Calculus III Advanced (Course) Final Exam

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Grade A

Calculus III Advanced (Course) Final Exam

Reprinted from UF Mac2313 Final Exam UF Mac2313 Final Exam Solutions.
Rules
The Good News

  • Open book.
  • Open notebook.
  • Open internet.
  • No time limit.
  • Collaboration is encouraged.
  • Calculator use is encouraged.

The Bad News

  • The student must get every single problem correct in order to pass.
  • Show your work.
  • Looking up where to find the answer is part of the problem.
  • If the student's notes are inadequate to answer the question, those notes must be updated and referred to in the answer.

Exam

Question 1 (10pt.)

The vector equation for a curve \(C\) is \( \overrightarrow{r}(t)=\left \langle 3 \cos t, 3 \sin t, 4t \right \rangle\).

(a) (5pt.) Find the arc length between the points \(P(3,0,0)\) and \(Q(3,0,8\pi)\) on the curve \(C\).
(b) (5pt.) Find the coordinates for a point \(R\) other than \(P(3,0,0)\) on the curve \(C\), such that the arc length between \(R\) and \(Q\) is the same length as the arc length between \(P\) and \(Q\).

Solution
Green check.svg.png(a) Here, \(t=0\) to \(t=2 \pi\). Letting \(x(t)=3 \cos t\), \(y(t)=3 \sin t\), and \(z(t)=4t\) and differentiation produces \(x^{\prime}(t)=-3 \sin t\), \(y^{\prime}(t)=3 \cos t\), and \(z^{\prime}(t)=4\). The arc length from \(t=0\) to \(t=2 \pi\) is given by

$$ \int_{0}^{2 \pi}| \overrightarrow{r}^{\prime}(t) | $$ $$=\int_{0}^{2 \pi} \sqrt{9 \sin^{2} t + 9 \cos^{2} t + 16 } \: dt$$
$$= \int_{0}^{2 \pi} \sqrt{9 \sin^{2} 0 + 9 \cos^{2} 1 + 16 } \: dt $$
$$=\int_{0}^{2 \pi} \sqrt{25 }\: dt $$
$$=\int_{0}^{2 \pi} 5\: dt = 10 \pi.$$

Green check.svg.png(b) Take the formula created in 1(a) and set \(5(t-2\pi) = 10 \pi\) to yield \(t=4\pi\), which corresponds to \(R(3,0,16\pi)\).

Question 2 (10pt.)

For \(f(x,y,z)=xe^{y}+ye^{z}+ze^{x}\).

(a) (3pt.) Find the gradient for \(f(x,y,z)\).
(b) (4pt.) Find the directional derivative for \(f\) at the point \((0,0,0)\) in \((0,2,1)\)'s direction.
(c) (3pt.) Find the maximum change rate for \(f\) at the point \((0,0,0)\). In which direction does it occur?

Solution
Green check.svg.png(a) Taking the partial derivatives yields

\(f_{x}(x,y,z)=e^{y}+e^{x}z\)
\(f_{y}(x,y,z)=e^{z}+e^{y}x\)
\(f_{z}(x,y,z)=e^{z}y+e^{x}\)

which produces

\(\nabla f(x,y,z) \) \(=f_{x}(x,y,z)\textbf{i}+f_{y}(x,y,z)\textbf{j}+f_{z}(x,y,z)\textbf{k} \)
\(=\left( e^{y}+e^{x}z\right)\textbf{i} +\left( e^{z}+e^{y}x\right) \textbf{j} +\left( e^{z}y+e^{x}\right) \textbf{k} \)

Green check.svg.png(b)

\(u \) $$= \frac{\langle 0,2,1 \rangle}{\sqrt{0^{2}+2^{2}+1^{2}}} = \frac{1}{\sqrt{5}}\langle 0,2,1 \rangle \text{Normalize the destination vector.}$$
\((\nabla f)(0,0,0) \) \(= \langle 1,1,1 \rangle \:\:\:\: e^{0}=1 \)
\( D_{u}f(0,0,0) \) \(= (\nabla f)(0,0,0) \cdot u\)
$$= \frac{1}{\sqrt{5}} \langle 1,1,1 \rangle \cdot \langle 0,2,1 \rangle = \frac{3}{\sqrt{5}}$$

Green check.svg.png(c) The maximum change rate occurs in the gradient's direction. Therefore,

\(\nabla f(0,0,0) = \langle 1,1,1 \rangle \),

and the change rate is given by the gradient's magnitude

\(|\nabla f(0,0,0)| = \sqrt{3} \).

Question 3 (10pt.)

Find the local maximum and minimum values at the saddle points for \(f(x,y)=x^{2}-xy+y^{2}+9x-6y+10\), if they exist.
SolutionGreen check.svg.png Take the partial derivatives

\(f_{x} \) \(=2x-y+9 \)
\(f_{y} \) \(=-x+2y-6 \)

Letting \(f_{x}=0 \) and \(f_{y}=0 \) yields \(x=-4\) and \(y=1\). Therefore, (-4,1) is the only critical point. The second partial derivatives are

\(f_{xx} \) \(= 2\)
\(f_{yy} \) \(=2 \)
\(f_{xy} \) \(=-1 \)

Since \(f_{xx}f_{yy}-f_{xx}^{2} > 0 \) at the point (-4,1), \(f(x,y)\) has a local minimum \(f(-4,1)=-11\) at the point (-4,1).

Question 4 (10pt.)

Sketch the region for the integral and calculate the iterated integral by first reversing the integration order.

$$ \int_{0}^{3} \int_{\sqrt{y/3}}^{1} e^{x^{3}} \: dx \: dy$$
Figure 1

SolutionGreen check.svg.png

$$ \int_{0}^{3} \int_{\sqrt{y/3}}^{1} e^{x^{3}} \: dx \: dy$$ $$= \int_{0}^{1} \int_{0}^{3x^{2}} e^{x^{3}}\: dy \: dx $$
$$= \int_{0}^{1} 3x^{2} e^{x^{3}} \: dx $$
$$=\int_{0}^{1} e^{u} \: du \:\:\:\: \text{substitute } u=x^{3}$$
\(= e-1\)

Question 5 (10pt.)

Find the volume for the solid bounded by the two parabolas

\(z=3x^{2}+3y^{2}\)

and

\(z=4-x^{2}-y^{2}\).

SolutionGreen check.svg.png Set \(3x^{2}+3y^{2} =4-x^{2}-y^{2}\) yields \(x^{2}+y^{2}=1\). This means the intersection is the unit disk.

The volume is given by

$$\int \int_{D}[(4-x^{2}-y^{2})-(3x^{2}+3y^{2})] \:dA $$ $$=\int \int_{D} (4-4x^{2}-4y^{2}) \:dA $$
$$=\int_{0}^{2\pi} \int_{0}^{1}(4-4r^{2})r \: dr \: d \theta $$
$$= \int_{0}^{2\pi} \int_{0}^{1}(4r-4r^{3}) \: dr \: d \theta $$
$$= \left. \int_{0}^{2\pi}(2r^{2}-r^{4}) \right]_{0}^{1} = 2 \pi$$

Question 6 (10pt.)

Find the volume for the solid that lies within the sphere

\(x^{2}+y^{2}+z^{2}=4\),

above the \(xy\)-plane and below the cone

\(z=\sqrt{x^{2}+y^{2}} .\)

Solution Green check.svg.png In spherical coordinates,

\(x\) \(= \rho \sin \phi \cos \theta\)
\( y\) \(= \rho \sin \phi \sin \theta \)
\( z\) \(= \rho \cos \phi. \)

Plug these into \(x^{2}+y^{2}+z^{2}=4\) yields \(\rho = 2\).
Plug these into \(z=\sqrt{x^{2}+y^{2}} \) yields the cone is \(\phi = \pi/4\).
The volume is

$$V $$ $$=\int_{0}^{2\pi} \int_{\pi/4}^{\pi/2} \int_{0}^{2} \rho^{2} \sin \phi \: d\rho \:d\phi \: d \theta $$
$$=\int_{0}^{2\pi} \: d \theta \int_{\pi/4}^{\pi/2} \sin \phi \: d\phi \int_{0}^{2} \rho^{2} \:d\rho $$
$$= (2\pi)\left( \frac{\sqrt{2}}{2} \right)\frac{8}{3} = \frac{8}{3}\sqrt{2}\pi.$$

Question 7 (10pt.)

Use Stokes's Theorem to evaluate

$$ \int \int_{S} \text{curl} \: \textbf{F} \cdot \: dS, $$

where \(\textbf{F}(x,y,z) = x^{2}z^{2}\textbf{i}+y^{2}z^{2}\textbf{j}+xyz\textbf{k} \) and \(S\) is in the paraboloid \(z=x^{2}+y^{2}\) that lies inside the cylinder \(x^{2}+y^{2}=1\), oriented upward.
Hint: Stokes's Theorem:

Figure 2

SolutionGreen check.svg.png The boundary for \(S\) is the intersection between the paraboloid and cylinder. The circle has radius 1 on the plane \(z = 1\). Thus, the boundary \(C\) has a parametric equation

\( r(\theta)\) \(= \langle \cos \theta, \sin \theta, 1 \rangle, \:\:\:\: 0 \leqslant \theta \leqslant 2\pi \)
\( r^{\prime}(\theta)\) \(= \langle - \sin \theta, \cos \theta, 0 \rangle. \)

Applying Stokes' Theorem produces

$$\int \int_{S} \text{curl} \textbf{F} \cdot \: dS $$ $$= \int_{C} \textbf{F} \cdot \:dr $$
$$= \int_{0}^{2\pi} \textbf{F} \cdot r^{\prime}(\theta) \: d\theta$$
$$= \int_{0}^{2\pi} \langle (\cos \theta)^{2}(1)^{2}, (\sin \theta)^{2}(1)^{2},(\cos \theta)(\sin \theta)(1) \rangle \cdot r^{\prime}(\theta) \: d\theta$$
$$= \int_{0}^{2\pi} (-\cos^{2}\theta \sin \theta+ \sin^{2} \theta \cos \theta )\: d\theta$$
$$= \left. \frac{\cos^{3} \theta}{3} \right]_{0}^{2\pi} +\left. \frac{\sin^{3} \theta}{3} \right]_{0}^{2\pi}= 0+0 =0.$$

Question 8 (10pt.)

Evaluate

$$\int \int_{S} \textbf{F} \cdot \: d S $$

where

\( \textbf{F} = x\textbf{i}+y\textbf{j}+\textbf{k}\)

and \(S\) is the cone

\( z^{2} = x^{2}+y^{2}\)

that lies between the planes \(z=1\) and \(z=2\), oriented upward.
SolutionGreen check.svg.png The parametric equation for \(S\) is \(r(a,\theta)= \langle a \cos \theta, a \sin \theta, a \rangle\), where \(1 \leqslant a \leqslant 2\) and \(0\leqslant \theta \leqslant 2\pi\).

\( r_{a}\) \(= \langle \cos \theta, \sin \theta, 1 \rangle\)
\( r_{\theta}\) \(= \langle -a \sin \theta, a \cos \theta, 0 \rangle\)
\( r_{a} \times r_{\theta} \) \(=\langle -a \cos \theta, -a \sin \theta, a \rangle \:\:\:\:\) upward orientation
$$ \int \int_{S} \textbf{F} \cdot \: dS$$ $$= \int \int_{D} \textbf{F} \cdot (r_{a} \times r_{\theta}) \:dA , \:\:\:\: D=[1,2] \times [0,2\pi]$$
$$=\int_{0}^{2\pi} \int_{1}^{2} \langle a\cos \theta, a\sin \theta, 1 \rangle \cdot \langle -a \cos \theta, -a \sin \theta, a \rangle \:da \:d\theta $$
$$=\int_{0}^{2\pi} \int_{1}^{2} (-a^{2}+a) \:da \:d\theta $$
$$= \left. 2 \pi \left( -\frac{a^{3}}{3} + \frac{a^{2}}{2} \right) \right]_{1}^{2} = -\frac{5\pi}{3} $$

Question 9 (10pt.)

Find the area for the surface \(3x+4y+z=6\) that lies in the first octant.
Solution Green check.svg.png A parameter about the surface plane \(S\) is

\( r(x,y)=\langle x,y,6-3x-4y\rangle\),

where

\(0 \leqslant y \leqslant \frac{6-3x}{4}\), \( 0 \leqslant x \leqslant 2\). The partial derivatives are
\( r_{x}\) \(= \langle 1,0,-3 \rangle \)
\( r_{y}\) \(= \langle 0,1,-4 \rangle \)
\( r_{x} \times r_{y}\) \(= \langle 3,4,1 \rangle \)
\( |r_{x} \times r_{y}|\) \(= \sqrt{26} \)

The area is given by

$$ \int \int_{S} \: dS $$ $$= \int \int_{S} |r_{x} \times r_{y}| \: dy \:dx $$
$$= \int_{0}^{2} \int_{0}^{\frac{6-3x}{4}} \sqrt{26} \: dy \:dx $$
$$= \sqrt{26} \int_{0}^{2} \frac{6-3x}{4} \: dx$$
$$= \sqrt{26} \left. \left( \frac{6}{4}x-\frac{3}{8}x^{2} \right) \right]_{0}^{2} = \frac{3}{2} \sqrt{26}.$$

Question 10 (10pt.)

\( \textbf{F}(x,y) = x^{2}\textbf{i}+y^{2}\textbf{j}\)
(a) (5pt.) Show that \(\textbf{F}\) is conservative and find \(f\) such that
\( \nabla f(x,y)=\textbf{F}(x,y).\)
(b) (5pt.) Use the result in part (a) and the Fundamental Theorem for line integral to calculate
$$\int_{C} \textbf{F} \cdot \: d\textbf{r},$$

where \(C\) is the arc for the parabola \(y=2x^{2}\) from \((0,0)\) to \((1,2)\).
Solution
Green check.svg.png(a) Set \(P=x^{2}\), \(Q=y^{2}\). Then

$$\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}.$$

Since \(\textbf{F}\) is well defined everywhere, \(\textbf{F}\) is conservative.

If \(\nabla f(x,y)=\textbf{F}(x,y)\), then

\( f_{x}\) \(=x^{2} \)
\( f_{y}\) \(=y^{2} \)

By partial integration,

$$ f=\frac{1}{3}x^{3} +\frac{1}{3}y^{3} +K$$

where \(K\) is a constant. Green check.svg.png(b)

$$\int_{C} \textbf{F} \cdot \: dr $$ \(=f(1,2)-f(0,0) \)
\(= 3.\)

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Parent Article: Calculus III Advanced (Course)