MATHEMATICS 200 April 2010 Final Exam Solutions

Transcription

1 MATHEMATICS April Final Eam Solutions. (a) A surface z(, y) is defined by zy y + ln(yz). (i) Compute z, z y (ii) Evaluate z and z y in terms of, y, z. at (, y, z) (,, /). (b) A surface z f(, y) has derivatives f and f y at (, y, z) (,, ). (i) If increases from to., and y decreases from to.6, find the change in z using a linear approimation. (ii) Find the equation of the tangent plane to the surface at the point (,, ). Solution. (a) (i) We are told that z(, y) obeys z(, y) y y + ln ( y z(, y) ) ( ) for all (, y) (near (, )). Differentiating ( ) with respect to gives y z (, y) + z + (, y) z(, y) z (, y) y z(,y) or, dropping the arguments (, y) and multiplying both the numerator and denominator by z, z z z yz Differentiating ( ) with respect to y gives z( ) (yz ) z(, y) + y z y (, y) z y + (, y) y z(, y) z y (, y) + z(, y) y y z(,y) or, dropping the arguments (, y) and multiplying both the numerator and denominator by yz, z y z + yz yz y z y z( + y yz) y(yz ) (a) (ii) When (, y, z) (,, /), z (, ) y z (,y,z)(,,/) z y (, ) + z y y z (,y,z)(,,/) +

2 (b) (i) The linear approimation to f(, y) at (, ) is f(, y) f(, ) + f (, ) ( ) + f y (, ) (y ) + ( ) (y ) So the change is z is approimately (ii) The equation of the tangent plane is or (. ) (.6 ).4 z f(, ) + f (, ) ( ) + f y (, ) (y ) + ( ) (y ) y z 4. (a) For the function z f(, y) + y + y 6 y 6. Find and classify as [local maima, local minima, or saddle points] all critical points of f(, y). (b) The images below depict level sets f(, y) c of the functions in the list at heights c,.,.,...,.9,. Label the pictures with the corresponding function and mark the critical points in each picture. (Note that in some cases, the critical points might not be drawn on the images already. In those cases you should add them to the picture.) (i) f(, y) ( + y )( y) + (ii) f(, y) + y (iii) f(, y) y( + y)( y) + (iv) f(, y) + y

3 Solution. (a) To find the critical points we will need the gradient of f and to apply the second derivative test of Theorem.9.6 in the CLP III tet we will need all second order partial derivatives. So we need all partial derivatives of f up to order two. Here they are. The critical points are the solutions of f + y + y 6 y 6 f + y 6 f 6 f y f y + 6y f yy 6 f y f + y 6 f y + 6y Subtracting the second equation from times the first equation gives 6 9 ( )( + ), Since y (from the second equation), the critical points are (, ), (, ) and the 4 classification is critical point f f yy fy f type (, ) 4 (9) (6) () > 9 local min (, ) ( 6) (6) () < saddle point (b) Both of the functions f(, y) + y (i.e. (ii)) and f(, y) + y (i.e. (iv)) are invariant under rotations around the (, ). So their level curves are circles centred on the origin. In polar coordinates + y is r. So the sketched level curves of the function in (ii) are r,.,.,...,.9,. They are equally spaced. So at this point, we know that the third picture goes with (iv) and the fourth picture goes with (ii). Notice that the lines y, y and y are all level curves of the function f(, y) y( + y)( y) + (i.e. of (iii)) with f. So the first picture goes with (iii). And the second picture goes with (i). Here are the pictures with critical points marked on them. There are saddle points where level curves cross and there are local ma s or min s at bull s eyes. (i) (ii)

4 (iii) (iv). The temperature T (, y) at a point of the y plane is given by T (, y) ye A bug travels from left to right along the curve y at a speed of.m/sec. The bug monitors T (, y) continuously. What is the rate of change of T as the bug passes through the point (, )? Solution. The slope of y at (, ) is d d. So a unit vector in the bug s direction of motion is, and the bug s velocity vector is v.,. The temperature gradient at (, ) is T (, ) ye, e (.y)(,) e, e and the rate of change of T (per unit time) that the bug feels as it passes through the point (, ) is T (, ) v. e, e,.4e 4. (a) Does the function f(, y) + y have a maimum or a minimum on the curve y? Eplain. (b) Find all maima and minima of f(, y) on the curve y. Solution. (a) f(, y) + y is the square of the distance from the point (, y) to the origin. There are points on the curve y that have either or y arbitrarily large and so whose distance from the origin is arbitrarily large. So f has no maimum on the curve. On the other hand f will have a minimum, achieved at the points of y that are closest to the origin. (b) On the curve y we have y and hence f +. As d ( + ) d (4 ) 4

5 and as no point of the curve has, the minimum is achieved when ±. So the minima are at ±(, ), where f takes the value.. Let G be the region in R given by (a) Sketch the region G. + y y y (b) Epress the integral G f(, y) da a sum of iterated integrals f(, y) ddy. (c) Epress the integral f(, y) da as an iterated integral in polar coordinates (r, θ) G where r cos(θ) and y r sin(θ). Solution. (a) Observe that the condition + y restricts G to the interior of the circle of radius centred on the origin, and the conditions y restricts G to, y, i.e. to the first quadrant, and the conditions y and y restrict y. So G lies below the (steep) line y and lies above the (not steep) line y. Here is a sketch of G y y ` y?,? G?,? y { (b) Observe that the line y crosses the circle + y at a point (, y) obeying + () + y and that the line y crosses the circle + y at a point (, y) obeying (y) + y + y y

6 ( So the intersection point of y and + y in the first octant is ( the intersection point of y and + y in the first octant is We ll set up the iterated integral using horizontal strips as in the sketch ),, ). and y y{ a y?,? y y {??,? Looking at that sketch, we see that, on G, y runs from to, and for each fied y between and, runs from y to y, and for each fied y between and runs from y to y. So f(, y) da G y dy d f(, y) + y/ y dy d f(, y) y/ (b) In polar coordinates the equation + y becomes r, and the equation y / becomes r sin θ r cos θ or tan θ, and the equation y becomes r sin θ r cos θ or tan θ. Looking at the sketch 6

7 y θ arctan r θ arctan we see that, on G, θ runs from arctan to arctan, and for each fied θ in that range, r runs from to. As da r dr, and r cos θ, y r sin θ, f(, y) da arctan G arctan dr r f(r cos θ, r sin θ) 6. For the integral I + y dy d (a) Sketch the region of integration. (b) Evaluate I. Solution. (a) On the domain of integration, runs from to, and for each fied in that range, y runs from to. We may rewrite y as y, which is a rightward opening parabola. Here are two sketches of the domain of integration, which we call D. The left hand sketch also shows a vertical slice, as was used in setting up the integral. 7

8 y y y p,q p,q y? y (b) The inside integral, + y dy, of the given integral looks pretty nasty. So let s reverse the order of integration, by using horizontal, rather than vertical, slices. Looking at the figure on the right above, we see that So y runs from to, and for each fied y in that range runs from to y. I y dy d + y dy y + y du u [ ] u / / ( ) 9 with u + y, du y dy. Looks pretty rigged! 7. A thin plate of uniform density k is bounded by the positive and y aes and the circle + y. Find its centre of mass. Solution. Call the plate P. By definition, the centre of mass is (, ȳ), with and ȳ being the weighted averages of the and y coordinates, respectively, over P. That is, ρ(, y) da y ρ(, y) da P P P ρ(, y) da ȳ ρ(, y) da with ρ(, y) k. Here is a sketch of P. P 8

9 y r P By symmetry under reflection in the line y, we have ȳ. So we just have to determine P da da P The denominator is just one quarter of the area of circular disk of radius. That is, da π. We ll evaluate the numerator using polar coordinates as in the figure above. P 4 Looking at that figure, we see that θ runs from to π, and for each fied θ in that range, r runs from to. As da r dr, and r cos θ, the numerator P da π/ [ sin θ ] π/ [ {}}{ ] π/ [ ] dr r r cos θ cos θ dr r [ r ] All together 8. Let ȳ / π/4 4 π I T ( + y ) dv where T is the solid region bounded below by the cone z + y and above by the sphere + y + z 9. 9

10 (a) Epress I as a triple integral in spherical coordinates. (b) Epress I as a triple integral in cylindrical coordinates. (c) Evaluate I by any method. Solution. (a) Recall that in spherical coordinates so that ρ sin ϕ cos θ y ρ sin ϕ sin θ z ρ cos ϕ + y + z 9 is ρ, and + y z is ρ sin ϕ ρ cos ϕ, or tan ϕ or ϕ π 6, and the integrand + y ρ sin ϕ, and dv is ρ sin ϕ dρ dϕ So T { (ρ sin ϕ cos θ, ρ sin ϕ sin θ, ρ cos ϕ) } ρ, θ π, ϕ π/6 and, I T dρ ( + y ) dv π (b) In cylindrical coordinates π/6 + y + z 9 is r + z 9 and + y z is r z and the integand + y r, and dv is r dr dz dϕ ρ 4 sin ϕ π dρ π/6 dϕ ρ sin ϕ +y {}}{ ρ sin ϕ Observe that r + z 9 and r z intersect when r + r 9 so that r and z. Here is a sketch of the y cross section of T. z z? r r, z? r ` z 9

11 So and T { (r cos θ, r sin θ, z) r, θ π, r z 9 r } I V ( + y ) dv / (c) We ll use the spherical coordinate form. I π π dρ dρ π π π/6 π/6 π 9 r +y {}}{ dr r dz r (r ) dϕ ρ 4 sin ϕ [ dρ ρ 4 cos ϕ + cos ϕ [ 8 ] dϕ ρ 4 sin ϕ [ cos ϕ ] 4 {}}{ 8 π [ 4 ] π/6 π [ 9 ].4 / π 9 r dr r dz r ] dρ ρ 4