Second Midterm Exam Math 212 Fall 2010

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1 Second Midterm Exam Math 22 Fall 2 Instructions: This is a 9 minute exam. You should work alone, without access to any book or notes. No calculators are allowed. Do not discuss this exam with anyone other than your instructor. hen you have completed the exam, write out and sign the Honor Code pledge on the front. The exam consists of 6 questions. You must show all of your work on each problem to receive full credit, and be sure to clearly indicate your final answer to each question.. 5 Points Let f(x, y, z) (xz, sin(xy)) and g(u, v) (u 2, e uv, ln v) (a) Compute the derivatives Df and Dg. Df f x f 2 x f y f 2 y Dg g u g 2 u g u f z f 2 z g v g 2 v g v z x, y cos xy x cos xy 2u ve ue uv. /v (b) Using the multivariable chain rule, compute the derivative D(f g). (Your final answer should be a matrix whose entries are functions of u and v.) D(f g) (Df)(Dg) 2u z x ve y cos xy x cos xy uv ue uv /v z(2u) x/v (y cos xy)(2u) + (x cos xy)(ve uv ) (x cos xy)(ue uv ) 2u ln v u 2 /v 2ue uv cos(u 2 e uv ) + u 2 ve uv cos(u 2 e uv ) u e uv cos(u 2 e uv )

2 (c) Using the multivariable chain rule, compute the derivative D(g f). (Your final answer should be a matrix whose entries are functions of x, y, and z.) D(g f) (Dg)(Df) 2u ve uv ue uv z x y cos xy x cos xy /v 2uz 2ux ve uv z + ue uv y cos cy ue uv x cos xy ve uv x y cos xy v x cos xy v 2xz 2 2x 2 z (sin xy)e xz sin xy z + xze xz sin xy y cos xy xze xz sin xy x cos xy (sin xy)e xz sin xy x y cos xy sin xy x cos xy sin xy 2. 2 Points Let f(x, y) 2x 2 + y 4 4xy. (a) Find the critical points of f. e set f (4x 4y, 4y 4x), so x y and 4y 4y 4y(y )(y + ), and we get three critical points, (, ), (, ), (, ). (b) Compute the Hessian matrix of f. Hf x 2 y x x y y y 2 (c) Using the multivariable second derivative test, classify the critical points of f. e have 2 f 4 > everywhere, and the determinant of Hf is 6(y 2 ), so by x 2 the second derivative test, (, ) and (, ) are local minima and (, ) is a saddle point. (d) Starting from the point (, 2), in which direction does the function f decrease most rapidly? The negative of the gradient points in the direction of greatest decrease: f(, 2) ( 4, 2), so the direction is ( 4, 2)

3 . 5 Points Let I 4 2 x x dy dx y 5 + (a) rite I as the double integral over a region D in the plane. Sketch the region D. (b) Reverse the order of integration, i.e. rewrite I as an iterated integral dx dy. 2 y 2 x dx dy y 5 + (c) Compute I. 2 y 2 x dx dy y ( 2 2 x2 2 ln y 5 + y 4 y 2 y 5 + dy u du x ) dy where u y5 +, du 5ydy 4. 5 Points Find the dimensions of a right circular cylinder of the maximum possible volume that can be inscribed in a sphere of radius 6. (Make sure to clearly explain what your variables mean.)

4 Let the sphere by given by x 2 +y 2 +z By symmetry, we may as well assume that the axis of the cylinder is the z-axis. Let (x, y, z) be a point on both the surface of the sphere and the surface of the cylinder. The volume of the inscribed cylinder is then (πr 2 )(2z), where r x 2 + y 2 is the radius and 2z is the height. Thus, we are trying to maximize the function f(r, z) r 2 z, subject to the constraint g(r, z) r 2 + z To do so, we solve the Lagrange multiplier equation f λ g, where f (2rz, r 2 ) and g (2r, 2z). This gives 2rz r2, so 2r 2z r2 2z 2. Substituting this into the constraint equation yields z 2 6 2, so z 6, and r 2z 6 2. Thus the largest volume cylinder has height 2 and radius Points Let be the region in R defined by the inequalities y x 2, z y, and z. Let f(x, y, z) be a continuous function on. (a) Sketch the region. 4

5 (b) rite f(x, y, z) dv as an iterated integral in the order dz dy dx. f(x, y, z) dv y x 2 f(x, y, z)dz dy dx. (c) rite f(x, y, z) dv as an iterated integral in the order dx dy dz. 5

6 f(x, y, z) dv (d) Evaluate y dv z y y f(x, y, z)dx dy dz. z y y y dx dy dz z z ( 4 5 y5/2 z y y y dx dy dz y 2y y dy dz ) dz 4 ( 5 z)5/2 dz 8 5 ( z)7/2 z 8 5 7/

7 6. 5 Points Let f(x, y) x 2 + y 2 2x 4y. Find the absolute maximum and minimum of f, subject to the constraints x 2 + y 2 2 and y. First, we set f (2x 2, 2y 4), which tells us that (, 2) is the only critical point of f, but it is outside the region defined by the constraints. Thus the minimum and maximum of f will be on the boundary. The curve bounding the region has two smooth components, meeting non-smoothly at the points ( 2, ) and ( 2, ). The first component is given implicitly by the equation g (x, y) x 2 +y 2 2. Thus we set f λ g, giving (2x 2, 2y 4) λ(2x, 2y). After checking that there are no solutions with x or y, we get 2x 2 2y 4, so 2, and y 2x. The line y 2x meets the 2x ( 2y ) x y ( ) circle x 2 + y in two points, ±, 2 2 2, but only, 2 2 satisfies y. The second boundary component is given implicitly by the equation g 2 (x, y) y. e set f λ g 2, giving (2x 2, 2y 4) λ(, ), so that x, giving us the point (, ). e now have four candidates for extrema of f on our region, the two points (± 2, ) where the boundary components ( meet, ) and the two constrained critical points from the Lagrange 2 multiplier calculations,, 2 2 and (, ). To find the absolute maximum and minimum, we just evaluate f at these four points: ( ) f 2, ( f ) 2, absolute maximum ( ) 2 f, absolute minimum f (, ) 7