Math 241, Final Exam. 12/11/12.

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1 Math, Final Exam. //. No notes, calculator, or text. There are points total. Partial credit may be given. ircle or otherwise clearly identify your final answer. Name:. (5 points): Equation of a line. Find parametric equations for the line through the point (,, ) that is parallel to the plane x + y + z 7 and perpendicular to the line x 3 + t, y t, z + t. Solution: The direction vector for the auxiliary line is v,,. The normal vector to the plane is n,,. The direction vector of the desired line is i j k v n ( ) i ( ) j + ( ( )) k 3 i + j + k. Hence, the line we seek has parametric equations x 3t, y t, z + t, t R.. (5 points): hain rule. Suppose that z F (x, y), x G(s, t), y H(u), u I(t). Write down the form of the chain rule that you would use to compute z/ t. (It may be helpful to use a tree diagram to show variable dependencies.) z t z x x t + z dy du y du dt. 3. ( points): Gradient vector and tangent planes. onsider the surface S : x y + z. (a) (5 points): Find a vector normal to S at the point P (x, y, z ). Solution: We write F (x, y, z) x y z. The normal is F (x, y, z ), y, z. (b) (5 points): Find the unique point P (x, y, z ) on S at which the tangent plane to S is parallel to the plane x + y z 3. Solution: The given plane has normal n,, ; the tangent plane to S at P has normal F (x, y, z ). It follows that these planes are parallel if and only if their normals are, which holds if and only if there exists k in R with, y, z k,,. omparing components of i gives k /. Hence, we have y k / and z k /; these give y / and z /. We use the equation for S to recover x y + z (/6) /8. We conclude that the desired point is (/8, /, /).

2 . (5 points): Directional derivative. Let f(x, y, z) x y + x + z. (a) ( points): ompute the directional derivative of f(x, y, z) at (,, ) in the direction of u i + j k. Solution: We compute v u/ u /3, /3, /3 and f xy + + z, x, x/ + z. Hence, we have f(,, ) 5,,. It follows that D v f(,, ) 5,, /3, /3, /3 /3 + /3 /3. (b) (5 points): Find the maximum rate of change of f(x, y, z) at the point (,, ). (You may use your work from part (a), if you choose.) Solution: From (a), we have f(,, ) 5,, (5 points): Absolute Maxima and Minima. Let f(x, y) 6x + 3y 6x 9, and let S {(x, y) : x + y }. Find the absolute maximum and the absolute minimum of f(x, y) in S (in its interior and on its boundary). (Note: Since I want the absolute max/min, it is not necessary to identify critical points in the interior as local max/min or saddles. Equations for the boundary of S are S : y x with x and S : y x with x.) Solution: We first look for critical points. We have f x x 6 x ; f y 6y y. Hence, the only critical point in the interior of S is (/, ), and f(/, ) /. Whether we work with S or S, we have y x with x. To find interior critical points on S or S, we study g(x) f(x, ± x ) 6x + 3( x ) 6x 9 3x 6x+3. We have g (x) 6x 6, so (, ± 3) is a critical point, and f(, ± 3). Lastly, we consider the common endpoints of S and S : f(, ) 3 and f(, ) 7. We conclude that the absolute max is f(, ) 7 and the absolute min is f(/, ) /. 6. (5 points): Lagrange multipliers. Find the maximum and minimum values of f(x, y, z) x + 8y + z subject to x + y + z 5. Solution: Let g(x, y, z) x + y + z 5. We compute f, 8,, g x, y, z. There exists λ in R with f λ g if and only if λx, λy, 5 λz. It follows that x /λ, y /λ, z 5/λ. We substitute in the constraint equation g to obtain ( ) ( ) ( ) λ λ λ

3 Simplifying, we obtain (+6+5)/λ 5/λ 5, so we have λ ±. The value λ gives (,, 5); λ gives (,, 5). We compute f(,, 5) is the maximum value, and f(,, 5) 9 is the minimum value, subject to the g. 7. (5 points): Double integral in rectangular coordinates. Evaluate the double integral ln 3 3 e y dx dy. (onsider drawing a quick sketch of the region of integration in the xy-plane.) Your answer should be a number. Solution: We compute: ln 3 3 e y dx dy 3 3 dy dx 3 dx sin 3 sin. ] y dx 8. ( points): Double integral in polar coordinates. onvert the double integral y (x + y) dx dy y to polar coordinates. (onsider drawing a quick sketch of the region in the xy-plane. Do not evaluate. y (x + y) dx dy y π/ π/ (r cos θ + r sin θ)(r dr dθ) r (cos θ + sin θ) dr dθ. 9. ( points): Triple integral in rectangular coordinates. onsider the triple integral y x f(x, y, z) dz dy dx (onsider drawing a quick sketch of the three-dimensional region of integration.) 3

4 (a) ( points): onvert the triple integral to a triple integral in rectangular coordinates with dv dx dy dz. y x f(x, y, z) dz dy, dx z y y f(x, y, z) dx dy dz. (b) ( points): onvert the triple integral to a triple integral in rectangular coordinates with dv dy dz dx. y x f(x, y, z) dz dy dx x z x f(x, y, z) dy dz dx.. (5 points): Spherical coordinates. onvert the triple integral x 8 x y x +y xz dz dy dx to spherical coordinates. (It will help to draw a quick sketch of the three-dimensional region of integration.) Do not evaluate. π/ π/ π/ π/ (ρ sin φ cos θ)(ρ cos φ)(ρ sin φ dρ dφ dθ) ρ sin φ cos φ cos θ dρ dφ dθ.. (5 points): hange of variables and the Jacobian. Find the area of the region R in the xy-plane bounded by xy, xy 8, xy 3 5, xy 3 5. (Use a change of variables to evaluate a double integral over a region S in the uv-plane.) Your answer should be a number. Solution: We make the substitution u xy with u 8, and v xy 3 with 5 v 5. We compute (u, v) (x, y) u/ x u/ y v/ x v/ y y x y 3 3xy 3xy3 xy 3 xy 3 v It follows that (x, y)/ (u, v) /v. We compute 8 5 da v dv du ( 8 R 5 ) ( 5 ) du 5 v dv (ln 5 ln 5) ln(5/5) ln 3.

5 . 6.3, # (5 points): Line integrals. Find the work done by the force field F (x, y) y 3/ i + 3x y j in moving an object from (, ) to (, ). Make sure to thoroughly justify your reasoning. Your answer should be a number. Solution: We observe that (y 3/ ) y (3/)(y / ) 3y / (3x y). x Therefore, F is conservative. To recover a potential f(x, y), we first compute f f(x, y) dx + g(y) y 3/ dx + g(y) xy 3/ + g(y). x We also require f y 3x y (xy3/ + g(y)) y 3x y + g (y). It follows that g (y), so we have g(y) K, a constant, and f(x, y) xy 3/ + K. Hence, the line integral over, a path from (, ) to (, ), depends only on the endpoints. We use the Fundamental Theorem of alculus for line integrals to obtain Work F d r (,) (,) f d r f(, ) f(, ) (3 + K) ( + K) , #5 (5 points): Green s Theorem. Evaluate the line integral xy dx + x y dy along the positively oriented boundary of the triangle with vertices (, ), (, ), (, ). Your answer should be a number. Solution: We apply Green s Theorem to obtain xy dx + x y dy D (x(x ) x(x )) dx 3 (xy xy) da x 3 dx 3 x x ] x. xy dy dx xy ] x x dx 5