MATH H53 : Final exam 11 May, 18 Name: You have 18 minutes to answer the questions. Use of calculators or any electronic items is not permitted. Answer the questions in the space provided. If you run out of room for an answer, continue on the back of the page. There is an extra sheet at the end for scratch work. Any writing on the extra sheet is strictly for scratch work and will be ineligible for grading. For questions with multiple parts, you can solve a part assuming the previous parts and get full credit for that particular problem. Question Points Score 1 1 8 3 8 4 8 5 8 6 1 7 1 8 14 Total: 8 1
1. Let L(x, y) = ln x + y, and define a function f : R \ {(, )} R by f(x, y) = x + y xy L(x, y). (a) (4 points) Calculate L and the Hessian L at any point (x, y) R \ {(, )}. Note. These calculations will also be used in the final problem. So do them carefully! Solution: we use chain and product rules throughout. We then have x L x = x + y, y L y = x + y L xx = y x (x + y ) L xy = L yx = = xy (x + y ) L yy = x y (x + y ).
(b) (4 points) Find all the critical points of the function f(x, y). Solution: Using the formulae from the first part, we see that f = x y x x + y, y x y x + y, and so at a critical point { x = x y x +y y = y x. x +y From this it easily follows that y = x, and we obtain two critical points ( 1, 1 ) and ( 1, 1 ). (c) (4 points) Classify the critical points as local min, local max, saddle points or points where the second derivative test fails. Solution: By symmetry, we actually only need to check one. Again, using the formulae from part(a), the Hessian of f is given by [ ] [ fxx f f := xy 1 y x 1 + xy (x = +y ) f yx f yy 1 + xy 1 x y (x +y ) (x +y ) So the Hessian at (1/, 1/) is given by f(1/, 1/) = [ ] 1 3. 3 1 (x +y ) So the determinant is D(1/, 1/) = 8 <, and the point is a saddle point. Similarly the other point is also a saddle. ] 3
. Let S be the surface z = 1 + x + y, and let Q be a point with coordinates (4, 4, ). (a) (5 points) Find the shortest distance from Q to points on S. Solution: We need to minimize f(x, y, z) = (x 4) + (y 4) + z subject to the constrain g(x, y, z) := z x y = 1. The Lagrange multiplier equations are (x 4) = λx (y 4) = λy z = λz z x y = 1. From the third equation either z = or λ = 1. If z =, then we cannot find any (x, y) satisfying the constrain (fourth) equation. So λ = 1 and solving for x and y we obtain our critical points (,, 3) and (,, 3). The distance from the given point is d = ( 4) + ( 4) + 3 = 17. (b) (3 points) Let P be a point on S that achieves the shortest distance in the above part. Show that the segment P Q is perpendicular to the tangent plane of S at P. Solution: The vector corresponding to the segment P Q is,, 3. The gradient of g at P is g(p ) = 4, 4, 6. Since g(p ) = P Q, and since g(p ) is perpendicular to the tangent plane to S at P, it follows that P Q is orthogonal to the tangent plane to S at P. 4
3. (8 points) Evaluate the surface integral S (x + y + z) ds, where S is the part of the sphere x + y + z = 1 inside the cone z = 3(x + y ). Solution: By symmetry, S (x + y) ds =. For the third integral, we use spherical coordinates. The cone in spherical coordinates is given by ϕ = π/6. So S can be parametrized by x = cos θ sin ϕ, y = sin θ sin ϕ, z = cos ϕ, where θ [, π) and ϕ (, π/6). Also the surface area element is ds = sin ϕ dθ dϕ. So the surface integral is S (x + y + z) ds = = π = π π/6 π = π 4. π/6 cos ϕ sin ϕ dθ dϕ sin ϕ dϕ [ cos ϕ ] π/6 5
4. (8 points) Let Ω be the region bounded by the planes x = y, x + y + z =, x = and z =, and let S be the boundary surface of this region. Calculate the flux of the vector field F = xzî yzĵ + zˆk across S with orientation given by the outward normal. Solution: Divergence theorem. It is easily seen that divf = 1, and so F ds = dv S = = = = 1 = Ω 1 x x y x 1 x 1 = 3. x 1 1 dz dy dx ( x y) dy dx [( x)y y ] x ( x) ( x)x + x (1 x) dx x 6
5. Let S denote the sphere of radius 3 centered at the origin. Let Γ denote the curve obtained by intersecting S by the plane x + z =. (a) (3 points) Write down a parametrization for Γ. Indicate clearly the interval over which your parameter varies. Solution: Project to the xz-plane. The equation of the projection is Completing squares this gives the ellipse which can be parametrized as x + y + ( x) = 9. x = 1 + (x 1) + y = 7, 7 cos t, y = 7 sin t, with t [, π). Since the curve also lies on the plane x + z =, solving for z we get 7 cos t z = 1. (b) (3 points) Show that the curvature at any point is a constant. Find the value of that constant. Solution: If we denote the trajectory by r(t), we compute r 7 sin t (t) =, 7 sin t 7 cos t,, r 7 cos t (t) =, 7 cos t 7 sin t,, and r (t) r (t) = 7î + 7ˆk. Finally we compute the curvature κ(t) = r (t) r (t) r (t) 3 = 1 7. (c) ( points) Using part(b), and no further integral calculations, compute the length of the curve Γ. Justify your answer. Solution: Recall that the radius of the osculation circle is 1/κ. In our case the trajectory itself is a circle, and so the radius of that circle will be also 7. But then the length of the trajectory is the circumference and hence is π 7. 7
6. Consider the parabolic coordinate change x = u v, y = uv. Let S be the square in the uv-plane with vertices (1, ), (, ), (, 1) and (1, 1), and let T be it s image in the xy-plane. (a) (4 points) Draw the region T in the xy-plane, and describe it by indicating it s corners, and writing down the equations of it s boundary curves. Solution: When v = and u [1, ], x = u with x [1, 4] and y =. So this gives a side of T on the x axis. (black curve in the figure) When u = and v [, 1], x = 4 v and y = v with y [, ]. Eliminating v, we get the parabola x = 4 (y/) with y (green curve in the figure). When v = 1 and u [1, ], x = u 1 and y = u. Eliminating u, we get the parabola x = y 1 with y [1, ] (red curve in the figure). When u = 1 and v [, 1], x = 1 v and y = v. Eliminating v we get the parabola x = 1 y with y [, 1] (the blue curve in the figure). The vertices are clearly (1, ), (4, ), (3, ) and (, 1). (b) (6 points) Find the area of T. Solution: We use change of variables. The Jacobian is (x, y) (u, v) = u v v u = (u + v ). So the area is A(T ) = T dx dy = 1 1 (u + v ) dv du = 16 3. 8
7. (a) (5 points) Find a smooth simple closed curve C that maximizes the line integral (y 3 y) dx x 3 dy. Justify your answer. Hint. Green s theorem. C Solution: By Green s theorem if C encloses the region D, then (y 3 y) dx x 3 dy = (1 6x 3y ) da. C To maximize this we need to integrate over the largest possible region where the integrand is positive. So C has to be the ellipse 6x + 3y = 1. (b) (3 points) For a fixed vector p, if we define the vector field F = p r, where r(x, y, z) = xî + yĵ + zˆk is the usual radial vector field, show that D F = p. Solution: Let p = a, b, c. We then compute that î ĵ ˆk F = a b c = (bz cy)î + (cx az)ĵ + (ay bx)ˆk. x y z Calculating the curl î ĵ ˆk F = x y z (bz cy) (cx az) (ay bx) = aî + bĵ + cˆk = p. 9
(c) (4 points) Use part(b) to show that if C is any smooth simple closed curve contained in a plane with unit normal vector ˆn = a, b, c, and S is the region on the plane enclosed by the curve, then the area of S is given by A(S) = 1 (bz cy) dx + (cx az) dy + (ay bx) dz. C Solution: Consider the vector field F = ˆn r, where r(x, y, z) = xî + yĵ + zˆk. Then by the previous part F = ˆn, and so F ds = F ˆn ds = ˆn ˆn ds = A(S). S S Then by Stokes theorem, if we parametrize the curve by γ(t) A(S) = 1 F ds S = 1 F d γ C = 1 (bz cy) dx + (cx az) dy + (ay bx) dz, where we use the computation in part(b) that C F = (bz cy)î + (cx az)ĵ + (ay bx)ˆk. S 1
8. The stream function Ψ for a particular flow is given by Ψ = F + G with F (r, θ, z) = (1 1 r ) r sin θ ˆk, G(r, θ, z) = ln r ˆk, where (r, θ, z) are the usual cylindrical coordinates. The velocity vector is then defined by u = Ψ. Also, let ϕ = (1 + 1 ) r r cos θ. (a) (4 points) Compute F and G. Hint. Easier to convert back to Cartesean coordinates. You can make use of Problem 1(a). Also note that G should be a familiar vector field. Solution: Note that in Cartesean coordinates F = y y x + y, G = ln x + y. One can check by direct computation (using the formulae for the gradient and second derivatives of L(x, y) in Problem 1(a)) that F = (1 + y x xy )î (x + y ) (x + y ) ĵ G y = x + y î + x x + y ĵ. 11
(b) ( points) Compute ϕ, and show that F = ϕ. Solution: In Cartesean coordinates, and so we compute that ϕ(x, y, z) = x + x x + y, ϕ = 1 + y x (x + y ), xy (x + y ). (c) (3 points) Show that u =. Give proper justification, and indicate any theorem you might be using. Solution: G is indeed a familiar vector field, and we have have shown in class (Lecture 4) that curl( G) =. On the other hand F = ϕ, and by a theorem proved in class, curl( ϕ) =. Since u = F + G, clearly u =. 1
(d) (3 points) Use the previous parts to compute the circulation C a u d r, around a circle C a in the xy-plane, centered at the origin with radius a, and traversed in the anti-clockwise direction. Solution: By the way u is defined, clearly u d r = F d r + G d r C a C a C a = ϕ d r + G d r. C a C a The first term is zero by the fundamental theorem for line integrals since C a is a closed curve. The second term is π by a calculation we did in class, and so C a u d r = π. (e) ( points) Now let C be the ellipse of intersection of the cylinder x + y = 1 and the plane z = 1 + x. Using the previous parts, and with no further calculations, compute the circulation u d r, C where C is traversed in the anti-clockwise direction as seen from the top. Justify your answer, and choices of sign, if any. Solution: Let S be the part of the cylinder between x + y = 1 and the ellipse C with orientation given by the outward normal. Then since u =, by Stokes theorem u d r =. S The boundary S consists of the circle x + y = 1, z = and the ellipse C. We call the bottom circle with anti-clockwise orientation by C 1, using notation from part(d). The key point is that the orientation induced from the surface S is anti-clockwise on the base circle and clockwise on the ellipse C. Since the problem statement has C with the anti-clockwise orientation, the above line integral translates to u d r = u d r u d r, S C 1 C and so by Stokes theorem and part(d), u d r = π. C 13
Scratch Work (not to be graded.) 14