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PAGE 1

Tangent Plane to a Surface

The equation of the tangent plane to the graph of the function \( f(x, y) = x - \frac{y^2}{2} \) at \( (1, 2, -1) \) is

define \( F = x - \frac{1}{2}y^2 - z \)

\( \vec{\nabla}F = \langle 1, -y, -1 \rangle \)

at \( (1, 2, -1) \): \( \vec{\nabla}F = \langle 1, -2, -1 \rangle \)

normal to surface

so use as normal of tangent plane

A. \( 2x + y + 4z = 0 \)
B. \( x + 4y = 9 \)
C. \( x - 2y - z = 2 \)
D. \( -x + 2y + z = 2 \)
E. \( x - y - 2z = 1 \)

Plane Equation Derivation

plane: \( a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \)

\( (1)(x - 1) - 2(y - 2) - (z + 1) = 0 \)

\( x - 1 - 2y + 4 - z - 1 = 0 \)

\( x - 2y - z = -2 \)

\( -x + 2y + z = 2 \)

\( \langle a, b, c \rangle \) : normal vector

\( (x_0, y_0, z_0) \) : point

PAGE 2

Line Integrals and Green's Theorem

If \( C \) goes from \( (1, 0) \) counterclockwise once around the circle \( x^2 + y^2 = 1 \), then

\[ \int_C (x^2 + 2xy)dx + (x^2 + 2x + y)dy = \]

Figure: Coordinate graph showing a unit circle centered at the origin with counterclockwise arrows and point (1,0).

A. \( 4\pi \)
B. \( -4\pi \)
C. \( -2\pi \)
D. 0
E. \( 2\pi \)

\( \vec{F} = \langle x^2 + 2xy, x^2 + 2x + y \rangle \)

\( d\vec{r} = \langle dx, dy \rangle \)

choices: parametrize \( C \), then compute \( \int_C \vec{F} \cdot d\vec{r} \)

Closed \( C \), so can use Green's Theorem

Green's: \( \int_C \vec{F} \cdot d\vec{r} = \iint_R (g_x - f_y) dA \)

\( g_x = 2x + 2 \)

\( f_y = 2x \)

\( g_x - f_y = 2 \)

\[ \int_C \vec{F} \cdot d\vec{r} = \iint_R 2 dA = 2 \iint_R dA = 2 \cdot \pi(1)^2 = 2\pi \]

area of \( R \): circle radius 1

PAGE 3

Surface Area of a Parametric Surface

Find the area of the parametric surface \( \vec{r}(u, v) = u \cos v \mathbf{i} + u \sin v \mathbf{j} + v \mathbf{k} \), \( 0 \le u \le \pi \), \( 0 \le v \le u \).

Area formula:

\[ \text{area: } \iint_R |\vec{r}_u \times \vec{r}_v| dA \]

Note: \( dS = |\vec{r}_u \times \vec{r}_v| dA \)

\[ \vec{r}_u = \langle \cos v, \sin v, 0 \rangle \]\[ \vec{r}_v = \langle -u \sin v, u \cos v, 1 \rangle \]\[ \vec{r}_u \times \vec{r}_v = \langle \sin v, -\cos v, u \rangle \]\[ |\vec{r}_u \times \vec{r}_v| = \sqrt{1 + u^2} \]

Limits: \( 0 \le u \le \pi, \quad 0 \le v \le u \)

Multiple Choice Options

A. \( \frac{2}{3}(1 + \pi^2)^{\frac{3}{2}} \)
B. \( \frac{1}{3}((1 + \pi^2)^{\frac{3}{2}} - 1) \) Correct
C. \( \frac{1}{2}(\sqrt{1 + \pi^2} - 1) \)
D. \( \frac{1}{5}((1 + \pi^2)^{\frac{3}{2}} - 1) \)
E. \( \frac{1}{5}(1 + \pi^2)^{\frac{3}{2}} \)

Calculation

\[ \int_0^\pi \int_0^u \sqrt{1 + u^2} \, dv \, du = \int_0^\pi u \sqrt{1 + u^2} \, du \]

Substitution: \( w = 1 + u^2 \)

\( dw = 2u \, du \)

\[ = \int_1^{1+\pi^2} \frac{1}{2} w^{1/2} \, dw = \left. \frac{1}{3} w^{3/2} \right|_1^{1+\pi^2} \]\[ = \frac{1}{3} \left[ (1 + \pi^2)^{3/2} - 1 \right] \]

PAGE 4

Line Integral over a Curve

Let \( C \) be the curve \( y = \frac{1}{3}x^3 \), \( 0 \le x \le 1 \). Compute \( \int_C 12y \, ds \).

Parametrize \( C \):

Let \( x = t \)

Then \( \vec{r}(t) = \langle t, \frac{1}{3}t^3 \rangle \), \( 0 \le t \le 1 \)

Then \( ds = |\vec{r}'| \, dt \)

\[ \vec{r}' = \langle 1, t^2 \rangle \]\[ |\vec{r}'| = \sqrt{1 + t^4} \]

Multiple Choice Options

A. \( 8(2^{3/2} - 1) \)
B. \( 12(2^{3/2} - 1) \)
C. \( \frac{2}{3}(2^{3/2} - 1) \) Correct
D. \( \frac{3}{2}(2^{3/2} - 1) \)
E. \( 2(2^{3/2} - 1) \)

Integration

\[ \int_C 12y \, ds = \int_0^1 12 \left( \frac{1}{3}t^3 \right) \sqrt{1 + t^4} \, dt \]\[ = \int_0^1 4t^3 \sqrt{1 + t^4} \, dt \]

Substitution: \( u = 1 + t^4 \)

\( du = 4t^3 \, dt \)

\[ = \int_1^2 u^{1/2} \, du = \left. \frac{2}{3} u^{3/2} \right|_1^2 \]\[ = \frac{2}{3} \left[ (2)^{3/2} - 1 \right] \]

PAGE 5

Tangent Line Intersection with the xy-plane

Let \( C \) be the curve given by \( \vec{r}(t) = \langle 4\sqrt{t}, t, 5 - t^2 \rangle \) for \( t > 0 \). At what point does the tangent line to \( C \) at \( (4, 1, 4) \) intersect the \( xy \) plane?

A. \( (0, 1, 0) \)
B. \( (4\sqrt{5}, \sqrt{5}, 0) \)
C. \( (2, 1, 0) \)
D. \( (8, 3, 0) \)
E. \( (0, -1, 0) \)

Figure: A 3D sketch of a curve r(t) with a tangent line at point (4,1,4) passing through a rectangular xy-plane where z=0.

Line: \( \vec{l}(t) = \vec{r}_0 + t \vec{v} \)

\[ \vec{r}' = \langle \frac{2}{\sqrt{t}}, 1, -2t \rangle \]

Now need \( t \) so the point is \( \langle 4, 1, 4 \rangle \). Given \( \vec{r}(t) = \langle 4\sqrt{t}, t, 5 - t^2 \rangle \), we see \( t = 1 \).

\[ \vec{r}'(1) = \langle 2, 1, -2 \rangle = \vec{v} \]

Tangent line:

\[ \vec{l}(t) = \langle 4, 1, 4 \rangle + t \langle 2, 1, -2 \rangle \]\[ = \langle 4 + 2t, 1 + t, 4 - 2t \rangle \]

\( z = 0 \) at intersection w/ \( xy \)-plane

\( t = 2 \)

Location at \( t = 2 \): \( \vec{l}(2) = \langle 8, 3, 0 \rangle \)

PAGE 6

Constrained Optimization: Lagrange Multipliers

Find the maximum value of \( x^2 + y^2 \) subject to the constraint \( x^2 - 2x + y^2 - 4y = 0 \).

Lagrange is usable

A. 2
B. 4
C. 10
D. 16
E. 20

\( f = x^2 + y^2 \)

\( g = x^2 - 2x + y^2 - 4y = 0 \)

\[ \vec{\nabla} f = \lambda \vec{\nabla} g \quad \text{solve for } x, y \]

\[ \langle 2x, 2y \rangle = \lambda \langle 2x - 2, 2y - 4 \rangle \]

\( 2x = \lambda · (2x - 2) \rightarrow \lambda = \frac{2x}{2x - 2} = \frac{x}{x - 1} \)

\( 2y = \lambda · (2y - 4) \rightarrow \lambda = \frac{2y}{2y - 4} = \frac{y}{y - 2} \)

\( \} \frac{x}{x - 1} = \frac{y}{y - 2} \)

\[ x(y - 2) = y(x - 1) \]\[ xy - 2x = xy - y \rightarrow y = 2x \]

PAGE 7

Substitution and Optimization

Sub into \(x^2 - 2x + y^2 - 4y = 0\)

\[x^2 - 2x + (2x)^2 - 4(2x) = 0\]\[x^2 - 2x + 4x^2 - 8x = 0\]\[5x^2 - 10x = 0\]\[5x(x - 2) = 0 \rightarrow x = 0, \quad x = 2\]\[y = 0, \quad y = 4\]

Points to check: \((0, 0), (2, 4)\)

Compare \(f = x^2 + y^2 \rightarrow\) max at \((2, 4), f = 20\)

PAGE 8

Triple Integral Conversion to Cylindrical Coordinates

The triple integral \(\int_{-2}^{2} \int_{0}^{\sqrt{4-x^2}} \int_{0}^{\sqrt{x^2+y^2}} (x^2+y^2) \, dz \, dy \, dx\) when converted to cylindrical coordinates becomes:

Analysis of Limits

\(dz \, dy \, dx\)

"floor" xy-plane

\(-2 \le x \le 2\)
\(0 \le y \le \sqrt{4-x^2}\)

top of circle radius 2

Figure: A semi-circle on the xy-plane with radius 2, centered at the origin, spanning from x = -2 to 2 above the x-axis.

\(0 \le r \le 2\)

\(0 \le \theta \le \pi\)

Z-Limit Analysis

\(0 \le z \le \sqrt{x^2+y^2}\)

xy-plane to cone

In cylindrical, \(z = r\)

Multiple Choice Options

A. \(\int_{0}^{\pi} \int_{0}^{4} \int_{0}^{r} z^2 r \, dz \, dr \, d\theta\)
B. \(\int_{0}^{\pi} \int_{0}^{2} \int_{0}^{r} z^2 r \, dz \, dr \, d\theta\)
C. \(\int_{0}^{\pi} \int_{0}^{2} \int_{0}^{r} r^3 \, dz \, dr \, d\theta\) (Correct)
D. \(\int_{0}^{2\pi} \int_{0}^{2} \int_{0}^{r} r^3 \, dz \, dr \, d\theta\)
E. \(\int_{0}^{2\pi} \int_{0}^{4} \int_{0}^{r} r^3 \, dz \, dr \, d\theta\)

Final Conversion

\[\int_{0}^{\pi} \int_{0}^{2} \int_{0}^{r} r^2 \cdot r \, dz \, dr \, d\theta\]\[= \int_{0}^{\pi} \int_{0}^{2} \int_{0}^{r} r^3 \, dz \, dr \, d\theta\]

PAGE 9

Line Integral of a Conservative Vector Field

Evaluate ∫_C F ∙ dr where C is the line segment from P(1, 0) to Q(0, 2) and F(x, y) = ⟨ye^xy, xe^xy + cos y⟩ is a conservative vector field.

A. 1 + sin 2
B sin 2
C. 1 − sin 2
D. −1 + sin 2
E. −1 − sin 2

Note: conservative vector field → curl F = 0

Fundamental Theorem of Line Integrals

If \( \vec{F} = \vec{\nabla} \phi \) then

\[ \int_C \vec{F} \cdot d\vec{r} = \phi(\text{end}) - \phi(\text{start}) \]

Finding the Potential Function \( \phi \)

We are given \( \vec{F} = \langle ye^{xy}, xe^{xy} + \cos y \rangle = \vec{\nabla} \phi = \langle \phi_x, \phi_y \rangle \).

\( \phi_x = ye^{xy} \implies \phi = \int ye^{xy} dx = e^{xy} + h(y) \)

\( \phi_y = xe^{xy} + \cos y \)

Differentiating our expression for \( \phi \) with respect to \( y \):

\[ \phi_y = xe^{xy} + \frac{dh}{dy} \]

Comparing the two expressions for \( \phi_y \):

So, \( \frac{dh}{dy} = \cos y \implies h = \sin y + C \)

Thus, the potential function is: \( \phi = e^{xy} + \sin y + C \)

Final Calculation

\[ \int_C \vec{F} \cdot d\vec{r} = \phi(\text{end}) - \phi(\text{start}) = \phi(0, 2) - \phi(1, 0) \]\[ = (1 + \sin 2 + C) - (1 + C) \]\[ = \sin 2 \]