MATHEMATICS 200 December 2011 Final Exam Solutions

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1 MATHEMATICS December 11 Final Eam Solutions 1. Consider the function f(, ) e +4. (a) Draw a contour map of f, showing all tpes of level curves that occur. (b) Find the equation of the tangent plane to the graph z f(, ) at the point where (, ) (, 1). (c) Find the tangent plane approimation to the value of f(1.99, 1.1) using the tangent plane from part (b). Solution. (a) Observe that, for an constant C, the curve + 4 C is the level curve f e C. If C, then + 4 C is the pair of lines ±. If C >, then + 4 C > is the hperbola ± 1 C +. If C <, then + 4 C < is the hperbola ± C + 4. f e 9 f e f e 9 f e 1 f 1 f e 1 f e 9 f e f e 9 (b), (c) Since f (, 1) e +4 (,)(,1) 4 f (, 1) 8e +4 (,)(,1) 8 The tangent plane to z f(, ) at (, 1) is z f(, 1) + f (, 1) ( ) + f (, 1) ( 1) 1 4( ) + 8( 1) and the tangent plane approimation to the value of f(1.99, 1.1) is f(1.99, 1.1) 1 4(1.99 ) + 8(1.1 1) 1.1 1

2 . Suppose z f(, ) has continuous second order partial derivatives, and r cos t, r sin t. Epress the following partial derivatives in terms r, t, and partial derivatives of f. (a) z t (b) z t Solution. B definition z(r, t) f(r cos t, r sin t). (a) B the chain rule z t (r, t) r sin t f f (r cos t, r sin t) + r cos t (r cos t, r sin t) (b) B linearit, the product rule and the chain rule z t (r, t) [ r sin t f ] (r cos t, r sin t) + [ r cos t f ] (r cos t, r sin t) t t r cos t f (r cos t, r sin t) + r sin t f (r cos t, r sin t) r sin t cos t f (r cos t, r sin t) r sin t f (r cos t, r sin t) r sin t cos t f (r cos t, r sin t) + r cos t f (r cos t, r sin t) r cos t f f r sin t + r sin t f r sin t cos t f + r cos t f with all of the partial derivatives of f evaluated at (r cos t, r sin t).. A bee is fling along the curve of intersection of the surfaces z + + and z in the direction for which z is increasing. At time t, the bee passes through the point (1, 1, ) at speed 6. (a) Find the velocit (vector) of the bee at time t. (b) The temperature T at position (,, z) at time t is given b T + t + z. Find the rate of change of temperature eperienced b the bee at time t. Solution. (a) Let s use v to denote the bee s velocit vector at time t.

3 The bee s direction of motion is tangent to the curve. That tangent is perpendicular to both the normal vector to z + + at (1, 1, ), which is,,,, (,,z)(1,1,) and the normal vector to z at (1, 1, ), which is,, 1,, 1 (,,z)(1,1,) So v has to be some constant times,,,, 1 det or, equivalentl, some constant times 1,,. îı ĵj ˆk 4, 8, 8 1 Since the z component of v has to be positive, v has to be a positive constant times 1,,. Since the speed has to be 6, v has to have length 6. As 1,, v 1,,, 4, 4 (b) Suppose that the bee is at ( (t), (t), z(t) ) at time t. Then the temperature that the bee feels at time t is T ( (t), (t), z(t), t ) (t)(t) (t) + (t)t + z(t) Then the rate of change of temperature (per unit time) felt b the bee at time t is d dt T ( (t), (t), z(t), t ) t ()() + () () () + () + () + z () Recalling that, at time t, the bee is at (1, 1, ) and has velocit, 4, 4 d dt T ( (t), (t), z(t), t ) t ( )(1) + (1)( 4) ( ) + ( 4) + (1) Find the radius of the largest sphere centred at the origin that can be inscribed inside (that is, enclosed inside) the ellipsoid ( + 1) + + (z 1) 8

4 Solution. In order for a sphere of radius r centred on the origin to be enclosed in the ellipsoid, ever point of the ellipsoid must be at least a distance r from the origin. So the largest allowed r is the distance from the origin to the nearest point on the ellipsoid. We have to minimize f(,, z) + + z subject to the constraint g(,, z) ( + 1) + + (z 1) 8. B Theorem.1. in the CLP III tet, an local minimum or maimum (,, z) must obe the Lagrange multiplier equations for some real number λ. f 4λ( + 1) λg (E1) f λ λg (E) f z z 4λ(z 1) λg z (E) ( + 1) + + (z 1) 8 (E4) B equation (E), (1 λ), which is obeed if and onl if at least one of, λ 1 is obeed. If, the remaining equations reduce to λ( + 1) z λ(z 1) (E1) (E) ( + 1) + (z 1) 4 (E4) Note that λ cannot be 1 if it were, (E1) would reduce to 1. So equation (E1) gives λ 1 λ or λ Equation (E) gives z λ 1 λ or z λ Substituting λ and z λ into (E4) gives 1 (1 λ) + 1 (1 λ) 4 1 (1 λ) 1 1 λ ± So we now have two candidates for the location of the ma and min, namel (,, z) ( 1 +,, 1 ) and (,, z) ( 1,, 1 + ). 4

5 If λ 1, the remaining equations reduce to ( + 1) z (z 1) (E1) (E) ( + 1) + + (z 1) 8 (E4) Equation (E1) gives and equation (E) gives z. Substituting these into (E4) gives ± So we have the following candidates for the locations of the min and ma point ( 1 +,, 1 ) ( 1,, 1 + ) (,, ) (,, ) value of f ( ) ( + ) 1 1 min ma ma Recalling that f(,, z) is the square of the distance from (,, z) to the origin, the maimum allowed radius for the enclosed sphere is (a) Consider the iterated integral 4 i. Draw the region of integration. ii. Evaluate the integral. (b) Evaluate the double integral D cos( ) d d + da over the region D { (, ) +, }. Solution. (a) i. On the domain of integration runs from 4 to and for each in that range, runs from (when ) to. The figure on the left below provides a sketch of the domain of integration. It also shows the generic horizontal slice that was used to set up the given iterated integral. 5

6 p,q p,q p, 4q p, 4q? (a) ii. The inside integral, cos( ) d looks nast. So let s reverse the order of integration and use vertical, rather than horizontal, slices. From the figure on the right above, on the domain of integration, runs from to and for each in that range, runs from to. So the integral 4 cos( ) d d sin(8) d d cos( ) [ sin( d cos( ) ) (b) Let s switch to polar coordinates. In polar coordinates, the circle + is r and the line is θ π 4. ] ` 6

7 In polar coordinates da r dr dθ, so the integral D + da π/4 π/4 dθ [ cos θ 1 1 dθ sin θ ] π/4 dr r [ r 4 4 {}}{ r sin θ ] + {}}{ r 6. Let R be the triangle with vertices (, ), (1, ), and (, ). Let R have densit ρ(, ). Find ȳ, the coordinate of the center of mass of R. You do not need to find. Solution. Here is a sketch of R. p,q p1,q R p,q Note that the equation of the straight line through (, ) and (, ) is, or. (As a check note that both points (, ) and (, ) are on. The equation of the straight line through (1, ) and (, ) is, or (As a check note that both points (, ) and (1, ) are on. B definition, the coordinate of the center of mass of R is the weighted average of over R, which is ρ(, ) da R ȳ ρ(, ) da R R da R da On R, runs from to. That is,.. 7

8 For each fied in that range, runs from to. In inequalities, that is. Thus { R (, ), For both n and n, we have n da So R ȳ d d n d n 1 [ n+1 n + 1 n+ n + 1 [ ] n+ n + 1 n+ n + n+1 (n + 1)(n + ) R da R da 4 (4)(5) ()(4) } 7. Evaluate the triple integral dv, where E is the region in the first octant bounded E b the parabolic clinder and the planes + z 1,, and z. Solution. First, we need to develop an understanding of what E looks like. Here are sketches of the parabolic clinder, on the left, and the plane + z 1, on the right. 6 5 ] z z ` z 1 8

9 E is constructed b using the plane + z 1 to chop the top off of the parabolic clinder. Here is a sketch. z z 1 1, z p1, 1, q So and the integral E { (,, z) 1, 1, z 1 } E dv d d 1 1 dz d d (1 ) ] 1 d [ [ ] d The bod of a snowman is formed b the snowballs + + z 1 (this is its bod) and + + (z 4) 4 (this is its head). (a) Find the volume of the snowman b subtracting the intersection of the two snow balls from the sum of the volumes of the snow balls. [Recall that the volume of a sphere of radius r is 4π r.] 9

10 (b) We can also calculate the volume of the snowman as a sum of the following triple integrals: 1... π π π 4 r π π π 6 ρ sin ϕ dρ dθ dϕ r r dz dr dθ ρ sin(ϕ) dρ dθ dϕ Circle the right answer from the underlined choices and fill in the blanks in the following descriptions of the region of integration for each integral. [Note: We have translated the aes in order to write down some of the integrals above. The equations ou specif should be those before the translation is performed.] i. The region of integration in (1) is a part of the snowman s bod/head/bod and head. It is the solid enclosed b the sphere/cone defined b the equation and the sphere/cone defined b the equation. ii. The region of integration in () is a part of the snowman s bod/head/bod and head. It is the solid enclosed b the sphere/cone defined b the equation and the sphere/cone defined b the equation. iii. The region of integration in () is a part of the snowman s bod/head/bod and head. It is the solid enclosed b the sphere/cone defined b the equation and the sphere/cone defined b the equation. Solution. (a) As a check, the bod of the snow man has radius 1.46, which is between (the low point of the head) and 4 (the center of the head). Here is a sketch of a side view of the snowman. 1

11 z ` ` pz 4q 4 z ` ` z 1 We want to determine the volume of the intersection of the bod and the head, whose side view is the darker shaded region in the sketch. The outer boundar of the bod and the outer boundar of the head intersect when both + + z 1 and + + (z 4) 4. Subtracting the second equation from the first gives z (z 4) 1 4 8z 16 8 z Then substituting z into either equation gives +. So the intersection of the outer boundaries of the head and bod (i.e. the neck) is the circle +, z. The top boundar of the intersection is part of the top half of the snowman s bod and so has equation z + 1. The bottom boundar of the intersection is part of the bottom half of the snowman s head, and so has equation z 4 4 The intersection of the head and bod is thus V { (,, z +, 4 4 z 1 } We ll compute the volume of V using clindrical coordinates Volume(V) π 1 r dr dθ dz r 4 4 r dr π r [ 1 r r ] [ π 1 ( ) 1 r / r 1 4 r ( ) /] [ π 1 ( ) / 1( ) / 1 ) / 9 () () ( + 1 ] ( ) / 4 11

12 [ π π ] ( ) / 8 1 [ 1 ] ( ) / So the volume of the snowman is [ ] 4π ( ) / 4π 1 ( ) / π 1 π [ (1 ) ] / + 54 (b) The figure on the left below is another side view of the snowman. This time it is divided into a lighter gra top part, a darker gra middle part and a lighter gra bottom part. The figure on the right below is an enlarged view of the central part of the figure on the left. ` ` pz 4q 4 z? π{ p,,4q p?,,q z? ` ` z 1 π{ p,,q i. The top part is the Pac Man π{ p,,4q? p?,,q 1

13 part of the snowman s head. It is the part of the sphere + + (z 4) 4 that is above the cone + z 4 (which contains the points (,, 4) and (,, )). ii. The middle part is the diamond shaped p,,4q p?,,q? p,,q part of the snowman s head and bod. It is bounded on the top b the cone + z 4 (which contains the points (,, 4) and (,, )) and is bounded on the bottom b the cone z ( + ) (which contains the points (,, ) and (,, )). iii. The bottom part is the Pac Man? p?,,q? π{ p,,q 1

14 part of the snowman s bod. It is the part of the sphere + + z 1 that is below the cone z ( + ) (which contains the points (,, ) and (,, )). 14