Free Calculus III: Multivariable Calculus Study Resources

To change the order of integration, we must first identify the region $D$ from the given limits:

1. Identify Boundaries: The original limits are $0 \le x \le 1$and$x \le y \le 1. This describes a triangular region in the xy$-plane.
2. Visualize the Region:
   • The left boundary is the $y$-axis ($x=0$).
   • The upper boundary is the horizontal line $y=1$.
   • The lower/right boundary is the line $y=x$.
3. Switch to Type II: To integrate with respect to $x first, we look at the horizontal extent for a fixed value of y$:
   • For any $y$ in the interval $[0, 1]$, the $x-values start at the y$-axis ($x=0) and end at the line y=x$ (which is $x=y$).
   • Therefore, the inner limits are $0 \le x \le y$.
   • The outer limits for $y$ cover the full vertical range of the triangle: $0 \le y \le 1$.

Combining these, we get: $\int_0^1 \int_0^y e^{y^2} \, dx \, dy$

Evaluating this is now possible because the inner integral \int_0^y e^{y^2}$ \, dx$ results in x e^{y^2} \Big|_0^y = y e^{y^2}$$, which can be integrated with respect to $y$ using a simple $u$-substitution ($u=y^2$).

Final Answer: A

To analyze the limit $\lim_{(x,y) \to (0,0)} \frac{x^2y}{x^4 + y^2}$, we first examine linear paths $y = mx$. Substituting $y = mx$ into the function gives:

$\lim_{x \to 0} \frac{x^2(mx)}{x^4 + (mx)^2} = \lim_{x \to 0} \frac{mx^3}{x^4 + m^2x^2} = \lim_{x \to 0} \frac{mx}{x^2 + m^2} = \frac{0}{0 + m^2} = 0$

While all linear paths yield 0, this is insufficient to prove the limit exists. In multivariable calculus, the limit must be the same along every possible path.

Next, we analyze the function using a parabolic path, $y = kx^2. Choosing this path is strategic because the powers in the denominator (x^4$ and $y^2) become comparable. Substituting y = kx^2$:

$\lim_{x \to 0} \frac{x^2(kx^2)}{x^4 + (kx^2)^2} = \lim_{x \to 0} \frac{kx^4}{x^4 + k^2x^4} = \lim_{x \to 0} \frac{kx^4}{x^4(1 + k^2)} = \frac{k}{1 + k^2}$

If we let $k=1$ (the path $y=x^2$), the limit is $1/2$. If we let$k=0$(the path$y=0$), the limit is 0. Since different paths yield different limits, the limit does not exist.

To describe the motion of a particle in space:

1. The velocity vector $\mathbf{v}(t)$ is the first derivative of the position vector with respect to time: $\mathbf{v}(t) = \frac{d}{dt}\mathbf{r}(t) = \mathbf{r}'(t)$ Geometrically, the velocity vector is always tangent to the trajectory (path) of the particle at any point $t$.

2. The acceleration vector $\mathbf{a}(t)$ is the first derivative of the velocity vector, which is the second derivative of the position vector: $\mathbf{a}(t) = \frac{d}{dt}\mathbf{v}(t) = \mathbf{r}''(t)$

Option B is incorrect because it swaps the order of derivatives and incorrectly states the direction of velocity. Option C defines velocity as a scalar (speed) rather than a vector and suggests acceleration is always parallel to velocity (which only happens in straight-line motion). Option D uses the wrong derivative for acceleration.

Correct Option: A

To determine if a vector field $\mathbf{F} = \langle P, Q, R \rangle$ is conservative on a simply connected domain like $\mathbb{R}^3$, we must verify that its curl is zero, which is equivalent to checking three pairs of mixed partial derivatives:

1. Check $\frac{\partial P}{\partial y}$ and $\frac{\partial Q}{\partial x}$: $\frac{\partial}{\partial y}(2xy + z^2) = 2x$ $\frac{\partial}{\partial x}(x^2 + \cos(z)) = 2x$ These are equal ($2x = 2x$).

2. Check $\frac{\partial P}{\partial z}$ and $\frac{\partial R}{\partial x}$: $\frac{\partial}{\partial z}(2xy + z^2) = 2z$ $\frac{\partial}{\partial x}(2xz - y\sin(z)) = 2z$ These are equal ($2z = 2z$).

3. Check $\frac{\partial Q}{\partial z}$ and $\frac{\partial R}{\partial y}$: $\frac{\partial}{\partial z}(x^2 + \cos(z)) = -\sin(z)$ $\frac{\partial}{\partial y}(2xz - y\sin(z)) = -\sin(z)$ These are equal ($-\sin(z) = -\sin(z)$).

Since all three conditions $(\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}, \frac{\partial P}{\partial z} = \frac{\partial R}{\partial x}, \frac{\partial Q}{\partial z} = \frac{\partial R}{\partial y})$ are satisfied everywhere in $\mathbb{R}^3$ and the domain is simply connected, the vector field is conservative. Option C is incorrect because divergence being zero (solenoidal) does not imply a field is conservative (irrotational). Options B and D are factually incorrect based on the calculations above. The vector field is conservative because its curl is zero.

To evaluate the integral, we analyze the geometry of the region and the integrand to convert the problem into spherical coordinates.

1. Analyze the Boundary: The boundary of the region $V$ is given by $x^2 + y^2 + z^2 = 2z$. In spherical coordinates, we use the transformations $x^2 + y^2 + z^2 = \rho^2$ and $z = \rho \cos \phi$. Substituting these into the boundary equation yields: $\rho^2 = 2\rho \cos \phi \implies \rho = 2 \cos \phi$ Since the sphere $x^2 + y^2 + (z-1)^2 = 1$ is tangent to the $xy-plane at the origin and lies entirely in the upper half-space (z \ge 0$), the polar angle $\phi$ ranges from 0 to $\pi/2. The region is symmetric about the z$-axis, so $0 \le $\theta$ \le $2\pi$$.

2. Transform the Integrand: The function to be integrated is $f(x, y, z)$= \sqrt{x^2 + y^2 + z^2}$, which simplifies to$\rho$in spherical coordinates. The volume element is dV$= \rho^2 \sin \phi$\,$d\rho$\,$d\phi$\,$d\theta$$.

3. Set Up and Evaluate the Integral: $I = \int_{0}^{2\pi} \int_{0}^{\pi/2} \int_{0}^{2\cos \phi} \rho \cdot (\rho^2 \sin \phi) \, d\rho \, d\phi \, d\theta = \int_{0}^{2\pi} \int_{0}^{\pi/2} \left[ \frac{\rho^4}{4} \sin \phi \right]_{0}^{2\cos \phi} d\phi \, d\theta$ $I = \int_{0}^{2\pi} \int_{0}^{\pi/2} 4 \cos^4 \phi \sin \phi \, d\phi \, d\theta$ Using the substitution $u = \cos \phi$, $du$= -\sin \phi$\,$d\phi$:$\int_{0}^{\pi/2} 4 \cos^4 \phi \sin \phi , d\phi = 4 \int_{0}^{1} u^4 , du = \frac{4}{5}$Finally, integrating over$\theta$from 0 to$2\pi$:$I = \int_{0}^{2\pi} \frac{4}{5} , d\theta = \frac{8\pi}{5}$The final answer is **$\frac{8\pi}{5}$$**.

To evaluate the line integral $\int_C \mathbf{F} \cdot d\mathbf{r}$ for a complex path $C, we must first determine if the vector field \mathbf{F} = \langle P, Q \rangle is conservative by verifying if the partial derivatives satisfy \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$.

1. Verify the Conservative Property: $\frac{\partial P}{\partial y} = \frac{\partial}{\partial y} \left( \frac{y}{1+x^2 y^2} + 2xy \right) = \frac{(1+x^2 y^2)(1) - y(2x^2 y)}{(1+x^2 y^2)^2} + 2x = \frac{1 - x^2 y^2}{(1+x^2 y^2)^2} + 2x$ $\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x} \left( \frac{x}{1+x^2 y^2} + x^2 \right) = \frac{(1+x^2 y^2)(1) - x(2xy^2)}{(1+x^2 y^2)^2} + 2x = \frac{1 - x^2 y^2}{(1+x^2 y^2)^2} + 2x$ Since $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$ and the domain $\mathbb{R}^2$ is simply connected, the vector field is conservative.

2. Identify the Potential Function $f(x, y)$: We seek a function $f$ such that $\nabla f = \mathbf{F}$. Integrating the first component $P$ with respect to $x$: $f(x, y) = \int \left( \frac{y}{1+x^2 y^2} + 2xy \right) dx = \arctan(xy) + x^2 y + g(y)$ Differentiating this result with respect to $y and setting it equal to Q$: $f_y = \frac{x}{1+x^2 y^2} + x^2 + g'(y) = \frac{x}{1+x^2 y^2} + x^2$ This implies $g'(y) = 0$, so we can choose $g(y) = 0. Thus, the potential function is f(x, y) = \arctan(xy) + x^2 y$.

3. Apply the Fundamental Theorem for Line Integrals: The integral is path-independent, meaning we only need to evaluate the potential function at the boundary points: $\int_C \mathbf{F} \cdot d\mathbf{r} = f(2, 0.5) - f(0, 5)$ $f(2, 0.5) = \arctan(2 \cdot 0.5) + 2^2(0.5) = \arctan(1) + 2 = \frac{\pi}{4} + 2$ $f(0, 5) = \arctan(0 \cdot 5) + 0^2(5) = 0 + 0 = 0$ $\text{Value} = \left( \frac{\pi}{4} + 2 \right) - 0 = \frac{\pi}{4} + 2$

The final answer is $\frac{\pi}{4} + 2$.

To determine the polar equation, we follow these analytical steps:

1. Identify the standard form: Since the focus is at the pole and the directrix is a vertical line $x = p$ (where $p > 0$), the major axis is horizontal. The standard polar form for such a conic is: $r = \frac{ep}{1 + e \cos \theta}$

2. Determine the semi-major axis: The length of the major axis is given as $2a = 10, which implies the semi-major axis a = 5$.

3. **Relate $a to the focal parameter p**: For an ellipse (e < 1) with a focus at the pole, the relationship between the semi-major axis and the distance to the directrix p$ is: $a = \frac{ep}{1 - e^2}$ Substituting the known values $a = 5$ and $e = 0.5$: $5 = \frac{0.5p}{1 - (0.5)^2} = \frac{0.5p}{0.75}$ $5 = \frac{2}{3}p \implies p = 7.5$

4. Construct the equation: Now, calculate the numerator $ep$: $ep = 0.5 \times 7.5 = 3.75 = \frac{15}{4}$ Substitute $ep$ and $e$ back into the polar form: $r = \frac{3.75}{1 + 0.5 \cos \theta}$ To simplify, multiply the numerator and denominator by 4: $r = \frac{15}{4 + 2 \cos \theta}$

Therefore, the correct equation is $r = \frac{15}{4 + 2 \cos \theta}$.

To solve this integral, we must first analyze the region of integration $E$ and recognize why the current order is problematic.

1. Identify the Problem: The innermost integral $\int \cos(x^2) \, dx$ is a non-elementary integral (it cannot be expressed in terms of standard functions). To evaluate the triple integral, we must change the order of integration so that we integrate with respect to $y$ or $z first, potentially producing an x term that allows for a u$-substitution in the $x$-integral.

2. Analyze the Region $E$: The given limits are:

   • $0 \le z \le 4$
   • $0 \le y \le 1$
   • $2y \le x \le 2$

   The projection of this solid onto the $xy-plane is a triangle bounded by y=0$, $x=2$, and $x=2y$.

3. **Change the Order of Integration ($dy \, dx)**: To switch from dx \, dy$ to $dy \, dx$ for this triangular region:

   • For a fixed $x$, $y ranges from the lower boundary y=0$ to the line $x=2y$, which is $y=x/2$.
   • The total range for $x$ is from 0 to 2.
   • The limits for $z$ remain 0 to 4.

   Thus, the new setup is $\int_{0}^{4} \int_{0}^{2} \int_{0}^{x/2} \cos(x^2) \, dy \, dx \, dz$.

4. Evaluate the Integral:

   • Inner ($dy$): $\int_{0}^{x/2} \cos(x^2) \, dy = [y \cos(x^2)]_{0}^{x/2} = \frac{x}{2} \cos(x^2)$.
   • Middle ($dx$): $\int_{0}^{2} \frac{x}{2} \cos(x^2) \, dx$. Let $u = x^2$, then $du = 2x \, dx \implies \frac{1}{4} du = \frac{x}{2} \, dx$. Changing limits: $x=0 \to u=0$; $x=2 \to u=4$. $\int_{0}^{4} \frac{1}{4} \cos(u) \, du = [\frac{1}{4} \sin(u)]_{0}^{4} = \frac{1}{4} \sin(4)$.
   • Outer ($dz$): $\int_{0}^{4} \frac{1}{4} \sin(4) \, dz = [\frac{z}{4} \sin(4)]_{0}^{4} = \sin(4)$.

Final Answer: $\sin(4)$

The acceleration vector $\mathbf{a} for a particle moving along a curve can be written as \mathbf{a} = a_T \mathbf{T} + a_N \mathbf{N}$, where $\mathbf{T} is the unit tangent vector and \mathbf{N}$ is the principal unit normal vector.

1. The tangential component ($a_T$) is given by $a_T = \frac{d}{dt}|\mathbf{v}(t)|$, which is the derivative of speed with respect to time. Thus, it accounts for the object's change in speed.
2. The normal component ($a_N$) is given by $a_N = \kappa v^2$ (where $\kappa$ is curvature and $v$ is speed). This component is responsible for the change in the direction of the velocity vector.

Therefore, Option A is correct.

To convert the polar equation $r = 6 \cos \theta - 8 \sin \theta$ to rectangular coordinates, follow these steps:

1. Multiply both sides of the equation by $r to create terms that can be easily substituted using the identities x = r \cos \theta$, $y = r \sin \theta$, and $r^2 = x^2 + y^2$: $r^2 = 6(r \cos \theta) - 8(r \sin \theta)$

2. Substitute the rectangular identities into the equation: $x^2 + y^2 = 6x - 8y$

3. Rearrange the equation to group $x$ and $y$ terms on one side: $x^2 - 6x + y^2 + 8y = 0$

4. Complete the square for both the $x$ and $y$ terms:

• For $x$: $(x^2 - 6x + 9)$ adding 9 to both sides.
• For $y$: $(y^2 + 8y + 16)$ adding 16 to both sides. $(x^2 - 6x + 9) + (y^2 + 8y + 16) = 9 + 16$

1. Simplify to the standard form of a circle equation $(x - h)^2 + (y - k)^2 = R^2$: $(x - 3)^2 + (y + 4)^2 = 25$

This represents a circle centered at $(3, -4) with a radius of 5. The correct rectangular form is **(x - 3)^2 + (y + 4)^2 = 25$**.

To find the resulting vector $\mathbf{u} + 2\mathbf{v}$, we apply scalar multiplication and then vector addition component-wise.

1. Scalar Multiplication: Multiply each component of vector $\mathbf{v}$ by the scalar 2: $2\mathbf{v} = 2\langle 1, 4, 0 \rangle = \langle 2(1), 2(4), 2(0) \rangle = \langle 2, 8, 0 \rangle$

2. Vector Addition: Add the components of $\mathbf{u} to the components of 2\mathbf{v}$: $\mathbf{u} + 2\mathbf{v} = \langle 3, -1, 2 \rangle + \langle 2, 8, 0 \rangle = \langle 3+2, -1+8, 2+0 \rangle = \langle 5, 7, 2 \rangle$

The final vector is $\langle 5, 7, 2 \rangle$.

To evaluate the integral, we first observe that the innermost integrand $\cos(y^3) has no elementary antiderivative with respect to y$. Therefore, we must change the order of integration.

Step 1: Identify the bounds of the region $E$. From the integral, the bounds are:

• $0 \le x \le 1$
• $0 \le z \le 1$
• $\sqrt{x} \le y \le 1$

Step 2: Change the order of integration. We want to integrate with respect to $x$ or $z first so that a factor of y^2 appears to facilitate a u-substitution later. We re-examine the xy-projection of the region E$. The bounds $0 \le x \le 1 and $$\sqrt{x} \le y \le 1$describe the area in the first quadrant below the line$y=1 and above the curve $y = \sqrt{x}$ (which is equivalent to x = y^2$).

Reversing the order for $x$ and $y$, we get:

• $0 \le y \le 1$
• $0 \le x \le y^2$

Since $z$ is independent of $x$ and $y in the original limits, we can place the dz integral anywhere. Let's use the order dx \, dz \, dy$: $I = \int_{0}^{1} \int_{0}^{1} \int_{0}^{y^2} \cos(y^3) \, dx \, dz \, dy$

Step 3: Evaluate the new integral. Inner integral ($dx$): $\int_{0}^{y^2} \cos(y^3) \, dx = [x \cos(y^3)]_0^{y^2} = y^2 \cos(y^3)$

Middle integral ($dz$): $\int_{0}^{1} y^2 \cos(y^3) \, dz = [z y^2 \cos(y^3)]_0^1 = y^2 \cos(y^3)$

Outer integral ($dy$): $\int_{0}^{1} y^2 \cos(y^3) \, dy$ Using $u$-substitution: Let $u = y^3$, so $du = 3y^2 \, dy$ or $y^2 \, dy$= \frac{1}{3}$du$. At $y=0, u=0$; at $y=1, u=1$. $\frac{1}{3} \int_{0}^{1} \cos(u) \, du = \frac{1}{3} [\sin(u)]_0^1 = \frac{1}{3} \sin(1)$

Thus, the final answer is $\frac{1}{3} \sin(1)$.

To find the arc length $L$ of a polar curve $r = f(\theta)$ from $\theta = \alpha$ to $\theta = \beta$, we use the formula:

$L = \int_{\alpha}^{\beta} \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} d\theta$

1. Find the derivative: Given $r = e^{3\theta}$, we calculate $\frac{dr}{d\theta}$ using the chain rule: $\frac{dr}{d\theta} = 3e^{3\theta}$

2. Set up the integrand: Substitute $r$ and $\frac{dr}{d\theta}$ into the radical expression: $\sqrt{(e^{3\theta})^2 + (3e^{3\theta})^2} = \sqrt{e^{6\theta} + 9e^{6\theta}} = \sqrt{10e^{6\theta}}$

3. Simplify the radical: Since $e^{6\theta} = (e^{3\theta})^2$ and $e^{3\theta} > 0$: $\sqrt{10e^{6\theta}} = \sqrt{10} e^{3\theta}$

4. Evaluate the integral: Integrate from 0 to $\pi$: $L = \int_{0}^{\pi} \sqrt{10} e^{3\theta} d\theta = \sqrt{10} \left[ \frac{1}{3}e^{3\theta} \right]_{0}^{\pi}$ $L = \frac{\sqrt{10}}{3} (e^{3\pi} - e^{0}) = \frac{\sqrt{10}}{3} (e^{3\pi} - 1)$

The exact arc length is $\frac{\sqrt{10}}{3}(e^{3\pi} - 1)$.

To identify the type of conic section from the general equation $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$, we look at the coefficients of the squared terms and the cross-product term.

1. Identify the coefficients: $A = 4$, $B = 0 (since there is no xy$ term), and $C = -9$.
2. Calculate the discriminant for conic classification: $B^2 - 4AC$.
3. Substitute the values: $0^2 - 4(4)(-9) = 144$.
4. Since the discriminant $144 > 0$, the equation represents a hyperbola.

Alternatively, when $B=0, we can simply look at the signs of A$ and $C$. Because $A$ and $C have opposite signs (4 is positive and -9$ is negative), the conic is a hyperbola.