Math 210, Final Exam, Spring 2012 Problem 1 Solution. (a) Find an equation of the plane passing through the tips of u, v, and w.
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1 Math, Final Exam, Spring Problem Solution. Consider three position vectors (tails are the origin): u,, v 4,, w,, (a) Find an equation of the plane passing through the tips of u, v, and w. (b) Find an equation of the line perpendicular to the plane from part (a) and passing through the origin. Solution: (a) Since the tails of the given vectors are at the origin, the tips of the vectors are the points U (,, ), V (4,, ), and W (,, ), respectively. The plane containing the tips has n UV UW as a normal vector. Since UV 3,, and UW,,, the normal vector is n UV UW, 5, 3 Using U (,, ) as a point on the plane, an equation for the plane is (x ) 5(y ) + 3(z ) (b) The line perpendicular to the plane in part (a) is parallel to the plane s normal vector. Thus, since, 5, 3 is parallel to the line and the origin (,, ) is on the line, the vector equation for the line is r (t),, + t, 5, 3
2 Math, Final Exam, Spring Problem Solution. Consider the curve r (t) t, t 3, < t <. (a) Find the curvature κ(t). (b) Find all values of t where κ(t). (c) Compute the limits lim κ(t), t lim κ(t) t (d) What do the limits in part (c) say about the curve r (t)? Solution: (a) By definition, the curvature of a curve parametrized by r (t) is given by the formula κ(t) r (t) r (t) r (t) 3 The first two derivatives of r (t) are r (t), 3t and r (t), 6t and their cross product is r (t) r (t) 6t ˆk. Thus, the curvature of r (t) is (b) The curvature is when t. κ(t) κ(t) κ(t) (c) The limits of κ(t) as t ± are lim κ(t) lim t ± t ± r (t) r (t) r (t) 3, 6t ˆk, 3t 3, 6 t ( + 9t 4 ) 3/ 6 t ( + 9t 4 ) 3/ (d) Lines are curves of zero curvature. behaves linearly as t ±. Thus, the limits in part (c) suggest that r (t)
3 Math, Final Exam, Spring Problem 3 Solution 3. Given the function of two variables G(x,y) y x (a) Determine the domain of G. (b) Sketch the level curves G, G, and G all on one coordinate grid. What kind of curves are they? (c) At the point (,), find the direction in which G has its maximum rate of increase. Also determine this maximum rate. Solution: (a) The domain of G is the set of all pairs (x,y) such that y x. y 6 C 5 C 4 3 C (b) x (c) Thedirectionofmaximumrateofincrease ofg(x,y)atthepoint(,)is, bydefinition, û G(,) G(,) The gradient of G is x G Gx,G y, y x y x The value of G at the point (,) is G(,), and its magnitude is G(,) 5. Thus, the direction of maximum rate of increase of G at (,) is
4 û, The maximum rate of increase, by definition, is G(,) 5 5.
5 Math, Final Exam, Spring Problem 4 Solution 4. Find absolute maximum and minimum of the function f(x, y) xy x over the region R {x + y 4}. Also, find the points where these extremes occur. Solution: First, the region R is closed and bounded (i.e. compact) and f is defined at every point in R. Therefore, we are guaranteed to find absolute extrema. Next, we look for all critical points of f in R. These will be points for which the first derivatives of f vanish. Thus, we must solve the system of equations: f x y, f y x which has x and y as the only solution. We must now determine the extreme values of f on the boundary of R which is the circle x +y 4. We will resort to using the method of Lagrange multipliers to find these values. The following system of equations must then be solved: f x λg x, f y λg y, g(x, y) where g(x, y) x + y 4. Evaluate the partial derivatives we then have y λ(x), () x λ(y), () x + y 4. (3) Dividing Equation () by Equation () and simplifying gives us y λ(x) x λ(y), y x x y, y(y ) x, x y y Substituting y y for x in Equation (3) and solving for x we get
6 x + y 4, y y + y 4, y y 4 which has the two solutions y, ± 33 4 Let y be the positive solution and y the negative one. If y y then the corresponding x-values are x, ± y y. Similarly, if y y then the corresponding x-values are x, ± y y. We must now evaluate f(x, y) at the critical point (, ) and at all critical points on the boundary of R. f(, ), f(x, y ) x (y ) (y ) y y y (y ) 3/ f(x, y ) x (y ) (y ) y y y (y ) 3/ f(x, y ) x (y ) (y ) y y y ( y ) 3/ f(x, y ) x (y ) (y ) y y y ( y ) 3/ A calculator would be useful here but isn t necessary. We can estimate 33 to be 5.75 using a linear approximation of F (x) x about x 36 giving us y.6875 and y.875. One can then show that f(x, y ) is the absolute maximum and f(x, y ) is the absolute minimum of f on R.
7 Math, Final Exam, Spring Problem 5 Solution 5. Consider the integral (a) Sketch the region of integration. (b) Reverse the order of integration properly. (c) Evaluate the integral from part (b). x x cos ( y ) dy dx. Solution: (a) The region of integration is sketched below (b) Upon switching the order of integration we obtain y (c) Evaluating the above double integral we get x cos ( y ) dx dy y x cos ( y ) dx dy [ x cos ( y ) ] y y cos ( y ) dy, [ sin ( y ) 4 sin() ], dy,
8 Math, Final Exam, Spring Problem 6 Solution 6. Consider the following vector field in space F x + y, x + z, y. (a) Check that this field is conservative. (b) Find a potential of F. (c) Evaluate the following line integral C F d r, where C is a contour originating at (,, ) and terminating at (,, ). Solution: (a) Let P x + y, Q x + z, and R y. Given that P y Q x, P z R x, and Q z R y we know that F is conservative by the cross-partials test. (b) By inspection, a potential function for F is ϕ(x, y, z) x + xy + yz. (c) Using the Fundamental Theorem of Line Integrals, we obtain C F d r ϕ(,, ) ϕ(,, ), ( ) ( ) () + + () + +,
9 Math, Final Exam, Spring Problem 7 Solution 7. Compute the circulation of the vector field H y 3, x 3 over the boundary of the region D {x + y, y }. Solution: The boundary of D is a simple, closed curve oriented counter clockwise. Therefore, we may use Green s Theorem to compute the circulation: H d r (Q x P y ) da D Letting P y 3 and Q x 3 we get Q x 3x and P 3y. Therefore, Q x P y 3(x + y ). Since D is a half-disk, we will use polar coordinates to evaluate the double integral above. The integrand then becomes 3r, da r dr dθ, and the region D can be described as { r, θ π}. Thus, the circulation is D H d r D D π π π 3π 4 (Q x P y ) da, [ 3 4 r4 3 4 dθ, 3r r dr dθ, ] dθ,
10 Math, Final Exam, Spring Problem 8 Solution 8. Compute the volume of the spherical wedge given in spherical coordinates by { W ρ, θ π, φ π } Solution: Using spherical coordinates, the volume of the wedge is computed as follows V dv, W π/ π/ π/ π/ π/ π/ π/ π/ 7π 6 ρ sin φ dρ dφ dθ, [ 3 ρ3 sin φ ] 7 sin φ dφ dθ, 3 [ 73 ] π/ cos φ dθ, 7 3 dθ, dφ dθ,
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