Math 210, Final Exam, Fall 2010 Problem 1 Solution. v cosθ = u. v Since the magnitudes of the vectors are positive, the sign of the dot product will
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1 Math, Final Exam, Fall Problem Solution. Let u,, and v,,3. (a) Is the angle between u and v acute, obtuse, or right? (b) Find an equation for the plane through (,,) containing u and v. Solution: (a) The cosine of the angle between u and v is: u v cosθ u v Since the magnitudes of the vectors are positive, the sign of the dot product will determine whether the angle is acute, obtuse, or right. The dot product is: u v,,,,3 Since the dot product is positive we know that cosθ > and, thus, the angle is acute. (b) A vector perpendicular to the plane is the cross product of u and v which both lie in the plane. n u v î ĵ ˆk n 3 n î 3 ĵ 3 + ˆk n î[( )(3) ()()] ĵ[()(3) ()()]+ ˆk[()() ( )()] n 3î 3ĵ+3ˆk n 3, 3,3 Using (,,) as a point on the plane, we have: 3(x ) 3(y +)+3(z )
2 Math, Final Exam, Fall Problem Solution. The curve r (t) sin(t),cos(t), t describes the movement of a particle in R 3. (a) Find the velocity and the acceleration of the particle as a function of t. (b) Find the tangent line to the curve at time t π/4. (c) Find the distance traveled between time t and t π. Solution: (a) The velocity and acceleration vectors are: v(t) r (t) cos(t), sin(t), a(t) v (t) sin(t), cos(t), (b) A vector equation for the line tangent to r (t) at t is: L(t) r (t )+ r (t )(t t ) Applying this formula to our vectors r (t) and r (t) at t π/4 we have: ( L(t) π ) r + ( π )( r t π ) ( π ( π L(t) sin,cos, 4) 4) π ( π ( π ( + cos, sin, t 4 4) 4) π ) 4 π ( L(t),, +,, t π ) 4 4 (c) The distance traveled by the particle is: L π π π π π π 5 r (t) dt (cos(t)) +( sin(t)) +( ) dt 4cos (t)+4sin (t)+dt 4+dt 5dt
3 3. Use Green s Theorem to compute (,), (,), (,) traversed in this order. Math, Final Exam, Fall Problem 3 Solution Solution: Green s Theorem states that F d s C C ydx+x ydy where C tracesthetrianglewithvertices D ( g x f ) da y where D is the region enclosed by C. The integrand of the double integral is: g x f y x x y y y xy Thus, the value of the integral is: ( F d g s C D x f ) da y (xy )da D y+ y [ x y x (xy )dxdy ] y+ y dy [( ( y +) y ( y +) ) ( ) y 3 4y +4y +y y 3 +y dy ( ) 4y +6y dy [ 4 3 y3 +3y y ] ( )] (y) y y dy 3
4 Math, Final Exam, Fall Problem 4 Solution 4. Find the critical points of z x 3 +x +y xy x and use the second derivative test to classify them as local maxima, local minima or saddles. Solution: By definition, an interior point (a,b) in the domain of f is a critical point of f if either () f x (a,b) f y (a,b), or () one (or both) of f x or f y does not exist at (a,b). The partial derivatives of f(x,y) x 3 +x + y xy x are f x 3x +x y and f y y x. These derivatives exist for all (x,y) in R. Thus, the critical points of f are the solutions to the system of equations: Solving Equation () for y we get: Substituting this into Equation () and solving for x we get: f x 3x +x y () f y y x () y x (3) 3x +x y 3x +x x 3x x 4 x or x We find the corresponding y-values using Equation (3): y x. If x, then y. If x, then y. Thus, the critical points are (,) and (, ). We now use the Second Derivative Test to classify the critical points. The second derivatives of f are: f xx 6x+, f yy, f xy The discriminant function D(x, y) is then: D(x,y) f xx f yy f xy D(x,y) (6x+)() ( ) D(x,y) x The values of D(x,y) at the critical points and the conclusions of the Second Derivative Test are shown in the table below.
5 (a,b) D(a,b) f xx (a,b) Conclusion (, ) 4 4 Local Minimum (, ) 4 Saddle Point Recall that (a,b) is a saddle point if D(a,b) < and that (a,b) corresponds to a local minimum of f if D(a,b) > and f xx (a,b) >.
6 Math, Final Exam, Fall Problem 5 Solution 5. Consider the vector field F cx y e y,x 3 y xe y on R where c is a constant. (a) Find the value for c that makes F a conservative vector field. (b) With c as in (a) find a function φ(x,y) so that F φ. Solution: (a) In order for the vector field F f(x,y),g(x,y) to be conservative, it must be the case that: f y g x Using f(x,y) cx y e y and g(x,y) x 3 y xe y we get: f y g x cx y e y 6x y e y (b) If F ϕ, then it must be the case that: cx y 6x y c 3 ϕ f(x,y) x () ϕ g(x,y) y () Using f(x,y) 3x y e y and integrating both sides of Equation () with respect to x we get: ϕ x f(x,y) ϕ x 3x y e y ϕ (3x x dx y e y) dx ϕ(x,y) x 3 y xe y +h(y) (3) We obtain the function h(y) using Equation (). Using g(x,y) x 3 y xe y we get the equation: ϕ y g(x,y) ϕ y x3 y xe y
7 WenowuseEquation(3)toobtainthelefthandsideoftheaboveequation. Simplifying we get: ( x 3 y xe y +h(y) ) x 3 y xe y y x 3 y xe y +h (y) x 3 y xe y h (y) which implies that h(y). Letting C, we find that a potential function for F is: ϕ(x,y) x 3 y xe y
8 Math, Final Exam, Fall Problem 6 Solution 6. Compute the volume of the region in R 3 bounded by the paraboloid z x + y, the cylinder x + y 9, and the plane z. Solution: The region R is plotted below. The volume can be computed using either a double or a triple integral. The double integral formula for computing the volume of a region R bounded above by the surface z f(x, y) and below by the surface z g(x, y) with projection D onto the xy-plane is: V (f(x, y) g(x, y)) da D In this case, the top surface is z x + y r in polar coordinates and the bottom surface is z. The projection of R onto the xy-plane is a disk of radius 3, described in polar coordinates as D {(r, θ) : r 3, θ π}. Thus, the volume formula is: V π 3 ( r ) r dr dθ () The triple integral formula for computing the volume of R is: ( ) f(x,y) V dz da Using cylindrical coordinates we have: D g(x,y) V π 3 r r dz dr dθ ()
9 Evaluating Equation () we get: V π 3 π π [ 8 4 8π [ 4 r4 8 4 dθ ] π ( r ) r dr dθ Note that Equation () will evaluate to the same answer. ] 3 dθ
10 Math, Final Exam, Fall Problem 7 Solution 7. Given the function f(x,y) xy +ycos(x) find: (a) the gradient f at the point P (,). (b) thedirectional derivative D v f(,), where f istheunit vector fromp (,)towards Q (,3). Solution: (a) The gradient of f is: At the point P (,) we have: f fx,f y y ysin(x),xy +cos(x) f(,) ()sin(),()()+cos(), (b) The unit vector v that points from P (,) towards Q (,3) is: PQ v PQ,,,, Thus, the directional derivative D v f(,) is: D v f(,) f(,) v,, +
11 Math, Final Exam, Fall Problem 8 Solution 8. Use the method of Lagrange multipliers to find points where f(x,y) xy attains its maximum and minimum subject to the constraint: x +4y. Solution: We find the minimum and maximum using the method of Lagrange Multipliers. First, we recognize that x + 4y is compact which guarantees the existence of absolute extrema of f. Then let g(x,y) x +4y. We look for solutions to the following system of equations: f x λg x, f y λg y, g(x,y) which, when applied to our functions f and g, give us: Dividing Equation () by Equation () gives us: y λ(x) () x λ(8y) () x +4y (3) y x λ(x) λ(8y) y x x 4y 4y x Combining this result with Equation (3) and solving for x gives us: When x we have: x +4y x +x x x x ± 4y x 4y y 4 y ± We obtain the same values of y when x. Therefore, the points of interest are (, ), (, ), (, ), and (, ).
12 We now evaluate f(x,y) xy at each point of interest. f(, ) ()() f(, ) ()( ) f(, ) ( )() f(, ) ( )( ) From the values above we observe that f attains an absolute maximum of minimum of. and an absolute
13 Math, Final Exam, Fall Problem 9 Solution 9. Given the function f(x,y) xe xy compute the partial derivatives: f x, f y, f x x, f x y, f y y Solution: The first partial derivatives of f(x,y) are The second derivatives are: f x exy +xye xy e xy (+xy) f y x e xy f x x x (exy (+xy)) ye xy (+xy)+ye xy f y y ( x e xy) x 3 e xy y f x y ( x e xy) xe xy +x ye xy x
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