve the relationship: if s(t)=3t2s(t) = 3t^2s(t)=3t2, what is the speed of the particle at time t=2t=2t=2?
▶Click to view answers
  1. The parameter ttt represents the arc length sss.
  2. Smoothness ensures the derivative is non-zero, preventing division by zero in the curvature formula.
  3. κ=1/R\kappa = 1/Rκ=1/R. If diameter is 10, R=5R=5R=5, so κ=1/5=0.2\kappa = 1/5 = 0.2κ=1/5=0.2.
  4. Speed v(t)v(t) v(t)= \frac{ds}{dt} = 6t$$. At t=2t=2t=2, speed is 12 units/sec.
Study Guide875 words

Calculus III: Area and Arc Length in Polar Coordinates Study Guide

Area and Arc Length in Polar Coordinates

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Area and Arc Length in Polar Coordinates

This study guide covers the foundational techniques for calculating the geometric properties of curves defined by polar equations, specifically focusing on the area of regions and the length of paths.

Learning Objectives

  • Apply the Formula for Area: Set up and evaluate integrals to find the area of regions bounded by polar curves r=f(θ)r = f(\theta)r=f(θ).
  • Calculate Area Between Curves: Identify intersection points of polar graphs and subtract inner areas from outer areas.
  • Determine Arc Length: Derive and apply the integral formula for the length of a polar curve over a specific interval of θ\thetaθ.
  • Account for Symmetry: Use symmetry to simplify calculations and identify all intersection points, including the pole (origin).

Key Terms & Glossary

  • Polar Sector: A wedge-shaped region bounded by two radii and a polar curve, analogous to a rectangle in Cartesian coordinates.
    • Real-World Example: A slice of pizza is a physical approximation of a polar sector.
  • The Pole: The origin (0,0)(0,0)(0,0) in the polar coordinate system.
    • Real-World Example: The center of a radar screen or the pivot point of a pendulum.
  • Cardioid: A heart-shaped polar curve defined by r=a(1±cos⁡θ)r = a(1 \pm \cos\theta)r=a(1±cosθ) or r=a(1±sin⁡θ)r = a(1 \pm \sin\theta)r=a(1±sinθ).
    • Real-World Example: The cross-section of certain microphone pickup patterns (cardioid microphones).
  • Radial Lines: Lines of constant angle θ=α\theta = \alphaθ=α.

The "Big Idea"

In rectangular coordinates, we approximate the area under a curve y=f(x)y=f(x)y=f(x) using thin rectangles (Area≈∑f(xi)ΔxArea \approx \sum f(x_i)\Delta xArea≈∑f(xi​)Δx). In polar coordinates, we approximate the area of a region bounded by r=f(θ)r=f(\theta)r=f(θ) using thin circular sectors. Since the area of a circular sector is A=12r2θA = \frac{1}{2}r^2\thetaA=21​r2θ, the total area becomes the integral of 12r2\frac{1}{2}r^221​r2 with respect to θ\thetaθ. Essentially, we are "sweeping" a ray from an initial angle to a final angle, accumulating area as we rotate.

Formula / Concept Box

ConceptFormulaNotes
Area of a Polar RegionA=∫αβ12[f(θ)]2dθA = \int_{\alpha}^{\beta} \frac{1}{2} [f(\theta)]^2 d\thetaA=∫αβ​21​[f(θ)]2dθβ−α\beta - \alphaβ−α must be ≤2π\leq 2\pi≤2π.
Area Between CurvesA=∫αβ12([router]2−[rinner]2)dθA = \int_{\alpha}^{\beta} \frac{1}{2} ([r_{outer}]^2 - [r_{inner}]^2) d\thetaA=∫αβ​21​([router​]2−[rinner​]2)dθAlways find intersection points first.
Arc Length of Polar CurveL=∫αβr2+(drdθ)2dθL = \int_{\alpha}^{\beta} \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} d\thetaL=∫αβ​r2+(dθdr​)2​dθDerived from parametric arc length.

Hierarchical Outline

  1. Area of Polar Regions
    • Derivation: Based on the area of a sector A=12r2ΔθA = \frac{1}{2}r^2\Delta\thetaA=21​r2Δθ.
    • Single Curve Area: Integrating 12r2\frac{1}{2}r^221​r2 over the interval [α,β][\alpha, \beta][α,β].
    • Multi-Curve Area: Finding the region trapped between routr_{out}rout​ and rinr_{in}rin​.
  2. Intersection Points in Polar Space
    • Algebraic Solutions: Solving r1(θ)=r2(θ)r_1(\theta) = r_2(\theta)r1​(θ)=r2​(θ).
    • The Pole Exception: The origin may be an intersection point even if r1(θ)=r2(θ)r_1(\theta) = r_2(\theta)r1​(θ)=r2​(θ) has no common solution, as curves may pass through the pole at different values of θ\thetaθ.
  3. Arc Length in Polar Coordinates
    • Parametric Transformation: x=rcos⁡θx = r\cos\thetax=rcosθ, y=rsin⁡θy = r\sin\thetay=rsinθ.
    • The Integrand: Simplifying (dx/dθ)2+(dy/dθ)2\sqrt{(dx/d\theta)^2 + (dy/d\theta)^2}(dx/dθ)2+(dy/dθ)2​ leads to r2+(r′)2\sqrt{r^2 + (r')^2}r2+(r′)2​.

Visual Anchors

Finding Area Between Curves Flowchart

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Figure 1 — Mermaid diagram

Polar Sector Visualization

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Figure 2 — TikZ diagram

Definition-Example Pairs

  • Term: Arc Length Integrand
    • Definition: The expression r2+(r′)2\sqrt{r^2 + (r')^2}r2+(r′)2​ derived from the Pythagorean theorem in differential form.
    • Example: For a circle r=3r = 3r=3, r′=0r' = 0r′=0. The integrand is 32+02=3\sqrt{3^2 + 0^2} = 332+02​=3. The length from 0 to 2π2\pi2π is ∫02π3dθ=6π\int_0^{2\pi} 3 d\theta = 6\pi∫02π​3dθ=6π (the circumference).
  • Term: Symmetry of Polar Curves
    • Definition: Property where a graph is identical across the polar axis (cos), the line θ=π/2\theta = \pi/2θ=π/2 (sin), or the pole.
    • Example: A cardioid r=1+cos⁡θr = 1 + \cos\thetar=1+cosθ is symmetric about the polar axis (xxx-axis). You can integrate from 0 to π\piπ and double the result.

Worked Examples

Example 1: Area of a Cardioid

Problem: Find the area of the region enclosed by r=2(1+cos⁡θ)r = 2(1 + \cos\theta)r=2(1+cosθ).

Solution:

  1. Identify Bounds: The cardioid is traced exactly once as θ\thetaθ goes from 0 to 2π2\pi2π.
  2. Set up Integral: A=∫02π12[2(1+cos⁡θ)]2dθ=2∫02π(1+2cos⁡θ+cos⁡2θ)dθA = \int_{0}^{2\pi} \frac{1}{2} [2(1+\cos\theta)]^2 d\theta = 2 \int_{0}^{2\pi} (1 + 2\cos\theta + \cos^2\theta) d\thetaA=∫02π​21​[2(1+cosθ)]2dθ=2∫02π​(1+2cosθ+cos2θ)dθ
  3. Simplify: Using cos⁡2θ=12(1+cos⁡2θ)\cos^2\theta = \frac{1}{2}(1 + \cos 2\theta)cos2θ=21​(1+cos2θ): A=2∫02π(1+2cos⁡θ+12+12cos⁡2θ)dθA = 2 \int_{0}^{2\pi} (1 + 2\cos\theta + \frac{1}{2} + \frac{1}{2}\cos 2\theta) d\thetaA=2∫02π​(1+2cosθ+21​+21​cos2θ)dθ
  4. Evaluate: A=2[32θ+2sin⁡θ+14sin⁡2θ]02π=2(32⋅2π)=6πA = 2 \left[ \frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta \right]_0^{2\pi} = 2(\frac{3}{2} \cdot 2\pi) = 6\piA=2[23​θ+2sinθ+41​sin2θ]02π​=2(23​⋅2π)=6π

Example 2: Arc Length of a Spiral

Problem: Find the arc length of the spiral r=θr = \thetar=θ for $$0 \leq \theta \leq 1$$.

Solution:

  1. Identify rrr and r′r'r′: r=θr = \thetar=θ, r′=1r' = 1r′=1.
  2. Formula: L=∫01θ2+12dθL = \int_{0}^{1} \sqrt{\theta^2 + 1^2} d\thetaL=∫01​θ2+12​dθ.
  3. Integration: Using the formula \int \sqrt{u^2 + a^2}$ du $= \frac{1}{2}(u\sqrt{u^2+a^2} + a^2\ln|u+\sqrt{u^2+a^2}|): L=[12(θθ2+1+ln⁡∣θ+θ2+1∣)]01=12(2+ln⁡(1+2))L = \left[ \frac{1}{2}(\theta\sqrt{\theta^2+1} + \ln|\theta+\sqrt{\theta^2+1}|) \right]_0^1 = \frac{1}{2}(\sqrt{2} + \ln(1+\sqrt{2}))L=[21​(θθ2+1​+ln∣θ+θ2+1​∣)]01​=21​(2​+ln(1+2​))

Checkpoint Questions

  1. Why do we use 12r2\frac{1}{2}r^221​r2 in the polar area integral instead of just rrr? (Hint: Think about circular sectors).
  2. Find the intersection points of r=1r = 1r=1 and r=2cos⁡θr = 2\cos\thetar=2cosθ. Do they intersect at the pole?
  3. If a curve is symmetric about the polar axis, what interval of integration can you use to find the total area?

[!IMPORTANT] When finding the area between two curves, always check if the curves intersect at the origin by solving r=0r=0r=0 for each equation separately. These points often don't appear in the simultaneous solution r1=r2r_1=r_2r1​=r2​.

Study Guide892 words

Calculus of Parametric Curves: Comprehensive Study Guide

Calculus of Parametric Curves

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Calculus of Parametric Curves: Comprehensive Study Guide

This guide covers the application of differential and integral calculus to curves defined by parametric equations. Instead of treating yyy as a direct function of xxx, we analyze both as functions of an independent parameter, typically ttt.

Learning Objectives

After studying this material, you should be able to:

  • 1.2.1 Determine derivatives and equations of tangents for parametric curves.
  • 1.2.2 Find the area under a parametric curve.
  • 1.2.3 Use the equation for arc length of a parametric curve.
  • 1.2.4 Apply the formula for surface area to a volume generated by a parametric curve.

Key Terms & Glossary

  • Parameter (ttt): An independent variable that determines the coordinates (x,y)(x, y)(x,y) of a curve simultaneously.
  • Parametric Curve: A set of points (x(t),y(t))(x(t), y(t))(x(t),y(t)) generated as the parameter ttt varies over an interval.
  • Tangent Vector: A vector representing the instantaneous direction of motion along a parametric curve.
  • Smooth Curve: A curve where x′(t)x'(t)x′(t) and y′(t)y'(t)y′(t) are continuous and not simultaneously zero.

The "Big Idea"

In standard Cartesian calculus, we describe what a path looks like (y=f(x)y=f(x)y=f(x)). In parametric calculus, we describe how a path is traversed. This allows us to model motion (like a baseball's trajectory) where time is the driving factor, and to analyze complex curves (like loops or vertical segments) that fail the Vertical Line Test.

Formula / Concept Box

ApplicationFormulaCondition
First Derivativedydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt}dxdy​=dx/dtdy/dt​dx/dt≠0dx/dt \neq 0dx/dt=0
Second Derivatived2ydx2=ddt(dydx)dx/dt\frac{d^2y}{dx^2} = \frac{\frac{d}{dt} \left( \frac{dy}{dx} \right)}{dx/dt}dx2d2y​=dx/dtdtd​(dxdy​)​Measures concavity
Arc LengthL=∫ab(dxdt)2+(dydt)2 dtL = \int_{a}^{b} \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \, dtL=∫ab​(dtdx​)2+(dtdy​)2​dtCurve traversed once
Area Under CurveA=∫aby(t)x′(t) dtA = \int_{a}^{b} y(t) x'(t) \, dtA=∫ab​y(t)x′(t)dtx(t)x(t)x(t) is monotonic
Surface Area (xxx-axis)S=∫ab2πy(t)(x′(t))2+(y′(t))2 dtS = \int_{a}^{b} 2\pi y(t) \sqrt{(x'(t))^2 + (y'(t))^2} \, dtS=∫ab​2πy(t)(x′(t))2+(y′(t))2​dty(t)≥0y(t) \geq 0y(t)≥0

Hierarchical Outline

  • I. Differentiation of Parametric Equations
    • A. Slope of Tangent Lines: Calculated by the ratio of vertical change to horizontal change relative to ttt.
    • B. Horizontal Tangents: Occur when dy/dt=0dy/dt = 0dy/dt=0 (and dx/dt≠0dx/dt \neq 0dx/dt=0).
    • C. Vertical Tangents: Occur when dx/dt=0dx/dt = 0dx/dt=0 (and dy/dt≠0dy/dt \neq 0dy/dt=0).
  • II. Integration of Parametric Equations
    • A. Area: Transitioning ∫y dx\int y \, dx∫ydx into parameter space using substitution dx=x′(t) dtdx = x'(t) \, dtdx=x′(t)dt.
    • B. Arc Length: Derived from the Pythagorean theorem applied to infinitesimal segments ds=dx2+dy2ds = \sqrt{dx^2 + dy^2}ds=dx2+dy2​.
  • III. Surface Area of Revolution
    • A. Rotation about x-axis: Uses 2πy2\pi y2πy as the circumference.
    • B. Rotation about y-axis: Uses 2πx2\pi x2πx as the circumference.

Visual Anchors

Differentiation Flowchart

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Parametric Geometry

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Definition-Example Pairs

  • Term: Arc Length Differential (dsdsds)
    • Definition: The infinitesimal distance along a curve, ds=(x′(t))2+(y′(t))2dtds = \sqrt{(x'(t))^2 + (y'(t))^2} dtds=(x′(t))2+(y′(t))2​dt.
    • Example: For a circle x=cos⁡t,y=sin⁡tx = \cos t, y = \sin tx=cost,y=sint, ds=(−sin⁡t)2+(cos⁡t)2dt=1 dtds = \sqrt{(-\sin t)^2 + (\cos t)^2} dt = 1 \, dtds=(−sint)2+(cost)2​dt=1dt.
  • Term: Second Derivative d2ydx2\frac{d^2y}{dx^2}dx2d2y​
    • Definition: The rate of change of the slope with respect to xxx, not ttt.
    • Example: If dy/dx=t2dy/dx = t^2dy/dx=t2 and x=t3x = t^3x=t3, the second derivative is (2t)/(3t2)=2/(3t)(2t) / (3t^2) = 2/(3t)(2t)/(3t2)=2/(3t).

Worked Examples

Example 1: Finding the Equation of a Tangent Line

Problem: Find the equation of the tangent line to the curve x=t2,y=t3−3tx = t^2, y = t^3 - 3tx=t2,y=t3−3t at t=2t = 2t=2.

  1. Find Point: x(2) = 4, y(2) = 8 - 6 = 2. Point is (4,2)(4, 2)(4,2).
  2. Find Derivatives: dx/dt=2tdx/dt = 2tdx/dt=2t and dy/dt=3t2−3dy/dt = 3t^2 - 3dy/dt=3t2−3.
  3. Calculate Slope: dydx=3t2−32t\frac{dy}{dx} = \frac{3t^2 - 3}{2t}dxdy​=2t3t2−3​. At t=2t=2t=2, m=3(4)−32(2)=94m = \frac{3(4)-3}{2(2)} = \frac{9}{4}m=2(2)3(4)−3​=49​.
  4. Equation: y−2=94(x−4)  ⟹  y=94x−7y - 2 = \frac{9}{4}(x - 4) \implies y = \frac{9}{4}x - 7y−2=49​(x−4)⟹y=49​x−7.

Example 2: Arc Length of a Circle

Problem: Use the arc length formula to find the circumference of a circle of radius rrr (x=rcos⁡t,y=rsin⁡tx = r\cos t, y = r\sin tx=rcost,y=rsint) for $$0 \leq t \leq 2\pi$$.

  1. Derivatives: x′(t)x'(t) x′(t)= -r\sin t, $y'(t) $= r\cos t.
  2. Integrand: (−rsin⁡t)2+(rcos⁡t)2=r2(sin⁡2t+cos⁡2t)=r\sqrt{(-r\sin t)^2 + (r\cos t)^2} = \sqrt{r^2(\sin^2 t + \cos^2 t)} = r(−rsint)2+(rcost)2​=r2(sin2t+cos2t)​=r.
  3. Integral: L = \int_{0}^{2\pi} r$ \, dt = [rt]$_0^{2\pi} = 2\pi r.

[!TIP] Always check if a curve is traversed more than once over the given interval. If ttt goes from 0 to 4π4\pi4π for a circle, the arc length formula will give 4πr4\pi r4πr, which is double the actual circumference.

Checkpoint Questions

  1. How do you find the values of ttt where a parametric curve has a horizontal tangent line?
  2. Why is the second derivative of a parametric curve NOT simply d2y/dt2d2x/dt2\frac{d^2y/dt^2}{d^2x/dt^2}d2x/dt2d2y/dt2​?
  3. Set up the integral for the area under the curve x=t+1,y=t2x = t + 1, y = t^2x=t+1,y=t2 from t=0t=0t=0 to t=2t=2t=2.
  4. What visual property of the curve does the sign of d2ydx2\frac{d^2y}{dx^2}dx2d2y​ determine?
Study Guide825 words

Calculus of Vector-Valued Functions: A Comprehensive Study Guide

Calculus of Vector-Valued Functions

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Calculus of Vector-Valued Functions: A Comprehensive Study Guide

Learning Objectives

After studying this material, you should be able to:

  • Compute the derivative of a vector-valued function by differentiating its component functions.
  • Apply derivative properties, including the sum, scalar multiple, and three types of product rules (scalar-vector, dot product, and cross product).
  • Calculate the tangent vector and the unit tangent vector T(t)\mathbf{T}(t)T(t) for a given curve at a specific point.
  • Evaluate definite and indefinite integrals of vector-valued functions component-wise.
  • Understand the physical interpretations of derivatives as velocity and acceleration vectors.

Key Terms & Glossary

  • Vector-Valued Function: A function of the form r(t)=f(t)i+g(t)j+h(t)k\mathbf{r}(t) = f(t)\mathbf{i} + g(t)\mathbf{j} + h(t)\mathbf{k}r(t)=f(t)i+g(t)j+h(t)k, where the output is a vector.
  • Component-wise Differentiation: The process of finding the derivative of a vector function by differentiating each scalar function (f,g,h)(f, g, h)(f,g,h) independently.
  • Tangent Vector: The derivative vector r′(t)\mathbf{r}'(t)r′(t), which points in the direction of the motion along the curve at time ttt.
  • Unit Tangent Vector: A vector T(t)\mathbf{T}(t)T(t) that has a magnitude of 1 and points in the direction of r′(t)\mathbf{r}'(t)r′(t).
  • Smooth Curve: A curve where r′(t)\mathbf{r}'(t)r′(t) is continuous and r′(t)≠0\mathbf{r}'(t) \neq \mathbf{0}r′(t)=0 for all ttt in the interval.

The "Big Idea"

The transition from single-variable calculus to vector-valued calculus is remarkably consistent: we treat each dimension (x,y,zx, y, zx,y,z) as an independent scalar function of a single parameter ttt. This allows us to describe motion and geometry in 3D space using the familiar tools of power rules, chain rules, and integration, provided we maintain the algebraic structure of vectors (like dot and cross products).

Formula / Concept Box

OperationFormula / Definition
Derivativer′(t)=⟨f′(t),g′(t),h′(t)⟩\mathbf{r}'(t) = \langle f'(t), g'(t), h'(t) \rangler′(t)=⟨f′(t),g′(t),h′(t)⟩
Unit Tangent VectorT(t)=r′(t)∥r′(t)∥\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|}T(t)=∥r′(t)∥r′(t)​
Indefinite Integral∫r(t)dt=⟨∫f(t)dt,∫g(t)dt,∫h(t)dt⟩+C\int \mathbf{r}(t) dt = \langle \int f(t) dt, \int g(t) dt, \int h(t) dt \rangle + \mathbf{C}∫r(t)dt=⟨∫f(t)dt,∫g(t)dt,∫h(t)dt⟩+C
Dot Product Ruleddt[u(t)⋅v(t)]=u′(t)⋅v(t)+u(t)⋅v′(t)\frac{d}{dt}[\mathbf{u}(t) \cdot \mathbf{v}(t)] = \mathbf{u}'(t) \cdot \mathbf{v}(t) + \mathbf{u}(t) \cdot \mathbf{v}'(t)dtd​[u(t)⋅v(t)]=u′(t)⋅v(t)+u(t)⋅v′(t)
Cross Product Ruleddt[u(t)×v(t)]=u′(t)×v(t)+u(t)×v′(t)\frac{d}{dt}[\mathbf{u}(t) \times \mathbf{v}(t)] = \mathbf{u}'(t) \times \mathbf{v}(t) + \mathbf{u}(t) \times \mathbf{v}'(t)dtd​[u(t)×v(t)]=u′(t)×v(t)+u(t)×v′(t)

Hierarchical Outline

  1. Differentiation of Vector-Valued Functions
    • Component-wise approach: r′(t)=lim⁡Δt→0r(t+Δt)−r(t)Δt\mathbf{r}'(t) = \lim_{\Delta t \to 0} \frac{\mathbf{r}(t+\Delta t) - \mathbf{r}(t)}{\Delta t}r′(t)=limΔt→0​Δtr(t+Δt)−r(t)​.
    • Calculus Rules:
      • Sum/Difference: (r±u)′=r′±u′(\mathbf{r} \pm \mathbf{u})' = \mathbf{r}' \pm \mathbf{u}'(r±u)′=r′±u′
      • Scalar Multiple: (cr)′=cr′(c\mathbf{r})' = c\mathbf{r}'(cr)′=cr′
      • Chain Rule: [r(f(t))]′=f′(t)r′(f(t))[\mathbf{r}(f(t))]' = f'(t)\mathbf{r}'(f(t))[r(f(t))]′=f′(t)r′(f(t))
  2. Geometric Interpretations
    • Tangent Vectors: r′(t)\mathbf{r}'(t)r′(t) represents the instantaneous direction of the curve.
    • Smoothness: A curve is smooth if its derivative is never the zero vector.
  3. Integration
    • Antiderivatives: Computed component by component.
    • Fundamental Theorem: ∫abr(t)dt=[R(t)]ab=R(b)−R(a)\int_a^b \mathbf{r}(t) dt = [\mathbf{R}(t)]_a^b = \mathbf{R}(b) - \mathbf{R}(a)∫ab​r(t)dt=[R(t)]ab​=R(b)−R(a).

Visual Anchors

Vector Differentiation Hierarchy

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Visualizing the Unit Tangent Vector

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Definition-Example Pairs

  • Term: Chain Rule for Vector Functions
  • Definition: The derivative of a vector function with a scalar function as its parameter is the derivative of the parameter times the derivative of the vector function evaluated at that parameter.
  • Real-World Example: If r(s)\mathbf{r}(s)r(s) describes a path in terms of distance sss, and s(t)s(t)s(t) describes distance as a function of time, then r′(t)\mathbf{r}'(t)r′(t) gives the velocity of an object moving along that path over time.

Worked Examples

Example 1: Finding the Unit Tangent Vector

Problem: Find the unit tangent vector for r(t)=⟨3cos⁡t,3sin⁡t,4t⟩\mathbf{r}(t) = \langle 3\cos t, 3\sin t, 4t \rangler(t)=⟨3cost,3sint,4t⟩.

Solution:

  1. Differentiate: r′(t)=⟨−3sin⁡t,3cos⁡t,4⟩\mathbf{r}'(t) = \langle -3\sin t, 3\cos t, 4 \rangler′(t)=⟨−3sint,3cost,4⟩.
  2. Find Magnitude: ∥r′(t)∥=(−3sin⁡t)2+(3cos⁡t)2+42\|\mathbf{r}'(t)\| = \sqrt{(-3\sin t)^2 + (3\cos t)^2 + 4^2}∥r′(t)∥=(−3sint)2+(3cost)2+42​ ∥r′(t)∥=9(sin⁡2t+cos⁡2t)+16=9(1)+16=25=5\|\mathbf{r}'(t)\| = \sqrt{9(\sin^2 t + \cos^2 t) + 16} = \sqrt{9(1) + 16} = \sqrt{25} = 5∥r′(t)∥=9(sin2t+cos2t)+16​=9(1)+16​=25​=5
  3. Normalize: T(t)=15⟨−3sin⁡t,3cos⁡t,4⟩=⟨−35sin⁡t,35cos⁡t,45⟩\mathbf{T}(t) = \frac{1}{5}\langle -3\sin t, 3\cos t, 4 \rangle = \langle -\frac{3}{5}\sin t, \frac{3}{5}\cos t, \frac{4}{5} \rangleT(t)=51​⟨−3sint,3cost,4⟩=⟨−53​sint,53​cost,54​⟩

Example 2: Definite Integral

Problem: Evaluate ∫01⟨t2,et⟩dt\int_0^1 \langle t^2, e^t \rangle dt∫01​⟨t2,et⟩dt.

Solution:

  1. Integrate each component: ∫01t2dt=[13t3]01=13\int_0^1 t^2 dt = [\frac{1}{3}t^3]_0^1 = \frac{1}{3}∫01​t2dt=[31​t3]01​=31​.
  2. Integrate the second component: ∫01etdt=[et]01=e−1\int_0^1 e^t dt = [e^t]_0^1 = e - 1∫01​etdt=[et]01​=e−1.
  3. Result: ⟨13,e−1⟩\langle \frac{1}{3}, e-1 \rangle⟨31​,e−1⟩.

Checkpoint Questions

  1. If ∥r(t)∥=c\|\mathbf{r}(t)\| = c∥r(t)∥=c (a constant), what can be said about the relationship between r(t)\mathbf{r}(t)r(t) and r′(t)\mathbf{r}'(t)r′(t)? Hint: Differentiate r(t)⋅r(t)=c2\mathbf{r}(t) \cdot \mathbf{r}(t) = c^2r(t)⋅r(t)=c2.
  2. Calculate the derivative of u(t)⋅v(t)\mathbf{u}(t) \cdot \mathbf{v}(t)u(t)⋅v(t) given u(t)=⟨t,t2⟩\mathbf{u}(t) = \langle t, t^2 \rangleu(t)=⟨t,t2⟩ and v(t)=⟨sin⁡t,1⟩\mathbf{v}(t) = \langle \sin t, 1 \ranglev(t)=⟨sint,1⟩.
  3. Explain why the magnitude of the unit tangent vector T(t)\mathbf{T}(t)T(t) is always 1 for any ttt where the derivative is non-zero.

[!TIP] Always simplify the magnitude of r′(t)\mathbf{r}'(t)r′(t) before dividing. Often, trigonometric identities like sin⁡2t+cos⁡2t=1\sin^2 t + \cos^2 t = 1sin2t+cos2t=1 will significantly reduce the complexity of your unit tangent vector expressions.

Study Guide845 words

Conic Sections: Geometric Foundations and Polar Representations

Conic Sections

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Conic Sections: Geometric Foundations and Polar Representations

This study guide explores the intersection of geometry and algebra through conic sections, focusing on their classification, standard forms, and polar coordinate representations.

Learning Objectives

After studying this chapter, you should be able to:

  • Identify the four types of conic sections based on the intersection of a plane and a cone.
  • Classify a conic section given its eccentricity (eee).
  • Convert between Cartesian and polar forms of parabolas, ellipses, and hyperbolas.
  • Determine key geometric features such as the focus, directrix, and focal parameter (ppp).
  • Analyze polar equations to determine the orientation (horizontal vs. vertical) of a conic.

Key Terms & Glossary

  • Nappe: One of the two halves of a double cone joined at the vertex.
  • Focus: A fixed point used to define the set of points forming a conic section.
  • Directrix: A fixed line used in conjunction with the focus to define conics.
  • Eccentricity (eee): A numerical value that describes the "flatness" or type of a conic section; the ratio of the distance to the focus over the distance to the directrix.
  • Focal Parameter (ppp): The distance from a focus to the nearest directrix.
  • Vertex: The point(s) where the conic section intersects its axis of symmetry.

The "Big Idea"

Conic sections are not just isolated shapes; they are a unified family of curves generated by slicing a three-dimensional double cone with a plane. This geometric unity is reflected in their shared algebraic structure in polar coordinates, where a single equation r=ep1±ecos⁡θr = \frac{ep}{1 \pm e \cos \theta}r=1±ecosθep​ can describe a circle, ellipse, parabola, or hyperbola simply by changing the value of eccentricity eee.

Formula / Concept Box

Conic TypeEccentricity (eee)Polar Equation Form (Focus at Pole)Relationship
Circlee=0e = 0e=0r=ar = ar=aa=radiusa = \text{radius}a=radius
Ellipse$0 < e < 1$r=ep1±ecos⁡θr = \frac{ep}{1 \pm e \cos \theta}r=1±ecosθep​e=c/ae = c/ae=c/a
Parabolae=1e = 1e=1r=p1±cos⁡θr = \frac{p}{1 \pm \cos \theta}r=1±cosθp​Dist(Focus) = Dist(Directrix)
Hyperbolae>1e > 1e>1r=ep1±ecos⁡θr = \frac{ep}{1 \pm e \cos \theta}r=1±ecosθep​e=c/ae = c/ae=c/a

[!TIP] If cos⁡θ\cos \thetacosθ is in the denominator, the major axis is horizontal. If sin⁡θ\sin \thetasinθ is in the denominator, the major axis is vertical.

Hierarchical Outline

  1. Geometric Generation
    • Circle: Plane perpendicular to cone axis.
    • Ellipse: Plane intersects one nappe at an angle.
    • Parabola: Plane parallel to the generating line (edge) of the cone.
    • Hyperbola: Plane intersects both nappes.
  2. Eccentricity and Classification
    • Definition: e=dist(P,F)dist(P,L)e = \frac{dist(P, F)}{dist(P, L)}e=dist(P,L)dist(P,F)​ where FFF is focus and LLL is directrix.
    • Values: e=1e=1e=1 (Parabola), e<1e<1e<1 (Ellipse), e>1e>1e>1 (Hyperbola).
  3. Polar Equations of Conics
    • Standard Polar Form: r=ed1±ecos⁡θr = \frac{ed}{1 \pm e \cos \theta}r=1±ecosθed​ or r=ed1±esin⁡θr = \frac{ed}{1 \pm e \sin \theta}r=1±esinθed​.
    • Normalization: The constant term in the denominator must be 1 to identify eee correctly.

Visual Anchors

Classification Flowchart

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Geometric Definition (Parabola)

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Definition-Example Pairs

  • Focal Parameter (ppp):
    • Definition: The distance from the focus to the directrix.
    • Example: In a parabola with focus at (2,0)(2, 0)(2,0) and directrix x=−2x = -2x=−2, the focal parameter p=4p = 4p=4.
  • Horizontal vs. Vertical Conic:
    • Definition: Determined by whether the directrix is vertical (x=kx=kx=k) or horizontal (y=ky=ky=k).
    • Example: The polar equation r=102+sin⁡θr = \frac{10}{2 + \sin \theta}r=2+sinθ10​ represents a vertical conic because of the sine term.

Worked Examples

Example 1: Determining Eccentricity

Problem: Find the eccentricity of the ellipse given by x225+y216=1\frac{x^2}{25} + \frac{y^2}{16} = 125x2​+16y2​=1. Solution:

  1. Identify a2=25a^2 = 25a2=25 and b2=16b^2 = 16b2=16, so a=5a = 5a=5 and b=4b = 4b=4.
  2. Use the relation c2=a2−b2c^2 = a^2 - b^2c2=a2−b2 for an ellipse: c2=25−16=9  ⟹  c=3c^2 = 25 - 16 = 9 \implies c = 3c2=25−16=9⟹c=3
  3. Calculate eccentricity e=cae = \frac{c}{a}e=ac​: e=35=0.6e = \frac{3}{5} = 0.6e=53​=0.6
  4. Since $0.6 < 1$, the classification as an ellipse is confirmed.

Example 2: Identifying Conics from Polar Form

Problem: Identify the conic r=123−6cos⁡θr = \frac{12}{3 - 6 \cos \theta}r=3−6cosθ12​ and find its eccentricity. Solution:

  1. We must make the constant term in the denominator equal to 1. Divide numerator and denominator by 3: r=12/3(3−6cos⁡θ)/3=41−2cos⁡θr = \frac{12/3}{(3 - 6 \cos \theta)/3} = \frac{4}{1 - 2 \cos \theta}r=(3−6cosθ)/312/3​=1−2cosθ4​
  2. Compare to the standard form r=ep1−ecos⁡θr = \frac{ep}{1 - e \cos \theta}r=1−ecosθep​.
  3. Here, e=2e = 2e=2.
  4. Since e>1e > 1e>1, the conic is a hyperbola.

Checkpoint Questions

  1. What happens to the shape of an ellipse as the eccentricity eee approaches 0?
  2. If a plane intersects both nappes of a cone, which conic section is formed?
  3. Given r=51+cos⁡θr = \frac{5}{1 + \cos \theta}r=1+cosθ5​, what is the eccentricity, and what type of conic is it?
  4. What is the distance from the vertex to the focus in a parabola where the distance from focus to directrix is ppp?
▶Click to expand answers
  1. It becomes more circular; when e=0e=0e=0, it is a circle.
  2. A hyperbola.
  3. e=1e=1e=1, so it is a parabola.
  4. p/2p/2p/2.
Study Guide925 words

Mastering Conservative Vector Fields: Path Independence and Potential Functions

Conservative Vector Fields

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Mastering Conservative Vector Fields

Learning Objectives

After studying this guide, you should be able to:

  • Identify a conservative vector field in a plane or in space.
  • Test for conservativeness using cross-partial derivatives (Py=QxP_y = Q_xPy​=Qx​) or the curl operator.
  • Find a potential function fff for a given conservative vector field F⃗\vec{F}F.
  • Apply the Fundamental Theorem for Line Integrals to evaluate work and line integrals efficiently.
  • Define topological concepts such as simple curves, closed curves, and simply connected regions.

Key Terms & Glossary

  • Conservative Vector Field: A vector field F⃗\vec{F}F that is the gradient of some scalar function fff (i.e., F⃗=∇f\vec{F} = \nabla fF=∇f).
  • Potential Function: The scalar function fff such that ∇f=F⃗\nabla f = \vec{F}∇f=F. Example: In physics, the gravitational potential energy is the potential function for the gravitational force field.
  • Path Independence: A property where the line integral of F⃗\vec{F}F depends only on the endpoints of the curve CCC, not the specific path taken.
  • Simply Connected Region: An open region DDD where every simple closed curve in DDD encloses only points that are also in DDD (no "holes"). Example: A disk is simply connected; an annulus (ring) is not.
  • Closed Curve: A curve where the terminal point is the same as the initial point (r(a)=r(b)r(a) = r(b)r(a)=r(b)).

The "Big Idea"

In many physical systems, the work done moving an object depends only on its starting and ending positions, not the route taken. These "conservative" systems (like gravity or electrostatic force) allow us to simplify complex path integrals into a simple subtraction of values from a Potential Function. This is the multivariable equivalent of the Fundamental Theorem of Calculus.

Formula / Concept Box

ConceptFormula / Rule
Gradient FieldF⃗=∇f=⟨fx,fy,fz⟩\vec{F} = \nabla f = \langle f_x, f_y, f_z \rangleF=∇f=⟨fx​,fy​,fz​⟩
FTLI∫C∇f⋅dr⃗=f(r⃗(b))−f(r⃗(a))\int_C \nabla f \cdot d\vec{r} = f(\vec{r}(b)) - f(\vec{r}(a))∫C​∇f⋅dr=f(r(b))−f(r(a))
2D TestIf F⃗=⟨P,Q⟩\vec{F} = \langle P, Q \rangleF=⟨P,Q⟩, check if ∂P∂y=∂Q∂x\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}∂y∂P​=∂x∂Q​
3D TestF⃗\vec{F}F is conservative if curl F⃗=∇⃗×F⃗=0⃗\text{curl } \vec{F} = \vec{\nabla} \times \vec{F} = \vec{0}curl F=∇×F=0
Independence of Path∮CF⃗⋅dr⃗=0\oint_C \vec{F} \cdot d\vec{r} = 0∮C​F⋅dr=0 for every closed loop CCC

Hierarchical Outline

  1. Definitions and Properties
    • Relationship between Gradient Fields and Path Independence.
    • The significance of simply connected domains for testing.
  2. Testing for Conservativeness
    • Component Test (2D): ∂P∂y=∂Q×x\frac{\partial P}{\partial y} = \frac{\partial Q}{\times x}∂y∂P​=×x∂Q​.
    • Curl Test (3D): Determinant method for cross products.
  3. Finding the Potential Function fff
    • Successive Partial Integration method.
    • Determining the constant of integration g(y)g(y)g(y) or h(y,z)h(y, z)h(y,z).
  4. Applications
    • Calculating Work using the Fundamental Theorem for Line Integrals.

Visual Anchors

Decision Tree: Is F⃗\vec{F}F Conservative?

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Figure 1 — Mermaid diagram

The Core Relationship

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Figure 2 — Mermaid diagram

Definition-Example Pairs

  • Term: Path Independence

  • Definition: The value of the line integral depends only on the start point AAA and end point BBB.

  • Real-World Example: Hiking up a mountain. If gravity is the only force, the change in your potential energy is the same whether you take the steep direct trail or the long winding switchbacks.

  • Term: Curl-Free Field

  • Definition: A vector field where the rotational component at every point is zero.

  • Real-World Example: A perfectly laminar flow in a pipe where a tiny paddle wheel placed in the fluid would not spin as it moves with the current.

Worked Examples

Example 1: Testing for Conservative Fields

Question: Is F⃗(x,y)=⟨2xy+ex,x2+1⟩\vec{F}(x,y) = \langle 2xy + e^x, x^2 + 1 \rangleF(x,y)=⟨2xy+ex,x2+1⟩ conservative?

Solution:

  1. Identify P=2xy+exP = 2xy + e^xP=2xy+ex and Q=x2+1Q = x^2 + 1Q=x2+1.
  2. Compute Py=∂∂y(2xy+ex)=2xP_y = \frac{\partial}{\partial y}(2xy + e^x) = 2xPy​=∂y∂​(2xy+ex)=2x.
  3. Compute Qx=∂∂x(x2+1)=2xQ_x = \frac{\partial}{\partial x}(x^2 + 1) = 2xQx​=∂x∂​(x2+1)=2x.
  4. Since Py=QxP_y = Q_xPy​=Qx​ and the domain R2\mathbb{R}^2R2 is simply connected, F⃗\vec{F}F is conservative.

Example 2: Finding a Potential Function

Question: Find fff for F⃗=⟨3x2y,x3+2y⟩\vec{F} = \langle 3x^2y, x^3 + 2y \rangleF=⟨3x2y,x3+2y⟩.

Solution:

  1. Set fx=3x2yf_x = 3x^2yfx​=3x2y. Integrate with respect to xxx: f(x,y)=x3y+g(y)f(x,y) = x^3y + g(y)f(x,y)=x3y+g(y).
  2. Differentiate our fff with respect to yyy: fy=x3+g′(y)f_y = x^3 + g'(y)fy​=x3+g′(y).
  3. Match this with the given QQQ: x3+g′(y)=x3+2yx^3 + g'(y) = x^3 + 2yx3+g′(y)=x3+2y.
  4. Solve for g′(y)g'(y)g′(y): g′(y)=2y→g(y)=y2+Kg'(y) = 2y \rightarrow g(y) = y^2 + Kg′(y)=2y→g(y)=y2+K.
  5. Final Potential Function: f(x,y)=x3y+y2+Kf(x,y) = x^3y + y^2 + Kf(x,y)=x3y+y2+K.

Checkpoint Questions

  1. True or False: If ∮CF⃗⋅dr⃗=0\oint_C \vec{F} \cdot d\vec{r} = 0∮C​F⋅dr=0 for one specific closed curve CCC, then F⃗\vec{F}F is guaranteed to be conservative. (Answer: False; it must be true for EVERY closed curve).
  2. What is the curl of a conservative vector field in R3\mathbb{R}^3R3? (Answer: The zero vector 0⃗\vec{0}0).
  3. If F⃗=∇f\vec{F} = \nabla fF=∇f, what is the value of ∫CF⃗⋅dr⃗\int_C \vec{F} \cdot d\vec{r}∫C​F⋅dr if CCC starts at (0,0)(0,0)(0,0) and ends at (1,1)(1,1)(1,1), given f(x,y)=x2+y2f(x,y) = x^2 + y^2f(x,y)=x2+y2? (Answer: f(1,1) - f(0,0) = (1+1) - (0) = 2).

[!TIP] Always check if the domain is simply connected before concluding a field is conservative based on the Py=QxP_y = Q_xPy​=Qx​ test. If there is a hole in the domain (like the origin in a vortex field), the test can yield "true" even if the field is not conservative!

Muddy Points & Cross-Refs

  • Confusing Curl and Gradient: Remember, a gradient turns a scalar into a vector. Curl turns a vector into another vector.
  • Need more on integration? See Section 5.1: Double Integrals over Rectangular Regions to review basic integration techniques used in finding potential functions.
Study Guide925 words

Mastering Cylindrical and Spherical Coordinates

Cylindrical and Spherical Coordinates

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Mastering Cylindrical and Spherical Coordinates

This guide provides a comprehensive overview of alternative 3D coordinate systems, focusing on how to represent points and surfaces using cylindrical and spherical coordinates to simplify complex multivariable calculus problems.

Learning Objectives

After studying this guide, you should be able to:

  • Locate points in space using cylindrical (r,θ,z)(r, \theta, z)(r,θ,z) and spherical (ρ,θ,ϕ)(\rho, \theta, \phi)(ρ,θ,ϕ) coordinates.
  • Convert coordinates between rectangular, cylindrical, and spherical systems.
  • Identify and describe surfaces defined by constant-coordinate equations (e.g., ρ=c\rho = cρ=c or r=cr = cr=c).
  • Recognize symmetry in physical systems to choose the most efficient coordinate system.

The "Big Idea"

In Cartesian coordinates (x,y,z)(x, y, z)(x,y,z), we describe the world using a grid of boxes. While intuitive, many physical phenomena—like the gravitational field of a planet, the flow of water in a pipe, or the heat distribution in a wire—exhibit radial or axial symmetry. Cylindrical and spherical coordinates allow us to "bend" our coordinate axes to match the geometry of the problem. This transformation often turns complex, multi-variable boundary equations into simple constants, making integration and physical modeling significantly easier.

Key Terms & Glossary

  • Azimuthal Angle (θ\thetaθ): The same angle used in polar coordinates, measured from the positive xxx-axis in the xyxyxy-plane. Example: The longitude on a globe.
  • Radial Distance (rrr): In cylindrical coordinates, the perpendicular distance from the zzz-axis to the point. Example: The radius of a soda can.
  • Rho (ρ\rhoρ): In spherical coordinates, the direct distance from the origin to the point (ρ≥0\rho \ge 0ρ≥0). Example: The distance from the center of the Earth to a satellite.
  • Phi (ϕ\phiϕ): The polar angle (or inclination) measured from the positive zzz-axis (0≤ϕ≤π0 \le \phi \le \pi0≤ϕ≤π). Example: High noon corresponds to a small ϕ\phiϕ if the sun is overhead.

Formula / Concept Box

SystemCoordinatesRelation to Rectangular (x,y,zx, y, zx,y,z)
Cylindrical(r,θ,z)(r, \theta, z)(r,θ,z)x=rcos⁡θx = r \cos \thetax=rcosθ, y=rsin⁡θy = r \sin \thetay=rsinθ, z=zz = zz=z
Spherical(ρ,θ,ϕ)(\rho, \theta, \phi)(ρ,θ,ϕ)x=ρsin⁡ϕcos⁡θx = \rho \sin \phi \cos \thetax=ρsinϕcosθ, y=ρsin⁡ϕsin⁡θy = \rho \sin \phi \sin \thetay=ρsinϕsinθ, z=ρcos⁡ϕz = \rho \cos \phiz=ρcosϕ
Inter-systemConversionr=ρsin⁡ϕr = \rho \sin \phir=ρsinϕ, z=ρcos⁡ϕz = \rho \cos \phiz=ρcosϕ, ρ2=r2+z2\rho^2 = r^2 + z^2ρ2=r2+z2

[!IMPORTANT] Always remember the domain constraints for spherical coordinates: ρ≥0\rho \ge 0ρ≥0, $0 \le \theta < 2\pi$$, and $0 \le \phi \le \pi.

Hierarchical Outline

  • I. Cylindrical Coordinates (r,θ,z)(r, \theta, z)(r,θ,z)
    • A. Definition: Extension of 2D polar coordinates into 3D by adding a zzz-axis height.
    • B. Constant Surfaces:
      • r=cr = cr=c: A vertical cylinder of radius ccc.
      • θ=c\theta = cθ=c: A vertical half-plane originating at the zzz-axis.
      • z=cz = cz=c: A horizontal plane.
  • II. Spherical Coordinates (ρ,θ,ϕ)(\rho, \theta, \phi)(ρ,θ,ϕ)
    • A. Definition: Describes points based on distance from origin and two angles.
    • B. Constant Surfaces:
      • ρ=c\rho = cρ=c: A sphere centered at the origin with radius ccc.
      • θ=c\theta = cθ=c: A vertical half-plane (same as cylindrical).
      • ϕ=c\phi = cϕ=c: A cone opening along the zzz-axis.
  • III. System Selection Strategy
    • Use Cylindrical for objects with rotational symmetry around an axis (pipes, wires).
    • Use Spherical for objects with symmetry around a central point (planets, point charges).

Visual Anchors

Coordinate Relationship Flow

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Figure 1 — Mermaid diagram

Visualizing Spherical Angles

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Figure 2 — TikZ diagram

Definition-Example Pairs

  • Surface ϕ=c\phi = cϕ=c

    • Definition: A set of points where the angle from the positive zzz-axis is constant.
    • Example: An ice cream cone. If ϕ=π/4\phi = \pi/4ϕ=π/4, the surface is a cone that opens at a 45-degree angle from the vertical.
  • Surface r=cr = cr=c

    • Definition: A set of points at a fixed horizontal distance from the zzz-axis.
    • Example: A skyscraper with a circular floor plan. No matter how high (z)(z)(z) or what direction (θ)(\theta)(θ) you look, the distance from the central elevator shaft (r)(r)(r) is the same.

Worked Examples

Example 1: Rectangular to Spherical

Problem: Convert the rectangular point P(1,1,2)P(1, 1, \sqrt{2})P(1,1,2​) to spherical coordinates.

Step-by-Step Solution:

  1. Find ρ\rhoρ: ρ=x2+y2+z2=12+12+(2)2=1+1+2=4=2\rho = \sqrt{x^2 + y^2 + z^2} = \sqrt{1^2 + 1^2 + (\sqrt{2})^2} = \sqrt{1 + 1 + 2} = \sqrt{4} = 2ρ=x2+y2+z2​=12+12+(2​)2​=1+1+2​=4​=2
  2. Find θ\thetaθ: tan⁡θ=yx=11=1  ⟹  θ=π4\tan \theta = \frac{y}{x} = \frac{1}{1} = 1 \implies \theta = \frac{\pi}{4}tanθ=xy​=11​=1⟹θ=4π​
  3. Find ϕ\phiϕ: Using z=ρcos⁡ϕz = \rho \cos \phiz=ρcosϕ: 2=2cos⁡ϕ  ⟹  cos⁡ϕ=22  ⟹  ϕ=π4\sqrt{2} = 2 \cos \phi \implies \cos \phi = \frac{\sqrt{2}}{2} \implies \phi = \frac{\pi}{4}2​=2cosϕ⟹cosϕ=22​​⟹ϕ=4π​ Result: The spherical coordinates are (2,π/4,π/4)(2, \pi/4, \pi/4)(2,π/4,π/4).

Example 2: Identifying a Surface

Problem: Describe the surface given by the equation ρ=4cos⁡ϕ\rho = 4 \cos \phiρ=4cosϕ.

Step-by-Step Solution:

  1. Multiply both sides by ρ\rhoρ: ρ2=4ρcos⁡ϕ\rho^2 = 4 \rho \cos \phiρ2=4ρcosϕ.
  2. Substitute rectangular identities: x2+y2+z2=4zx^2 + y^2 + z^2 = 4zx2+y2+z2=4z.
  3. Complete the square for zzz: x2+y2+(z−2)2=4x^2 + y^2 + (z - 2)^2 = 4x2+y2+(z−2)2=4. Result: This is a sphere centered at (0,0,2)(0, 0, 2)(0,0,2) with a radius of 2.

Checkpoint Questions

  1. What surface is described by the equation r=5r = 5r=5 in cylindrical coordinates?
    • Answer: A cylinder centered on the z-axis with radius 5.
  2. If a point is on the xyxyxy-plane, what is its ϕ\phiϕ value in spherical coordinates?
    • Answer: ϕ=π/2\phi = \pi/2ϕ=π/2.
  3. True or False: The value of θ\thetaθ is calculated the same way in both cylindrical and spherical systems.
    • Answer: True.
  4. Convert the spherical point (4,π/2,π)(4, \pi/2, \pi)(4,π/2,π) to rectangular coordinates.
    • Answer: x=0,y=0,z=−4x=0, y=0, z=-4x=0,y=0,z=−4. since ϕ=π\phi = \piϕ=π points directly down the negative z-axis.
Study Guide985 words

Study Guide: Directional Derivatives and the Gradient

Directional Derivatives and the Gradient

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Directional Derivatives and the Gradient

Learning Objectives

After studying this chapter, you should be able to:

  • Calculate the directional derivative of a function f(x,y)f(x, y)f(x,y) or f(x,y,z)f(x, y, z)f(x,y,z) in any given direction.
  • Compute the gradient vector ∇f\nabla f∇f for scalar functions.
  • Interpret the geometric significance of the gradient (direction of steepest ascent).
  • Apply the gradient to find tangent lines to level curves and tangent planes to level surfaces.
  • Determine the maximum and minimum rates of change at a specific point.

Key Terms & Glossary

  • Gradient (∇f\nabla f∇f): A vector field whose components are the partial derivatives of fff. It points in the direction of the greatest rate of increase.
  • Directional Derivative (DufD_{\mathbf{u}}fDu​f): The rate at which a function changes at a point in the direction of a unit vector u\mathbf{u}u.
  • Unit Vector: A vector with a magnitude (length) of 1. In directional derivatives, the direction must be normalized to a unit vector.
  • Level Curve: A curve defined by f(x,y)=cf(x, y) = cf(x,y)=c. The gradient ∇f\nabla f∇f at a point (x0,y0)(x_0, y_0)(x0​,y0​) is always perpendicular (orthogonal) to the level curve passing through that point.
  • Normal Vector: A vector perpendicular to a surface or curve; the gradient acts as the normal vector to level sets.

The "Big Idea"

In single-variable calculus, the derivative f′(x)f'(x)f′(x) gives the slope in one dimension. In multivariable calculus, partial derivatives fxf_xfx​ and fyf_yfy​ only give the slopes along the axes. The Gradient is the "master key" that unlocks the slope in any direction. It represents the local "uphill" direction on a topographic map. If you are standing on a mountain, the gradient tells you which way is straight up, and the directional derivative tells you how steep the path is if you choose to walk at a specific angle.


Formula / Concept Box

ConceptFormula / Property
The Gradient (2D)∇f(x,y)=⟨fx,fy⟩=∂f∂xi+∂f∂yj\nabla f(x, y) = \langle f_x, f_y \rangle = \frac{\partial f}{\partial x}\mathbf{i} + \frac{\partial f}{\partial y}\mathbf{j}∇f(x,y)=⟨fx​,fy​⟩=∂x∂f​i+∂y∂f​j
Directional DerivativeDuf(P)=∇f(P)⋅uD_{\mathbf{u}}f(P) = \nabla f(P) \cdot \mathbf{u}Du​f(P)=∇f(P)⋅u (where ∥u∥=1\|\mathbf{u}\| = 1∥u∥=1)
Maximum Rate of Change∥∇f(P)∥\|\nabla f(P)\|∥∇f(P)∥ (occurs when u\mathbf{u}u is in the direction of ∇f\nabla f∇f)
Minimum Rate of Change−∥∇f(P)∥-\|\nabla f(P)\|−∥∇f(P)∥ (occurs when u\mathbf{u}u is in the direction of −∇f-\nabla f−∇f)
Orthogonality∇f(P)⋅v=0\nabla f(P) \cdot \mathbf{v} = 0∇f(P)⋅v=0 for any vector v\mathbf{v}v tangent to the level curve

[!IMPORTANT] Always verify that your direction vector is a unit vector. If you are given a vector v\mathbf{v}v, you must use u=v∥v∥\mathbf{u} = \frac{\mathbf{v}}{\|\mathbf{v}\|}u=∥v∥v​ before calculating the dot product.


Hierarchical Outline

  1. The Gradient Vector
    • Definition: ∇f=⟨fx,fy,… ⟩\nabla f = \langle f_x, f_y, \dots \rangle∇f=⟨fx​,fy​,…⟩.
    • Computation: Requires finding all first-order partial derivatives.
  2. The Directional Derivative
    • Limit definition vs. the Dot Product Theorem.
    • Relationship to partials: fxf_xfx​ is DifD_{\mathbf{i}}fDi​f and fyf_yfy​ is DjfD_{\mathbf{j}}fDj​f.
  3. Properties of the Gradient
    • Steepest Ascent: Points in the direction of maximum increase.
    • Steepest Descent: Points in the direction of maximum decrease (−∇f-\nabla f−∇f).
    • Zero Change: Directions orthogonal to ∇f\nabla f∇f result in zero change (moving along a level curve).
  4. Applications
    • Tangent lines to level curves f(x,y)=cf(x, y) = cf(x,y)=c.
    • Normal lines to surfaces f(x,y,z)=cf(x, y, z) = cf(x,y,z)=c.

Visual Anchors

The Gradient Relationship

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Figure 1 — Mermaid diagram

Geometric Interpretation in 2D

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Figure 2 — TikZ diagram

Definition-Example Pairs

1. Directional Cosines

  • Definition: The components of a unit vector in 3D, expressed as cos⁡α,cos⁡β,cos⁡γ\cos \alpha, \cos \beta, \cos \gammacosα,cosβ,cosγ, representing the angles made with the x,y,x, y,x,y, and zzz axes.
  • Example: A unit vector pointing equally between the axes would have cos⁡α=cos⁡β=cos⁡γ=1/3\cos \alpha = \cos \beta = \cos \gamma = 1/\sqrt{3}cosα=cosβ=cosγ=1/3​.

2. Steepest Descent

  • Definition: The direction exactly opposite the gradient vector where the function's value decreases most rapidly.
  • Example: If ∇f=⟨3,4⟩\nabla f = \langle 3, 4 \rangle∇f=⟨3,4⟩ at a point, the direction of steepest descent is ⟨−3,−4⟩\langle -3, -4 \rangle⟨−3,−4⟩ with a rate of −5-5−5.

Worked Examples

Example 1: Finding the Directional Derivative

Problem: Find Duf(1,2)D_{\mathbf{u}}f(1, 2)Du​f(1,2) for f(x,y)=x2yf(x, y) = x^2yf(x,y)=x2y in the direction of v=⟨3,−4⟩\mathbf{v} = \langle 3, -4 \ranglev=⟨3,−4⟩.

Solution:

  1. Find the Gradient: fx=2xy  ⟹  fx(1,2)=2(1)(2)=4f_x = 2xy \implies f_x(1, 2) = 2(1)(2) = 4fx​=2xy⟹fx​(1,2)=2(1)(2)=4 fy=x2  ⟹  fy(1,2)=(1)2=1f_y = x^2 \implies f_y(1, 2) = (1)^2 = 1fy​=x2⟹fy​(1,2)=(1)2=1 So, ∇f(1,2)=⟨4,1⟩\nabla f(1, 2) = \langle 4, 1 \rangle∇f(1,2)=⟨4,1⟩.
  2. Normalize the Direction Vector: ∥v∥=32+(−4)2=5\|\mathbf{v}\| = \sqrt{3^2 + (-4)^2} = 5∥v∥=32+(−4)2​=5 u=v∥v∥=⟨35,−45⟩\mathbf{u} = \frac{\mathbf{v}}{\|\mathbf{v}\|} = \langle \frac{3}{5}, -\frac{4}{5} \rangleu=∥v∥v​=⟨53​,−54​⟩.
  3. Compute the Dot Product: Duf=∇f⋅u=⟨4,1⟩⋅⟨0.6,−0.8⟩D_{\mathbf{u}}f = \nabla f \cdot \mathbf{u} = \langle 4, 1 \rangle \cdot \langle 0.6, -0.8 \rangleDu​f=∇f⋅u=⟨4,1⟩⋅⟨0.6,−0.8⟩ Duf=(4)(0.6)+(1)(−0.8)=2.4−0.8=1.6D_{\mathbf{u}}f = (4)(0.6) + (1)(-0.8) = 2.4 - 0.8 = 1.6Du​f=(4)(0.6)+(1)(−0.8)=2.4−0.8=1.6.

Example 2: Direction of Maximum Increase

Problem: At the point (0,0,0)(0, 0, 0)(0,0,0), in what direction does f(x,y,z)=ex+yzf(x, y, z) = e^x + yzf(x,y,z)=ex+yz increase most rapidly?

Solution:

  1. Calculate the Gradient: ∇f=⟨ex,z,y⟩\nabla f = \langle e^x, z, y \rangle∇f=⟨ex,z,y⟩
  2. Evaluate at (0,0,0)(0, 0, 0)(0,0,0): ∇f(0,0,0)=⟨e0,0,0⟩=⟨1,0,0⟩\nabla f(0, 0, 0) = \langle e^0, 0, 0 \rangle = \langle 1, 0, 0 \rangle∇f(0,0,0)=⟨e0,0,0⟩=⟨1,0,0⟩.
  3. Result: The function increases most rapidly in the direction of the positive xxx-axis (i\mathbf{i}i). The maximum rate of change is ∥⟨1,0,0⟩∥=1\|\langle 1, 0, 0 \rangle\| = 1∥⟨1,0,0⟩∥=1.

Checkpoint Questions

  1. If ∇f(P)=⟨0,0⟩\nabla f(P) = \langle 0, 0 \rangle∇f(P)=⟨0,0⟩, what is the directional derivative in any direction u\mathbf{u}u?
  2. True or False: The directional derivative can be larger than the magnitude of the gradient.
  3. Find the gradient of f(x,y)=ln⁡(x2+y2)f(x, y) = \ln(x^2 + y^2)f(x,y)=ln(x2+y2) at the point (1,1)(1, 1)(1,1).
  4. If you are walking along a level curve of a mountain, what is the value of your directional derivative in the direction of your path?
▶Click to see Answers
  1. 0. Because Duf=⟨0,0⟩⋅u=0D_{\mathbf{u}}f = \langle 0, 0 \rangle \cdot \mathbf{u} = 0Du​f=⟨0,0⟩⋅u=0.
  2. False. The maximum value is ∥∇f∥\|\nabla f\|∥∇f∥.
  3. ∇f=⟨2xx2+y2,2yx2+y2⟩\nabla f = \langle \frac{2x}{x^2+y^2}, \frac{2y}{x^2+y^2} \rangle∇f=⟨x2+y22x​,x2+y22y​⟩. At (1,1)(1, 1)(1,1), ∇f=⟨1,1⟩\nabla f = \langle 1, 1 \rangle∇f=⟨1,1⟩.
  4. 0. Level curves represent constant height, so the rate of change is zero.
Study Guide860 words

Divergence and Curl: Vector Calculus Study Guide

Divergence and Curl

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Divergence and Curl: Vector Calculus Study Guide

This guide covers the fundamental derivatives of vector fields: Divergence and Curl. These operators allow us to measure the rate of expansion and the rotation of a vector field at any given point in space.


Learning Objectives

After studying this guide, you should be able to:

  • Calculate the divergence of a 2D or 3D vector field.
  • Calculate the curl of a 3D vector field using the determinant method.
  • Interpret the physical significance of divergence (flux density) and curl (rotation).
  • Use the curl test to determine if a vector field is conservative.
  • Understand the identity div(curl F)=0\text{div}(\text{curl } \mathbf{F}) = 0div(curl F)=0.

Key Terms & Glossary

  • Del Operator (∇\nabla∇): The vector differential operator defined as ⟨∂∂x,∂∂y,∂∂z⟩\langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \rangle⟨∂x∂​,∂y∂​,∂z∂​⟩.
  • Divergence: A scalar field that represents the volume density of the outward flux of a vector field from an infinitesimal point.
  • Curl: A vector field that represents the infinitesimal rotation of a 3D vector field.
  • Solenoidal: A vector field where div F=0\text{div } \mathbf{F} = 0div F=0 (incompressible).
  • Irrotational: A vector field where curl F=0\text{curl } \mathbf{F} = \mathbf{0}curl F=0.
  • Conservative Field: A field that is the gradient of some scalar potential function ϕ\phiϕ (i.e., F=∇ϕ\mathbf{F} = \nabla \phiF=∇ϕ).

The "Big Idea"

Just as the derivative of a single-variable function measures the rate of change, Divergence and Curl measure the "structural" changes of a vector field.

  • Divergence asks: "Is the fluid at this point expanding or compressing?" (Source vs. Sink).
  • Curl asks: "If I placed a tiny paddle wheel here, would it spin, and in which direction?" (Vorticity).

Formula / Concept Box

OperatorNotationDefinition (for F=⟨P,Q,R⟩\mathbf{F} = \langle P, Q, R \rangleF=⟨P,Q,R⟩)Result Type
Gradient∇f\nabla f∇f⟨fx,fy,fz⟩\langle f_x, f_y, f_z \rangle⟨fx​,fy​,fz​⟩Vector
Divergence∇⋅F\nabla \cdot \mathbf{F}∇⋅FPx+Qy+RzP_x + Q_y + R_zPx​+Qy​+Rz​Scalar
Curl∇×F\nabla \times \mathbf{F}∇×F∣ijk∂x∂y∂zPQR∣\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \partial_x & \partial_y & \partial_z \\ P & Q & R \end{vmatrix}​i∂x​P​j∂y​Q​k∂z​R​​Vector

[!IMPORTANT] The Conservative Test: For a simply connected region, F\mathbf{F}F is conservative if and only if curl F=0\text{curl } \mathbf{F} = \mathbf{0}curl F=0.


Hierarchical Outline

  1. The Del Operator (∇\nabla∇)
    • Foundational tool for multivariable differentiation.
  2. Divergence (∇⋅F\nabla \cdot \mathbf{F}∇⋅F)
    • Computation: Dot product of ∇\nabla∇ and F\mathbf{F}F.
    • Interpretation: Positive = Source, Negative = Sink, Zero = Incompressible.
  3. Curl (∇×F\nabla \times \mathbf{F}∇×F)
    • Computation: Cross product of ∇\nabla∇ and F\mathbf{F}F.
    • Interpretation: Vector points along the axis of rotation; magnitude is the speed of rotation.
  4. Second-Order Identities
    • curl(∇f)=0\text{curl}(\nabla f) = \mathbf{0}curl(∇f)=0 (Gradients are irrotational).
    • div(curl F)=0\text{div}(\text{curl } \mathbf{F}) = 0div(curl F)=0 (The rotation of a field has no net expansion).

Visual Anchors

Analyzing a Vector Field Flowchart

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Figure 1 — Mermaid diagram

The Geometry of Curl

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Figure 2 — TikZ diagram

Definition-Example Pairs

1. Positive Divergence

  • Definition: A point where the net flow of the vector field is outward.
  • Example: Air blowing out of a ventilation duct into a room.

2. Irrotational Field

  • Definition: A field where the curl is zero at every point.
  • Example: A static electric field produced by a point charge; a paddle wheel placed in this field would not rotate.

Worked Examples

Example 1: Basic Calculation

Task: Find the divergence and curl of F(x,y,z)=⟨x2,xy,z⟩\mathbf{F}(x,y,z) = \langle x^2, xy, z \rangleF(x,y,z)=⟨x2,xy,z⟩.

Solution:

  1. Divergence: div F=∂∂x(x2)+∂∂y(xy)+∂∂z(z)\text{div } \mathbf{F} = \frac{\partial}{\partial x}(x^2) + \frac{\partial}{\partial y}(xy) + \frac{\partial}{\partial z}(z)div F=∂x∂​(x2)+∂y∂​(xy)+∂z∂​(z) div F=2x+x+1=3x+1\text{div } \mathbf{F} = 2x + x + 1 = 3x + 1div F=2x+x+1=3x+1

  2. Curl: curl F=∣ijk∂x∂y∂zx2xyz∣\text{curl } \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \partial_x & \partial_y & \partial_z \\ x^2 & xy & z \end{vmatrix}curl F=​i∂x​x2​j∂y​xy​k∂z​z​​ i(∂yz−∂zxy)−j(∂xz−∂zx2)+k(∂xxy−∂yx2)\mathbf{i}(\partial_y z - \partial_z xy) - \mathbf{j}(\partial_x z - \partial_z x^2) + \mathbf{k}(\partial_x xy - \partial_y x^2)i(∂y​z−∂z​xy)−j(∂x​z−∂z​x2)+k(∂x​xy−∂y​x2) i(0−0)−j(0−0)+k(y−0)=⟨0,0,y⟩\mathbf{i}(0 - 0) - \mathbf{j}(0 - 0) + \mathbf{k}(y - 0) = \langle 0, 0, y \ranglei(0−0)−j(0−0)+k(y−0)=⟨0,0,y⟩

Example 2: Conservative Test

Task: Is F=⟨exsin⁡y,excos⁡y,0⟩\mathbf{F} = \langle e^x \sin y, e^x \cos y, 0 \rangleF=⟨exsiny,excosy,0⟩ conservative?

Solution: Calculate the curl: curl F=⟨(0−0),−(0−0),(excos⁡y−excos⁡y)⟩=⟨0,0,0⟩\text{curl } \mathbf{F} = \langle (0 - 0), -(0 - 0), (e^x \cos y - e^x \cos y) \rangle = \langle 0, 0, 0 \ranglecurl F=⟨(0−0),−(0−0),(excosy−excosy)⟩=⟨0,0,0⟩ Since curl F=0\text{curl } \mathbf{F} = \mathbf{0}curl F=0 and the domain is R3\mathbb{R}^3R3 (simply connected), the field is conservative.


Checkpoint Questions

  1. What is the divergence of the curl of any smooth vector field F\mathbf{F}F?
    • (Answer: 0)
  2. If div F=5\text{div } \mathbf{F} = 5div F=5 at a point PPP, is fluid moving toward or away from PPP?
    • (Answer: Away from P; it is a source)
  3. Calculate curl ⟨y,−x,0⟩\text{curl } \langle y, -x, 0 \ranglecurl ⟨y,−x,0⟩. What does the result tell you about the rotation?
    • (Answer: ⟨0,0,−2⟩\langle 0, 0, -2 \rangle⟨0,0,−2⟩. The rotation is clockwise around the z-axis.)
  4. True or False: If a field is conservative, its curl must be zero.
    • (Answer: True)

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Calculus III: Multivariable Calculus Practice Questions

Try 15 sample questions from a bank of 653. Answers and detailed explanations included.

Q1easy

Stokes' Theorem provides a fundamental relationship between the surface integral of the curl of a vector field and a line integral around the boundary of that surface. Which of the following formulas correctly represents this theorem for a vector field F\mathbf{F}F, an oriented surface SSS, and its closed boundary curve CCC?

A.

∬S(∇×F)⋅dS=∮CF⋅dr\iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = \oint_C \mathbf{F} \cdot d\mathbf{r}∬S​(∇×F)⋅dS=∮C​F⋅dr

B.

∬SF⋅dS=∭V(∇⋅F) dV\iint_S \mathbf{F} \cdot d\mathbf{S} = \iiint_V (\nabla \cdot \mathbf{F}) \, dV∬S​F⋅dS=∭V​(∇⋅F)dV

C.

∮C(∇×F)⋅dr=∬SF⋅dS\oint_C (\nabla \times \mathbf{F}) \cdot d\mathbf{r} = \iint_S \mathbf{F} \cdot d\mathbf{S}∮C​(∇×F)⋅dr=∬S​F⋅dS

D.

∬SF⋅dS=∮CF⋅dr\iint_S \mathbf{F} \cdot d\mathbf{S} = \oint_C \mathbf{F} \cdot d\mathbf{r}∬S​F⋅dS=∮C​F⋅dr

Show answer & explanation

Correct Answer: A

Stokes' Theorem relates the surface integral of the curl of a vector field over an oriented surface to the line integral of the vector field along its boundary.

  1. Formula: The theorem is expressed as ∬S(∇×F)⋅dS=∮CF⋅dr\iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = \oint_C \mathbf{F} \cdot d\mathbf{r}∬S​(∇×F)⋅dS=∮C​F⋅dr.
  2. Left Side: Represents the flux of the curl of F\mathbf{F}F through the surface SSS.
  3. Right Side: Represents the circulation of F\mathbf{F}F around the boundary curve CCC.
  4. Distractors: Option B is the Divergence (Gauss) Theorem. Option C incorrectly places the curl in the line integral. Option D incorrectly equates the flux of F\mathbf{F}F directly to its circulation.

Answer A correctly states the theorem.

Q2hard

Consider the vector w⃗\vec{w}w in the coordinate plane that originates at point A(4,−1)A(4, -1)A(4,−1) and terminates at point B(−2,3)B(-2, 3)B(−2,3). Which of the following sets of equations correctly represents the vector w⃗\vec{w}w in component form, linear combination form, and magnitude-direction form?

A.

w⃗=⟨−6,4⟩=−6i+4j=213(cos⁡θi+sin⁡θj)\vec{w} = \langle -6, 4 \rangle = -6\mathbf{i} + 4\mathbf{j} = 2\sqrt{13}(\cos \theta \mathbf{i} + \sin \theta \mathbf{j})w=⟨−6,4⟩=−6i+4j=213​(cosθi+sinθj), where θ=π−arctan⁡(23)\theta = \pi - \arctan\left(\frac{2}{3}\right)θ=π−arctan(32​)

B.

w⃗=⟨−2,3⟩=−2i+3j=13(cos⁡θi+sin⁡θj)\vec{w} = \langle -2, 3 \rangle = -2\mathbf{i} + 3\mathbf{j} = \sqrt{13}(\cos \theta \mathbf{i} + \sin \theta \mathbf{j})w=⟨−2,3⟩=−2i+3j=13​(cosθi+sinθj), where θ=arctan⁡(−32)\theta = \arctan\left(-\frac{3}{2}\right)θ=arctan(−23​)

C.

w⃗=⟨−6,4⟩=4i−6j=213(cos⁡θi+sin⁡θj)\vec{w} = \langle -6, 4 \rangle = 4\mathbf{i} - 6\mathbf{j} = 2\sqrt{13}(\cos \theta \mathbf{i} + \sin \theta \mathbf{j})w=⟨−6,4⟩=4i−6j=213​(cosθi+sinθj), where θ=arctan⁡(−23)\theta = \arctan\left(-\frac{2}{3}\right)θ=arctan(−32​)

D.

w⃗=⟨6,−4⟩=6i−4j=213(cos⁡θi+sin⁡θj)\vec{w} = \langle 6, -4 \rangle = 6\mathbf{i} - 4\mathbf{j} = 2\sqrt{13}(\cos \theta \mathbf{i} + \sin \theta \mathbf{j})w=⟨6,−4⟩=6i−4j=213​(cosθi+sinθj), where θ=2π−arctan⁡(23)\theta = 2\pi - \arctan\left(\frac{2}{3}\right)θ=2π−arctan(32​)

Show answer & explanation

Correct Answer: A

To represent the vector correctly, we must analyze its components, magnitude, and direction angle relative to the positive xxx-axis.

  1. Component Form: The components of a vector from initial point P(x1,y1)P(x_1, y_1)P(x1​,y1​) to terminal point Q(x2,y2)Q(x_2, y_2)Q(x2​,y2​) are given by ⟨x2−x1,y2−y1⟩\langle x_2 - x_1, y_2 - y_1 \rangle⟨x2​−x1​,y2​−y1​⟩. For points A(4,−1)A(4, -1)A(4,−1) and B(−2,3)B(-2, 3)B(−2,3): vx=−2−4=−6v_x = -2 - 4 = -6vx​=−2−4=−6 vy=3−(−1)=4v_y = 3 - (-1) = 4vy​=3−(−1)=4 Thus, w⃗=⟨−6,4⟩\vec{w} = \langle -6, 4 \ranglew=⟨−6,4⟩.

  2. Linear Combination Form: Using standard unit vectors i=⟨1,0⟩\mathbf{i} = \langle 1, 0 \ranglei=⟨1,0⟩ and j=⟨0,1⟩\mathbf{j} = \langle 0, 1 \ranglej=⟨0,1⟩: w⃗=−6i+4j\vec{w} = -6\mathbf{i} + 4\mathbf{j}w=−6i+4j

  3. Magnitude: ∥w⃗∥=(−6)2+42=36+16=52=213\|\vec{w}\| = \sqrt{(-6)^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}∥w∥=(−6)2+42​=36+16​=52​=213​

  4. Direction Angle: The vector lies in Quadrant II because vx<0v_x < 0vx​<0 and vy>0v_y > 0vy​>0. The reference angle α\alphaα is: α=arctan⁡(∣vy∣∣vx∣)=arctan⁡(46)=arctan⁡(23)\alpha = \arctan\left(\frac{|v_y|}{|v_x|}\right) = \arctan\left(\frac{4}{6}\right) = \arctan\left(\frac{2}{3}\right)α=arctan(∣vx​∣∣vy​∣​)=arctan(64​)=arctan(32​) Since it is in Quadrant II, the direction angle θ\thetaθ from the positive xxx-axis is: θ=π−α=π−arctan⁡(23)\theta = \pi - \alpha = \pi - \arctan\left(\frac{2}{3}\right)θ=π−α=π−arctan(32​)

Option A correctly identifies all three representations. Option B uses the terminal point as the vector. Option C swaps the i\mathbf{i}i and j\mathbf{j}j components. Option D incorrectly calculates the vector as A−BA-BA−B, reversing its direction. Therefore, the correct answer is A.

Q3hard

Consider the limit of the multivariable function f(x,y)=2x2yx4+y2f(x,y) = \frac{2x^2y}{x^4+y^2}f(x,y)=x4+y22x2y​ as (x,y)(x,y)(x,y) approaches the origin (0,0)(0,0)(0,0). By analyzing the function's behavior along different approach paths, determine which of the following statements correctly identifies the existence and value of this limit.

A.

The limit does not exist because the function approaches 0 along all linear paths y=mxy = mxy=mx, but it approaches 1 along the parabolic path y=x2y = x^2y=x2.

B.

The limit exists and is equal to 0 because the limit along every straight line y=mxy = mxy=mx through the origin is 0.

C.

The limit exists and is equal to 1 because the parabolic path y=x2y = x^2y=x2 provides the most direct approach to simplify the indeterminate form.

D.

The limit cannot be determined because it results in an indeterminate 00\frac{0}{0}00​ form that cannot be simplified using standard algebraic techniques for multivariable functions.

Show answer & explanation

Correct Answer: A

To determine if the limit L=lim⁡(x,y)→(0,0)2x2yx4+y2L = \lim_{(x,y) \to (0,0)} \frac{2x^2y}{x^4+y^2}L=lim(x,y)→(0,0)​x4+y22x2y​ exists, we analyze the function along different paths toward the origin:

  1. Linear Paths: Let (x,y)(x,y)(x,y) approach (0,0)(0,0)(0,0) along any line y=mxy = mxy=mx. Substituting y=mxy = mxy=mx into the function gives: f(x,mx)=2x2(mx)x4+(mx)2=2mx3x4+m2x2=2mxx2+m2f(x, mx) = \frac{2x^2(mx)}{x^4 + (mx)^2} = \frac{2mx^3}{x^4 + m^2x^2} = \frac{2mx}{x^2 + m^2}f(x,mx)=x4+(mx)22x2(mx)​=x4+m2x22mx3​=x2+m22mx​\nAs x→0x \to 0x→0, the limit becomes 00+m2=0\frac{0}{0 + m^2} = 00+m20​=0. Note that along the yyy-axis (x=0x=0x=0), the function is also identically 0 for all y≠0y \neq 0y=0. Since the limit is 0 along every straight line through the origin, one might prematurely conclude the limit is 0.

  2. Parabolic Path: Now consider the path y=x2y = x^2y=x2. Substituting y=x2y = x^2y=x2 into the function gives: f(x,x2)=2x2(x2)x4+(x2)2=2x4x4+x4=2x42x4=1f(x, x^2) = \frac{2x^2(x^2)}{x^4 + (x^2)^2} = \frac{2x^4}{x^4 + x^4} = \frac{2x^4}{2x^4} = 1f(x,x2)=x4+(x2)22x2(x2)​=x4+x42x4​=2x42x4​=1\nAs x→0x \to 0x→0, the limit along this path is 1.

  3. Conclusion: Because the function approaches different values (0 and 1) along two different paths toward the origin, the limit does not exist (DNE). The fact that all linear paths agree is not sufficient to prove existence. The correct answer is A.

Q4easy

Given the function f(x,y)=x2+3xy2f(x, y) = x^2 + 3xy^2f(x,y)=x2+3xy2, which of the following represents the gradient vector ∇f(x,y)\nabla f(x, y)∇f(x,y)?

A.

⟨2x+3y2,6xy⟩\langle 2x + 3y^2, 6xy \rangle⟨2x+3y2,6xy⟩

B.

⟨6xy,2x+3y2⟩\langle 6xy, 2x + 3y^2 \rangle⟨6xy,2x+3y2⟩

C.

2x+3y2+6xy2x + 3y^2 + 6xy2x+3y2+6xy

D.

⟨2x,6y⟩\langle 2x, 6y \rangle⟨2x,6y⟩

Show answer & explanation

Correct Answer: A

To find the gradient vector ∇f(x,y)\nabla f(x, y)∇f(x,y), we must compute the partial derivatives of the function with respect to each variable. First, find the partial derivative with respect to xxx (fxf_xfx​) by differentiating f(x,y)=x2+3xy2f(x, y) = x^2 + 3xy^2f(x,y)=x2+3xy2 while treating yyy as a constant: fx=∂∂x(x2)+∂∂x(3xy2)=2x+3y2f_x = \frac{\partial}{\partial x}(x^2) + \frac{\partial}{\partial x}(3xy^2) = 2x + 3y^2fx​=∂x∂​(x2)+∂x∂​(3xy2)=2x+3y2. Next, find the partial derivative with respect to yyy (fyf_yfy​) by differentiating while treating xxx as a constant: fy=∂∂y(x2)+∂∂y(3xy2)=0+3x(2y)=6xyf_y = \frac{\partial}{\partial y}(x^2) + \frac{\partial}{\partial y}(3xy^2) = 0 + 3x(2y) = 6xyfy​=∂y∂​(x2)+∂y∂​(3xy2)=0+3x(2y)=6xy. The gradient vector is the vector of these partial derivatives: ∇f(x,y)=⟨fx,fy⟩=⟨2x+3y2,6xy⟩\nabla f(x, y) = \langle f_x, f_y \rangle = \langle 2x + 3y^2, 6xy \rangle∇f(x,y)=⟨fx​,fy​⟩=⟨2x+3y2,6xy⟩. Thus, the correct option is A.

Q5easy

Consider the transformation from the uvuvuv-plane to the xyxyxy-plane defined by the following linear equations:

x=5u−2vx = 5u - 2vx=5u−2v y=−u+3vy = -u + 3vy=−u+3v

What is the value of the Jacobian determinant J(u,v)=∂(x,y)∂(u,v)J(u, v) = \frac{\partial(x, y)}{\partial(u, v)}J(u,v)=∂(u,v)∂(x,y)​?

A.

13

B.

17

C.

5

D.

15

Show answer & explanation

Correct Answer: A

To identify the Jacobian determinant J(u,v)=∂(x,y)∂(u,v)J(u, v) = \frac{\partial(x, y)}{\partial(u, v)}J(u,v)=∂(u,v)∂(x,y)​, we first compute the partial derivatives of the coordinate functions x(u,v)x(u, v)x(u,v) and y(u,v)y(u, v)y(u,v):

  1. ∂x∂u=5\frac{\partial x}{\partial u} = 5∂u∂x​=5
  2. ∂x∂v=−2\frac{\partial x}{\partial v} = -2∂v∂x​=−2
  3. ∂y∂u=−1\frac{\partial y}{\partial u} = -1∂u∂y​=−1
  4. ∂y∂v=3\frac{\partial y}{\partial v} = 3∂v∂y​=3

The Jacobian is defined as the determinant of the matrix of these partial derivatives:

J(u,v)=det⁡(∂x∂u∂x∂v∂y∂u∂y∂v)=det⁡(5−2−13)J(u, v) = \det \begin{pmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{pmatrix} = \det \begin{pmatrix} 5 & -2 \\ -1 & 3 \end{pmatrix}J(u,v)=det(∂u∂x​∂u∂y​​∂v∂x​∂v∂y​​)=det(5−1​−23​)

Applying the determinant formula for a $2 \times 2$$ matrix, $$\det \begin{pmatrix} a & b \ c & d \end{pmatrix} = ad - bc$:

J(u, v) = (5)(3) - (-2)(-1) J(u,v)=15−2J(u, v) = 15 - 2J(u,v)=15−2 J(u,v)=13J(u, v) = 13J(u,v)=13

Thus, the correct Jacobian determinant is 13.

Q6easy

Which of the following represents the Cartesian point (1,3,5)(1, \sqrt{3}, 5)(1,3​,5) in cylindrical coordinates (r,θ,z)(r, \theta, z)(r,θ,z)?

A.

(2,π3,5)(2, \frac{\pi}{3}, 5)(2,3π​,5)

B.

(2,π6,5)(2, \frac{\pi}{6}, 5)(2,6π​,5)

C.

(29,π3,5)(\sqrt{29}, \frac{\pi}{3}, 5)(29​,3π​,5)

D.

(4,π3,5)(4, \frac{\pi}{3}, 5)(4,3π​,5)

Show answer & explanation

Correct Answer: A

To convert from Cartesian coordinates (x,y,z)(x, y, z)(x,y,z) to cylindrical coordinates (r,θ,z)(r, \theta, z)(r,θ,z), we use the following formulas:

  1. Calculate the radial distance rrr: r=x2+y2=12+(3)2=1+3=4=2r = \sqrt{x^2 + y^2} = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2r=x2+y2​=12+(3​)2​=1+3​=4​=2

  2. Calculate the angular coordinate θ\thetaθ: tan⁡θ=yx=31=3\tan \theta = \frac{y}{x} = \frac{\sqrt{3}}{1} = \sqrt{3}tanθ=xy​=13​​=3​ Since the point (1,3)(1, \sqrt{3})(1,3​) is in the first quadrant of the xyxyxy-plane, θ=arctan⁡(3)=π3\theta = \arctan(\sqrt{3}) = \frac{\pi}{3}θ=arctan(3​)=3π​.

  3. Identify the vertical coordinate zzz: In cylindrical coordinates, the zzz-coordinate remains the same as in Cartesian coordinates, so z=5z = 5z=5.

Combining these, we get the cylindrical coordinates (2,π3,5)(2, \frac{\pi}{3}, 5)(2,3π​,5).

Q7easy

The modeling of an RLC series circuit (consisting of an inductor, resistor, and capacitor) results in a second-order linear differential equation for the charge q(t)q(t)q(t) on the capacitor. Suppose a specific circuit is governed by the equation:

0.2q′′(t)+4q′(t)+50q(t)=12sin⁡(t)0.2 q''(t) + 4 q'(t) + 50 q(t) = 12 \sin(t)0.2q′′(t)+4q′(t)+50q(t)=12sin(t)

Based on the standard form of this differential equation, what is the capacitance CCC of this circuit in farads?

A.

0.02 F

B.

50 F

C.

0.2 F

D.

4 F

Show answer & explanation

Correct Answer: A

To find the capacitance, we compare the given equation to the standard second-order differential equation for a series RLC circuit:

Lq′′(t)+Rq′(t)+1Cq(t)=E(t)L q''(t) + R q'(t) + \frac{1}{C} q(t) = E(t)Lq′′(t)+Rq′(t)+C1​q(t)=E(t)

  1. Identify the coefficients: By comparing the given equation $0.2 q''(t) + 4 q'(t) + 50 q(t) =12sin⁡(t)= 12 \sin(t)=12sin(t) to the standard form, we see that the coefficient of the q(t)$ term is 50.
  2. Relate the coefficient to capacitance: In the standard form, the coefficient of q(t)q(t)q(t) is 1C\frac{1}{C}C1​, where CCC is the capacitance.
  3. Solve for C: 1C=50\frac{1}{C} = 50C1​=50 C=150=0.02C = \frac{1}{50} = 0.02C=501​=0.02

Thus, the capacitance of the circuit is 0.02 F. The other coefficients represent the inductance (L=0.2L = 0.2L=0.2 H) and the resistance (R=4R = 4R=4 Ω\OmegaΩ).

Q8easy

Consider the point PPP plotted on the polar coordinate grid below. What are the polar coordinates of point PPP in the form (r,θ)(r, \theta)(r,θ)?

A.

(3,120∘)(3, 120^\circ)(3,120∘)

B.

(4,120∘)(4, 120^\circ)(4,120∘)

C.

(3,60∘)(3, 60^\circ)(3,60∘)

D.

(120,3)(120, 3)(120,3)

Show answer & explanation

Correct Answer: A

To identify the polar coordinates (r,θ)(r, \theta)(r,θ) of point PPP, follow these steps:

  1. Determine the radial coordinate (rrr): The value of rrr represents the distance from the pole (the origin) to the point. On a polar grid, this is found by counting the concentric circles. Point PPP is located on the third circle from the center, so r=3r = 3r=3.
  2. Determine the angular coordinate (θ\thetaθ): The value of θ\thetaθ represents the angle measured counterclockwise from the polar axis (the positive x-axis). Looking at the rays on the grid, point PPP lies on the ray labeled 120∘120^\circ120∘.

Combining these, the coordinates of point PPP are (3,120∘)(3, 120^\circ)(3,120∘).

Q9hard

Suppose u\mathbf{u}u and v\mathbf{v}v are non-zero vectors in R3\mathbb{R}^3R3 such that ∥u∥=3\|\mathbf{u}\| = 3∥u∥=3, ∥v∥=4\|\mathbf{v}\| = 4∥v∥=4, and their dot product is u⋅v=6\mathbf{u} \cdot \mathbf{v} = 6u⋅v=6. Analyze the geometric and algebraic properties of the cross product vector w=u×v\mathbf{w} = \mathbf{u} \times \mathbf{v}w=u×v and determine which of the following statements is true.

A.

The vector w\mathbf{w}w has a magnitude of 6, and it lies within the plane spanned by u\mathbf{u}u and v\mathbf{v}v.

B.

The area of the parallelogram formed by u\mathbf{u}u and v\mathbf{v}v is 636\sqrt{3}63​ square units, and w\mathbf{w}w is orthogonal to the vector 5u+πv5\mathbf{u} + \pi \mathbf{v}5u+πv.

C.

The magnitude of w\mathbf{w}w is 12 because the vectors are orthogonal, and reversing the operation to v×u\mathbf{v} \times \mathbf{u}v×u results in the same vector w\mathbf{w}w.

D.

The vector w\mathbf{w}w is parallel to both u\mathbf{u}u and v\mathbf{v}v, and its magnitude is calculated using the formula ∥u∥∥v∥cos⁡(θ)\|\mathbf{u}\|\|\mathbf{v}\|\cos(\theta)∥u∥∥v∥cos(θ).

Show answer & explanation

Correct Answer: B

To analyze the cross product w=u×v\mathbf{w} = \mathbf{u} \times \mathbf{v}w=u×v, we first determine the angle θ\thetaθ between the vectors using the dot product formula:

  1. Find cos⁡(θ)\cos(\theta)cos(θ): u⋅v=∥u∥∥v∥cos⁡(θ)\mathbf{u} \cdot \mathbf{v} = \|\mathbf{u}\| \|\mathbf{v}\| \cos(\theta)u⋅v=∥u∥∥v∥cos(θ) 6=(3)(4)cos⁡(θ)  ⟹  cos⁡(θ)=612=126 = (3)(4) \cos(\theta) \implies \cos(\theta) = \frac{6}{12} = \frac{1}{2}6=(3)(4)cos(θ)⟹cos(θ)=126​=21​ This implies θ=60∘\theta = 60^\circθ=60∘ or π3\frac{\pi}{3}3π​ radians.

  2. Calculate the magnitude of the cross product: The magnitude represents the area of the parallelogram formed by u\mathbf{u}u and v\mathbf{v}v. ∥w∥=∥u×v∥=∥u∥∥v∥sin⁡(θ)\|\mathbf{w}\| = \|\mathbf{u} \times \mathbf{v}\| = \|\mathbf{u}\| \|\mathbf{v}\| \sin(\theta)∥w∥=∥u×v∥=∥u∥∥v∥sin(θ) ∥w∥=(3)(4)sin⁡(60∘)=12(32)=63\|\mathbf{w}\| = (3)(4) \sin(60^\circ) = 12 \left( \frac{\sqrt{3}}{2} \right) = 6\sqrt{3}∥w∥=(3)(4)sin(60∘)=12(23​​)=63​

  3. Analyze orthogonality: The cross product w\mathbf{w}w is, by definition, orthogonal to both u\mathbf{u}u and v\mathbf{v}v. Therefore, it is orthogonal to any linear combination of u\mathbf{u}u and v\mathbf{v}v (any vector in the plane spanned by them). w⋅(5u+πv)=5(w⋅u)+π(w⋅v)=5(0)+π(0)=0\mathbf{w} \cdot (5\mathbf{u} + \pi\mathbf{v}) = 5(\mathbf{w} \cdot \mathbf{u}) + \pi(\mathbf{w} \cdot \mathbf{v}) = 5(0) + \pi(0) = 0w⋅(5u+πv)=5(w⋅u)+π(w⋅v)=5(0)+π(0)=0

Thus, the area is 636\sqrt{3}63​ and the vector is orthogonal to any linear combination of the original vectors. Option B is correct.

Q10easy

To find the angle θ\thetaθ between two planes, one must first identify the normal vectors of each plane. Given the equations of two planes:

P1:2x−3y+z−5=0P_1: 2x - 3y + z - 5 = 0P1​:2x−3y+z−5=0 P2:x+4y−2z+8=0P_2: x + 4y - 2z + 8 = 0P2​:x+4y−2z+8=0

Which pair of normal vectors n1\mathbf{n_1}n1​ and n2\mathbf{n_2}n2​ should be used in the angle calculation formula?

A.

n1=⟨2,−3,1⟩\mathbf{n_1} = \langle 2, -3, 1 \ranglen1​=⟨2,−3,1⟩ and n2=⟨1,4,−2⟩\mathbf{n_2} = \langle 1, 4, -2 \ranglen2​=⟨1,4,−2⟩

B.

n1=⟨2,−3,−5⟩\mathbf{n_1} = \langle 2, -3, -5 \ranglen1​=⟨2,−3,−5⟩ and n2=⟨1,4,8⟩\mathbf{n_2} = \langle 1, 4, 8 \ranglen2​=⟨1,4,8⟩

C.

n1=⟨12,−13,1⟩\mathbf{n_1} = \langle \frac{1}{2}, -\frac{1}{3}, 1 \ranglen1​=⟨21​,−31​,1⟩ and n2=⟨1,14,−12⟩\mathbf{n_2} = \langle 1, \frac{1}{4}, -\frac{1}{2} \ranglen2​=⟨1,41​,−21​⟩

D.

n1=⟨2,3,1⟩\mathbf{n_1} = \langle 2, 3, 1 \ranglen1​=⟨2,3,1⟩ and n2=⟨1,4,2⟩\mathbf{n_2} = \langle 1, 4, 2 \ranglen2​=⟨1,4,2⟩

Show answer & explanation

Correct Answer: A

To identify the normal vector of a plane from its scalar equation ax + by + cz + d = 0, we look at the coefficients of the variables xxx, yyy, and zzz. The normal vector is given by n=⟨a,b,c⟩\mathbf{n} = \langle a, b, c \ranglen=⟨a,b,c⟩.

  1. For Plane P1:2x−3y+z−5=0P_1: 2x - 3y + z - 5 = 0P1​:2x−3y+z−5=0, the coefficients are a=2a=2a=2, b=−3b=-3b=−3, and c=1c=1c=1. Thus, n1=⟨2,−3,1⟩\mathbf{n_1} = \langle 2, -3, 1 \ranglen1​=⟨2,−3,1⟩.
  2. For Plane P2:x+4y−2z+8=0P_2: x + 4y - 2z + 8 = 0P2​:x+4y−2z+8=0, the coefficients are a=1a=1a=1, b=4b=4b=4, and c=−2c=-2c=−2. Thus, n2=⟨1,4,−2⟩\mathbf{n_2} = \langle 1, 4, -2 \ranglen2​=⟨1,4,−2⟩.

Option B incorrectly includes the constant term ddd. Option C uses reciprocals of the coefficients. Option D fails to account for the negative signs in the coefficients. Therefore, Option A is the correct identification of the normal vectors.

Q11easy

When transforming a triple integral from rectangular coordinates (x,y,z)(x, y, z)(x,y,z) to spherical coordinates (ρ,θ,ϕ)(\rho, \theta, \phi)(ρ,θ,ϕ), which of the following expressions represents the correct differential volume element dVdVdV?

A.

dV=ρ2sin⁡ϕ dρ dϕ dθdV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\thetadV=ρ2sinϕdρdϕdθ

B.

dV=r dz dr dθdV = r \, dz \, dr \, d\thetadV=rdzdrdθ

C.

dV=ρsin⁡ϕ dρ dϕ dθdV = \rho \sin \phi \, d\rho \, d\phi \, d\thetadV=ρsinϕdρdϕdθ

D.

dV=ρ2cos⁡ϕ dρ dϕ dθdV = \rho^2 \cos \phi \, d\rho \, d\phi \, d\thetadV=ρ2cosϕdρdϕdθ

Show answer & explanation

Correct Answer: A

To transform an integral into spherical coordinates, we must account for the change in volume scales using the Jacobian determinant. The transformation equations are:

x=ρsin⁡ϕcos⁡θx = \rho \sin \phi \cos \thetax=ρsinϕcosθ y=ρsin⁡ϕsin⁡θy = \rho \sin \phi \sin \thetay=ρsinϕsinθ z=ρcos⁡ϕz = \rho \cos \phiz=ρcosϕ

The absolute value of the Jacobian for this transformation is ∣J∣=ρ2sin⁡ϕ|J| = \rho^2 \sin \phi∣J∣=ρ2sinϕ. Consequently, the differential volume element is:

dV=ρ2sin⁡ϕ dρ dϕ dθdV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\thetadV=ρ2sinϕdρdϕdθ

  • Option B is the differential volume element for cylindrical coordinates.
  • Option C incorrectly omits the square on the radial distance (ρ2\rho^2ρ2).
  • Option D incorrectly uses the cosine function; the projection onto the xyxyxy-plane requires the sine of the polar angle ϕ\phiϕ.

The correct choice is A.

Q12easy

What is the divergence of the two-dimensional vector field F(x,y)=⟨x2,y2⟩\mathbf{F}(x, y) = \langle x^2, y^2 \rangleF(x,y)=⟨x2,y2⟩?

A.

2x+2y2x + 2y2x+2y

B.

⟨2x,2y⟩\langle 2x, 2y \rangle⟨2x,2y⟩

C.

2x−2y2x - 2y2x−2y

D.

0

Show answer & explanation

Correct Answer: A

To find the divergence of a two-dimensional vector field F(x,y)=⟨P,Q⟩\mathbf{F}(x, y) = \langle P, Q \rangleF(x,y)=⟨P,Q⟩, we use the standard formula: div F=∇⋅F=∂P∂x+∂Q∂y\text{div } \mathbf{F} = \nabla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y}div F=∇⋅F=∂x∂P​+∂y∂Q​

  1. Identify the components: In the field F(x,y)=⟨x2,y2⟩\mathbf{F}(x, y) = \langle x^2, y^2 \rangleF(x,y)=⟨x2,y2⟩, we have P(x,y)=x2P(x, y) = x^2P(x,y)=x2 and Q(x,y)=y2Q(x, y) = y^2Q(x,y)=y2.
  2. Calculate the partial derivatives:
    • ∂P∂x=∂∂x(x2)=2x\frac{\partial P}{\partial x} = \frac{\partial}{\partial x}(x^2) = 2x∂x∂P​=∂x∂​(x2)=2x
    • ∂Q∂y=∂∂y(y2)=2y\frac{\partial Q}{\partial y} = \frac{\partial}{\partial y}(y^2) = 2y∂y∂Q​=∂y∂​(y2)=2y
  3. Sum the derivatives: Adding these together gives div F=2x+2y\text{div } \mathbf{F} = 2x + 2ydiv F=2x+2y.

Note that the divergence is a scalar function, so option B is incorrect. Option C incorrectly uses subtraction, and option D would be the result if we used mixed partial derivatives or calculated the curl for this specific field.

The correct answer is 2x+2y2x + 2y2x+2y.

Q13easy

Which of the following represents the direction cosines of the vector v⃗=⟨2,1,2⟩\vec{v} = \langle 2, 1, 2 \ranglev=⟨2,1,2⟩?

A.

cos⁡α=23,cos⁡β=13,cos⁡γ=23\cos \alpha = \frac{2}{3}, \cos \beta = \frac{1}{3}, \cos \gamma = \frac{2}{3}cosα=32​,cosβ=31​,cosγ=32​

B.

cos⁡α=2,cos⁡β=1,cos⁡γ=2\cos \alpha = 2, \cos \beta = 1, \cos \gamma = 2cosα=2,cosβ=1,cosγ=2

C.

cos⁡α=25,cos⁡β=15,cos⁡γ=25\cos \alpha = \frac{2}{5}, \cos \beta = \frac{1}{5}, \cos \gamma = \frac{2}{5}cosα=52​,cosβ=51​,cosγ=52​

D.

cos⁡α=49,cos⁡β=19,cos⁡γ=49\cos \alpha = \frac{4}{9}, \cos \beta = \frac{1}{9}, \cos \gamma = \frac{4}{9}cosα=94​,cosβ=91​,cosγ=94​

Show answer & explanation

Correct Answer: A

To find the direction cosines of a vector v⃗=⟨a,b,c⟩\vec{v} = \langle a, b, c \ranglev=⟨a,b,c⟩, we must first calculate the magnitude of the vector, ∣v⃗∣|\vec{v}|∣v∣.

  1. Calculate the magnitude: ∣v⃗∣=a2+b2+c2=22+12+22|\vec{v}| = \sqrt{a^2 + b^2 + c^2} = \sqrt{2^2 + 1^2 + 2^2}∣v∣=a2+b2+c2​=22+12+22​ ∣v⃗∣=4+1+4=9=3|\vec{v}| = \sqrt{4 + 1 + 4} = \sqrt{9} = 3∣v∣=4+1+4​=9​=3

  2. Identify the direction cosines: The direction cosines are the ratios of each component to the magnitude:

  • cos⁡α=a∣v⃗∣=23\cos \alpha = \frac{a}{|\vec{v}|} = \frac{2}{3}cosα=∣v∣a​=32​
  • cos⁡β=b∣v⃗∣=13\cos \beta = \frac{b}{|\vec{v}|} = \frac{1}{3}cosβ=∣v∣b​=31​
  • cos⁡γ=c∣v⃗∣=23\cos \gamma = \frac{c}{|\vec{v}|} = \frac{2}{3}cosγ=∣v∣c​=32​

Therefore, the set of direction cosines is cos⁡α=23,cos⁡β=13,cos⁡γ=23\cos \alpha = \frac{2}{3}, \cos \beta = \frac{1}{3}, \cos \gamma = \frac{2}{3}cosα=32​,cosβ=31​,cosγ=32​.

Q14hard

A particle moves in space according to the position vector r(t)=⟨cos⁡(2t),sin⁡(2t),t3/2⟩\mathbf{r}(t) = \langle \cos(2t), \sin(2t), t^{3/2} \rangler(t)=⟨cos(2t),sin(2t),t3/2⟩ for t≥0t \ge 0t≥0. Analyze the particle's trajectory to determine the arc-length function s(t)s(t)s(t) for the path starting at t=0t=0t=0, and find the total distance traveled by the particle as ttt increases from 0 to 4.

A.

127(10413−64)\frac{1}{27}(104\sqrt{13} - 64)271​(10413​−64)

B.

127(10413−4)\frac{1}{27}(104\sqrt{13} - 4)271​(10413​−4)

C.

118(10413−64)\frac{1}{18}(104\sqrt{13} - 64)181​(10413​−64)

D.

127(10413)\frac{1}{27}(104\sqrt{13})271​(10413​)

Show answer & explanation

Correct Answer: A

To find the total distance using the arc-length function, we follow these steps:

  1. Find the velocity vector r′(t)\mathbf{r}'(t)r′(t): r′(t)=ddt⟨cos⁡(2t),sin⁡(2t),t3/2⟩=⟨−2sin⁡(2t),2cos⁡(2t),32t1/2⟩\mathbf{r}'(t) = \frac{d}{dt} \langle \cos(2t), \sin(2t), t^{3/2} \rangle = \langle -2\sin(2t), 2\cos(2t), \frac{3}{2}t^{1/2} \rangler′(t)=dtd​⟨cos(2t),sin(2t),t3/2⟩=⟨−2sin(2t),2cos(2t),23​t1/2⟩

  2. Calculate the speed ∥r′(t)∥\|\mathbf{r}'(t)\|∥r′(t)∥: ∥r′(t)∥=(−2sin⁡(2t))2+(2cos⁡(2t))2+(32t1/2)2\|\mathbf{r}'(t)\| = \sqrt{(-2\sin(2t))^2 + (2\cos(2t))^2 + \left(\frac{3}{2}t^{1/2}\right)^2}∥r′(t)∥=(−2sin(2t))2+(2cos(2t))2+(23​t1/2)2​ ∥r′(t)∥=4sin⁡2(2t)+4cos⁡2(2t)+94t=4(1)+94t=16+9t4=1216+9t\|\mathbf{r}'(t)\| = \sqrt{4\sin^2(2t) + 4\cos^2(2t) + \frac{9}{4}t} = \sqrt{4(1) + \frac{9}{4}t} = \sqrt{\frac{16+9t}{4}} = \frac{1}{2}\sqrt{16+9t}∥r′(t)∥=4sin2(2t)+4cos2(2t)+49​t​=4(1)+49​t​=416+9t​​=21​16+9t​

  3. Set up the arc-length function s(t)s(t)s(t): s(t)=∫0t∥r′(u)∥du=∫0t12(16+9u)1/2dus(t) = \int_{0}^{t} \|\mathbf{r}'(u)\| du = \int_{0}^{t} \frac{1}{2}(16 + 9u)^{1/2} dus(t)=∫0t​∥r′(u)∥du=∫0t​21​(16+9u)1/2du

  4. Integrate: Using uuu-substitution with v=16+9uv = 16+9uv=16+9u and dv=9dudv=9dudv=9du: s(t)=12⋅19[23(16+9u)3/2]0t=127[(16+9t)3/2−163/2]s(t) = \frac{1}{2} \cdot \frac{1}{9} \left[ \frac{2}{3}(16+9u)^{3/2} \right]_0^t = \frac{1}{27} \left[ (16+9t)^{3/2} - 16^{3/2} \right]s(t)=21​⋅91​[32​(16+9u)3/2]0t​=271​[(16+9t)3/2−163/2] s(t)=127[(16+9t)3/2−64]s(t) = \frac{1}{27} \left[ (16+9t)^{3/2} - 64 \right]s(t)=271​[(16+9t)3/2−64]

  5. Evaluate at t=4t=4t=4: s(4)=127[(16+36)3/2−64]=127[523/2−64]s(4) = \frac{1}{27} \left[ (16+36)^{3/2} - 64 \right] = \frac{1}{27} \left[ 52^{3/2} - 64 \right]s(4)=271​[(16+36)3/2−64]=271​[523/2−64] Since 523/2=5252=52(213)=1041352^{3/2} = 52\sqrt{52} = 52(2\sqrt{13}) = 104\sqrt{13}523/2=5252​=52(213​)=10413​: s(4)=127(10413−64)s(4) = \frac{1}{27}(104\sqrt{13} - 64)s(4)=271​(10413​−64)

The total distance traveled is 127(10413−64)\frac{1}{27}(104\sqrt{13} - 64)271​(10413​−64).

Q15easy

Which of the following is the derivative r′(t)\mathbf{r}'(t)r′(t) of the vector-valued function r(t)=⟨t2,cos⁡(t),5t⟩\mathbf{r}(t) = \langle t^2, \cos(t), 5t \rangler(t)=⟨t2,cos(t),5t⟩?

A.

r′(t)=⟨2t,−sin⁡(t),5⟩\mathbf{r}'(t) = \langle 2t, -\sin(t), 5 \rangler′(t)=⟨2t,−sin(t),5⟩

B.

r′(t)=⟨2t,sin⁡(t),5⟩\mathbf{r}'(t) = \langle 2t, \sin(t), 5 \rangler′(t)=⟨2t,sin(t),5⟩

C.

r′(t)=⟨13t3,sin⁡(t),52t2⟩\mathbf{r}'(t) = \langle \frac{1}{3}t^3, \sin(t), \frac{5}{2}t^2 \rangler′(t)=⟨31​t3,sin(t),25​t2⟩

D.

r′(t)=2t−sin⁡(t)+5\mathbf{r}'(t) = 2t - \sin(t) + 5r′(t)=2t−sin(t)+5

Show answer & explanation

Correct Answer: A

To find the derivative of a vector-valued function r(t)=⟨f(t),g(t),h(t)⟩\mathbf{r}(t) = \langle f(t), g(t), h(t) \rangler(t)=⟨f(t),g(t),h(t)⟩, we differentiate each component function independently with respect to ttt. This is expressed as:

r′(t)=⟨f′(t),g′(t),h′(t)⟩\mathbf{r}'(t) = \langle f'(t), g'(t), h'(t) \rangler′(t)=⟨f′(t),g′(t),h′(t)⟩

Given r(t)=⟨t2,cos⁡(t),5t⟩\mathbf{r}(t) = \langle t^2, \cos(t), 5t \rangler(t)=⟨t2,cos(t),5t⟩, we differentiate each part:

  1. The derivative of the first component, f(t)=t2f(t) = t^2f(t)=t2, is f′(t)=2tf'(t) = 2tf′(t)=2t using the power rule.
  2. The derivative of the second component, g(t)=cos⁡(t)g(t) = \cos(t)g(t)=cos(t), is g′(t)=−sin⁡(t)g'(t) = -\sin(t)g′(t)=−sin(t).
  3. The derivative of the third component, h(t)=5th(t) = 5th(t)=5t, is h′(t)=5h'(t) = 5h′(t)=5.

Combining these results back into a vector, we get: r′(t)=⟨2t,−sin⁡(t),5⟩\mathbf{r}'(t) = \langle 2t, -\sin(t), 5 \rangler′(t)=⟨2t,−sin(t),5⟩

Choice A is the correct derivative vector.

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Flowchart, left to right. Force 1: Mag, Angle connects to Component Form. Force 2: Mag, Angle connects to C. C connects to Sum Components: i_total, j_total. D connects to Calculate Resultant Magnitude. E connects to Calculate Resultant Angle.
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Flowchart, top to bottom. Start: r(t) connects to Calculate Velocity: r'(t). B connects to Find Speed: ||r'(t)||. C connects to Integrate: s = Integral of Speed. D connects to Invert: Solve t in terms of s. E connects to Result: r(s) Arc-Length Parameterization.
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Flowchart, top to bottom. Start: Find Area Between Two Curves connects to Sketch the Graphs. B connects to Set r1 = r2 to find Intersection Points. C connects to Is the Pole an Intersection?. D -- Yes connects to Check if r=0 for both curves at any theta. D -- No connects to Determine Outer and Inner Radii. E connects to F. F connects to Set up Integral: 1/2 Integral r_out^2 - r_in^2 d_theta. G connects to Evaluate and Apply Symmetry if applicable.
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Flowchart, top to bottom. Start: Find dy/dx connects to Are x and y functions of t?. B -- Yes connects to Calculate dx/dt and dy/dt. C connects to Divide: dy/dt / dx/dt. D connects to Is dx/dt = 0?. E -- No connects to Result: Slope of Tangent. E -- Yes connects to Vertical Tangent or Undefined.
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Flowchart, top to bottom. Differentiation of Vector Functions connects to Component-wise Calculus. Differentiation of Vector Functions] --> B[Component-wise Calculus connects to Derivative Rules. B connects to dx/dt, dy/dt, dz/dt. C connects to Dot Product Rule. C connects to Cross Product Rule. C connects to Chain Rule. B1 connects to Result: Tangent Vector r'(t).
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Flowchart, top to bottom. Start: Examine Eccentricity 'e connects to Is e = 1?. B -- Yes connects to Parabola. B -- No connects to Is e < 1?. D -- Yes connects to Ellipse. D -- No connects to Hyperbola. E connects to Is e = 0?. G -- Yes connects to Circle.
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Flowchart, top to bottom. Start: Vector Field F connects to Is Domain Simply Connected?. B -- No connects to Test may be inconclusive/require more analysis. B -- Yes connects to Is it 2D or 3D?. D -- 2D: <P,Q> connects to Does Py = Qx?. D -- 3D: <P,Q,R> connects to Does curl F = 0?. E -- Yes connects to F is Conservative. E -- No connects to F is NOT Conservative. F -- Yes connects to G. 1 more statements.
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Mermaid diagram. root Conservative Field. Properties. Path Independence. Zero Work on Closed Loops. Mathematics. F = Gradient of f. Curl is Zero. Topological Needs. 2 more statements.
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Flowchart, left to right. A[Rectangular: x, y, z] -- "r = sqrt(x^2+y^2) connects to Cylindrical: r, theta, z. B -- "rho = sqrt(r^2+z^2) connects to Spherical: rho, theta, phi. C -- "z = rho cos(phi) connects to Rectangular: x, y, z] -- "r = sqrt(x^2+y^2)" --> B[Cylindrical: r, theta, z. C -- "x = rho sin(phi) cos(theta) connects to Rectangular: x, y, z] -- "r = sqrt(x^2+y^2)" --> B[Cylindrical: r, theta, z.
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Mermaid diagram. root Gradient Vector. Direction. Steepest Ascent. Orthogonal to Level Sets. Magnitude. Maximum Rate of Change. Components. Partial Derivative wrt x. 2 more statements.
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Flowchart, top to bottom. Vector Field F connects to Calculate Curl F. B -- Curl F = 0 connects to Irrotational / Potentially Conservative. B -- Curl F != 0 connects to Rotational. C connects to Simply Connected Region?. E -- Yes connects to F is Conservative. E -- No connects to Further Investigation Needed.