DO NOT BEGIN THIS TEST UNTIL INSTRUCTED TO START

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1 Math 265 Student name: KEY Final Exam Fall 23 Instructor & Section: This test is closed book and closed notes. A (graphing) calculator is allowed for this test but cannot also be a communication device (i.e., your cellphones or tablets are not calculators). Answer each question completely using exact values unless otherwise indicated. Show your work (legibly); answers without work and/or justifications will not receive credit. Place your final answer in the provided box. Each problem is worth points for a total of 8 points Σ DO NOT BEGIN THIS TEST UNTIL INSTRUCTED TO START

2 . Find the arc length of the curve r(t) e 2t, 2te t 2e t + 7, 3 t3, for t 3. r (t) 2e 2t, 2e t + 2te t 2e t, t 2 2e 2t, 2te t, t 2. Length b a r (t) dt (2e 2t ) 2 + ( 2te t ) 2 + ( t 2 ) 2 dt 4e 4t + 4t 2 e 2t + t 4 dt (2e 2t + t 2) 2 dt (2e 2t + t 2 ) dt ( e 2t t3) (e 6 + 9) (e + ) e 6 + 8

3 2. The surfaces xz 4 x 3 y 2 + 2yz 2 and 3x 2 z yz 3 x 2 y 2 both pass through the point (,, ). Find the unique line in parametric form which is tangent to both surfaces at the point (,, ). (Hint: the direction of this line must be perpendicular to the gradients of the surfaces at (,, ).) Following the hint we will first compute the gradients at these points. F(x, y, z) z 4 3x 2 y 2, 2x 3 y + 2z, 4xz 3 + 2y F(,, ) 2,, 6 G(x, y, z) 6xz 2xy 2, z 3 2x 2 y, 3x 2 3yz 2 G(,, ) 4, 3, The direction of this line must be perpendicular to both of these and so now we take the cross product to get the direction. i j k 2,, 6 4, 3, ( 8), 24, 6 8, 24, 6. This is the direction of the line, and since this only matters up to scale for convenience we will use the vector 3, 4, (though not needed). Finally, since we know the line also passes through the point (,, ), we have both point and direction and so the parametric form is as follows: x + 3t y + 4t z + t

4 3. Find the directional derivative of g(x, y, z) x 2 y 3xz + 5y + 7z at the point (,, ) in the direction which goes from (,, ) towards the point (5,, 8). The formula for the directional derivative is D u g(p) g(p) u, where u is a unit vector giving the direction of the derivative and P is the point at which the derivative will occur (in our case (,, )). So we need to find u and the gradient evaluated at the point. First, for u we note that we are given points A (,, ) and B (5,, 8) so that AB 5 ( ), ( ), 8 ( ) 6, 2, 9. This however is not a unit vector, so next we find the magnitude and scale, i.e., AB AB 2 so u AB 6, 2, 9. Second, we now work out the gradient. g(x, y, z) 2xy 3z, x 2 + 5, 3x + 7 g(,, ) 5, 6,. Combining we finally get the following: D u g(p) g(,, ) u 5, 6, , 2,

5 4. Find and classify all of the critical points for the function f(x, y) e y (y 2 x 2 ). First we find the critical points which means we need to know when the partial derivatives are zero, so we first compute the partial derivatives. f x (x, y) 2xe y f y (x, y) e y (y 2 x 2 ) + e y (2y) e y (y 2 x 2 + 2y) The first equation is only when x, thus the second equation is only when y 2 +2y or y(y + 2) or y, 2. Therefore the critical points are at (, ) and (, 2). To classify the points we need to look at D(x, y) f xx (x, y)f yy (x, y) ( f xy (x, y) ) 2, so we compute the second order partial derivatives. So that f xx (x, y) 2e y f xy (x, y) f yx (x, y) 2xe y f yy (x, y) e y (y 2 x 2 + 2y) + e y (2y + 2) e y (y 2 x 2 + 4y + 2) D(x, y) ( 2e y ) ( e y (y 2 x 2 + 4y + 2) ) ( 2xe y) 2 e 2y ( 2(y 2 x 2 + 4y + 2) 4x 2). Putting in the critical points D(, ) 4 < so that (, ) is a saddle, while D(, 2) 4e 4 > so is a local max or min. In the latter case we note that f xx (, 2) 2e 2 < so that (, 2) is a local max. Therefore we have the following: Critical Pt. Type (, ) Saddle (, 2) Local max

6 5. Let R be the region between the origin and the curve r θ for θ π (show to the right). A lamina based on R has density δ(x, y). Find (x 2 + y 2 ) δ(x, y) da, R i.e., the moment of inertia of R with respect to the z-axis. The requested integral is given to us and we note that x 2 + y 2 r 2, similarly we can use the description of the region to set up the integral. And so we have the following: π θ r 2 r dr dθ π 4 r4 rθ r dθ π 4 θ4 dθ 2 θ5 θπ θ 2 π5

7 6. Compute the integral y 3 y 3 (x + y 3 )e y(x+y3) dx dy by carrying out the following change of variables: u y and v x + y 3. As the problem indicates, this is a change of variables problem. First we can solve for x and y in terms of u and v to get the following: x v y 3 v u 3 y u With this we need to work on three parts: () Jacobian; (2) function; (3) bounds. For the Jacobian we have x y J(u, v) u u 3u 2. x v y v For the function we note using the form of u and v given in the statement of the problem we get (x + y 3 )e y(x+y3) ve uv. Finally the bounds of our original integral are y and y 3 x y 3. Or in other words the curves y, y, x + y 3 and x + y 3 which translate (respectively) to the curves u, u, v and v. Therefore our next bounds are u and v (i.e., a square). Combining all three of these parts together we can conclude y 3 y 3 (x + y 3 )e y(x+y3) dx dy ve uv du dv or ve uv dv du. The latter of which depends on which order we want to do the integration in. Now for our case it is easiest to first integrate with respect to u and then to v (i.e., if we did v first we would get into an unpleasant integration by parts problem). So finally we are ready to compute the integral and get the following: y 3 y 3 (x + y 3 )e y(x+y3) dx dy ve uv du dv u e uv dv u (e v ) dv v (e v v) v (e ) (e ) e 2

8 7. For t let C be the parametric curve ( t cos(2πt), sin( πt)). Find 2 ( ) ( 2xy + 2x dx + x 2 + e y) dy. C This involves a line integral. We note that the endpoints are (, ) (at t ) and (, ) (at t ). Since the end and starting points do not match we cannot use Green s Theorem, but it might be conservative, so let s check: ( x 2 + e y) 2x x ( ) 2xy + 2x 2x. y These are the same so it is conservative, woohoo! In particular we now need to determine f(x, y) so that f(x, y, z) 2xy + 2x.x 2 + e y. From the first term (i.e., f ) we can conclude that x f(x, y) x 2 y + x 2 + C(y) for some function C(y). On the other hand we have x 2 + C (y) f y x2 + e y. So that C (y) e y or C(y) e y. So we can conclude that f(x, y) x 2 y + x 2 + e y. Finally, since C f dr f(b) f(a) (i.e., line integrals of conservative functions are easy) we conclude the following: C (2xy + 2x) dx + (x 2 + e y ) dy ( x 2 y + x 2 + e y) (,) ( + + e) ( + + ) e + (,)

9 8. Let T be the solid which consists of the points satisfying z, x 2 + y 2, and x 2 + y 2 + z 2 4 (i.e., this is the top half of a sphere of radius 2 which has had a hole drilled through the middle). Find the following flux across the boundary of the solid T, e x + xz, arctan ( x + z), ze x + sin(y 2 ) n ds. T (Hint: express the solid in terms of cylindrical coordinates.) Since this involves the flux across the boundary we can use Gauss Divergence Theorem. So we have that this is equal to T ( ( e x + xz ) + ( ) ( arctan(x + z) + ze x + sin(y 2 ) )) dv x y z ( ( e x + z) + () + (e x + ) ) dv T T z dv. Next we need to describe this region. By the hint we should go with cylindrical coordinates. This is symmetrical around the z-axis and so we have θ 2π. We also have that r 2 (i.e., r ) and that we are inside the sphere of radius 2 so that we can conclude that r 2. For z we note that we go from z up to z 2 4 (x 2 + y 2 ) 4 r 2, i.e., z 4 r 2. Combining this altogether (and remembering dv r dz dr dθ) we have the following: T z dv 2π 2 4 r 2 2π 2 2π 2 2π 2 2π 2π 9 8 θ θ2π 2 z2 r zr dz dr dθ z 4 r 2 z dr dθ 2 r(4 r2 ) dr dθ ( 2r 2 r3) dr dθ ( r 2 r2 8 r4) dθ r ( (4 2) ( 8 )) dθ θ 9 8 (2π) 9 4 π

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