« Math 324 Mid-Term 1 Solution
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Mid-Term 2 Solution

五月 19, 2009

1) (a) \int_{0}^{2 \pi}(\frac{1}{\cos{t} + 2} + \sin^2{t}, 2\cos{t}\sin{t}) \cdot (-\sin{t}, \cos{t}) dt.

(b) \textbf{F} = (P, Q), P_y = 2y, Q_x = 2y. Hence \textbf{F} is conservative. Therefore, by the fundamental theorem of line integrals, since the curve C is closed. The answer is 0.

2(a) div(\textbf{F})(x, y, z) = 3x^2 + 3y^2 + 3z^2 - 1. Therefore, div(\textbf{F})(1, 1, 1) = 3 + 3 + 3 - 1 = 8. Therefore, $\textbf{F}$ is pretty cool at (1, 1, 1).

(b) We want the region in which div(\textbf{F}) < 0. This is the same as 3x^2 + 3y^2 + 3z^2 \le 1, which is a sphere with radius \frac{1}{\sqrt{3}}.

(c) The volume is \frac{4}{3(\sqrt{3})^3}\pi = \frac{4\sqrt{3}}{27} \pi

3) \textbf{F} = (2x + y, x) = (P, Q). Now P_y = Q_x = 1. Hence \textbf{F} is conservative. Suppose \textbf{F} = \triangledown f, solving f_x = 2x + y yields f(x,y) = x^2 + xy + g(y). Now f_y = x + g'(y) = x yields g'(y) = 0 meaning g(y) is constant. Hence we may choose f(x, y) = x^2 + xy.

Finally, \textbf{r}(0) = (0, 0), \textbf{r}(2) = (4, 0). Hence by the fundamental theorem of line integrals, we have \int_{C} \textbf{F} \cdot d\textbf{r} = f(4, 0) - f(0, 0) = 16.

4) (a) \frac{\partial g}{\partial x} = \frac{\partial g}{\partial r}\frac{\partial r}{\partial x} + \frac{\partial g}{\partial s}\frac{\partial s}{\partial x}.

Hence \frac{\partial g}{\partial x}(0, 0) = 5*1 + 6*1 = 11.

\frac{\partial g}{\partial y} = \frac{\partial g}{\partial r}\frac{\partial r}{\partial y} + \frac{\partial g}{\partial s}\frac{\partial s}{\partial y}.

Hence \frac{\partial g}{\partial y}(0, 0) = 5*0 + 6*1 = 6.

Therefore, \triangledown(g \circ \textbf{F})(0, 0) = (11, 6).

(b) \frac{d}{dt}(g \circ \textbf{c}) = (\frac{\partial g}{\partial r}, \frac{\partial g}{\partial s}) \cdot \textbf{c}'(t). Hence \frac{d}{dt}(g \circ \textbf{c})(0) = (3, 4) \cdot (2, 1) = 6 + 4 = 10.

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