MATHEMATICS 200 December 2013 Final Exam Solutions

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1 MATHEMATICS 2 December 21 Final Eam Solutions 1. Short Answer Problems. Show our work. Not all questions are of equal difficult. Simplif our answers as much as possible in this question. (a) The line L has vector parametric equation r(t) = (2 + t)îı + 4tĵj ˆk. (b) i. Write the smmetric equations for L. ii. Let α be the angle between the line L and the plane given b the equation + 2z =. Find α. i. Find the equation of the tangent plane to the surface 2 z + sin(π) = 2 at the point P = (1, 1, 1). ii. Let z be defined implicitl b 2 z + sin(π) = 2. Find z at the point P = (1, 1, 1). iii. Let z be the same implicit function as in part (ii), defined b the equation 2 z + sin(π) = 2. Let =.97, and = 1. Find the approimate value of z. (c) Suppose that u = 2 + z, = ρr cos(θ), = ρr sin(θ) and z = ρr. Find u r point (ρ, r, θ ) = (2,, π/2). at the (d) Let f() be a differentiable function, and suppose it is given that f () = 1. Let g(s, t) = f(as bt), where a and b are constants. Evaluate g at the point (s, t) = s (b, a), that is, find g (b,a). s (e) Suppose it is known that the direction of the fastest increase of the function f(, ) at the origin is given b the vector 1, 2. Find a unit vector u that is tangent to the level curve of f(, ) that passes through the origin. (f) Find all the points on the surface z 2 = 17 where the tangent plane is parallel to the plane 8z =. (g) Find the total mass of the rectangular bo [, 1] [, 2] [, ] (that is, the bo defined b the inequalities 1, 2, z ), with densit function h(,, z) =. Solution. (a) i. Since = 2 + t = t = 2 = 4t = t = 4 we have 2 = 4 z = 1 (a) ii. The direction vector for the line r(t) = 2îı ˆk + t(îı + 4ĵj) is d = îı + 4ĵj. A normal vector for the plane + 2z = is n = ± ( îı ĵj + 2 ˆk ). The angle θ between d 1

2 and n obes cos θ = d n d n = 1 6 = θ = arccos radians (We picked n = îı + ĵj 2ˆk to make θ π.) Then the angle between d and the 2 plane is α = π 2 arccos 1.8 radians 6 n θ α line L plane ` 2z (b) i. The surface has equation G(,, z) = 2 z + sin(π) + 2 =. So a normal vector to the surface at (1, 1 1) is G(1, 1, 1) = [( 2z + π cos(π) ) îı + ( sin(π) + 2 ) ĵj + z 2 2 ˆk](,,z)=(1,1, 1) = ( 2 π ) îı + 2ĵj + ˆk So the equation of the tangent plane is ( ) 2 π ( 1) + 2( 1) + (z + 1) = or (2 + π) z = π (b) ii. The functions z(, ) obes 2 z(, ) + sin(π) + 2 = for all and. Differentiating this equation with respect to gives Evaluating at (1, 1, 1) gives 2z(, ) + 2 z(, ) 2 z (, ) + π cos(π) = 2 + z z π + 2 (1, 1) π = = (1, 1) = (b) iii. Using the linear approimation about (, ) = (1, 1), z(, 1) z(1, 1) + z (1, 1) ( 1) 2

3 gives z(.97, 1) 1 + π + 2 (.) = 1 π = π (c) The function u(ρ, r, θ) = [ ρr cos θ ] 2 + [ ρr sin θ ] ρr = ρ 2 r 2 cos 2 θ + ρ 2 r 2 sin θ So and u r (ρ, r, θ) = 2ρ2 r cos 2 θ + 2ρ 2 r sin θ u r (2,, π/2) = 2(22 )()() 2 + 2(2 2 )()(1) = 24 (d) B the chain rule In particular g s (s, t) = [ ] f(as bt) = af (as bt) s g s (b, a) = af (ab ba) = af () = 1a (e) We are told that the direction of fastest increase for the function f(, ) at the origin is given b the vector 1, 2. This implies that f(, ) is parallel to 1, 2. This in turn implies that 1, 2 is normal to the level curve of f(, ) that passes through the origin. So 2, 1, being perpendicular to 1, 2, is tangent to the level curve of f(, ) that passes through the origin. The unit vectors that are parallel to 2, 1 are ± 1 2, 1. (f) A normal vector to the surface z 2 = 17 at the point (,, z) is 2, 18, 8z. A normal vector to the plane 8z = is 1,, 8. So we want 2, 18, 8z to be parallel to 1,, 8, i.e. to be a nonzero constant times 1,, 8. This is the case whenever = and z = 2 with. In addition, we want (,, z) to lie on the surface z 2 = 17. So we want =, z = 2 and 17 = z 2 = 2 + 4( 2) 2 = 17 2 = = ±1 So the allowed points are ±(1,, 2). (g) The mass is 1 d 2 d dz = 6 1 d =

4 2. The shape of a hill is given b z = Assume that the ais is pointing East, and the ais is pointing North, and all distances are in metres. (a) What is the direction of the steepest ascent at the point (, 1, 9)? (The answer should be in terms of directions of the compass). (b) What is the slope of the hill at the point (, 1, 9) in the direction from (a)? (c) If ou ride a biccle on this hill in the direction of the steepest descent at m/s, what is the rate of change of our altitude (with respect to time) as ou pass through the point (, 1, 9)? Solution. Write h(, ) = so that the hill is z = h(, ). (a) The direction of steepest ascent at (, 1, 9) is the direction of maimum rate of increase of h(, ) at (, 1) which is h(, 1) =,.1(2)(1) =, 2. In compass directions that is South. (b) The slope of the hill there is h(, 1), 1 = h (, 1) = 2 (c) Denote b ( (t), (t), z(t) ) our position at time t and suppose that ou are at (, 1, 9) at time t =. Then we know z(t) = 1.2 (t) 2.1(t) 2, so that z (t) =.4 (t) (t).2(t) (t), since ou are on the hill and () = and () > since ou are going in the direction of steepest descent and () 2 + () 2 + z () 2 = 2 since ou are moving at speed. Since () and () = 1, we have z () =.2(1) () = 2 (). So 2 = () 2 + () 2 + z () 2 = () 2 = () = and our rate of change of altitude is = (), (), z () =,, 2 d dt h( (t), (t) ) t= = h(, 1) (), () =, 2, = 2. (a) Find the minimum of the function f(,, z) = ( 2) 2 + ( 1) 2 + z 2 subject to the constraint z 2 = 1, using the method of Lagrange multipliers. 4

5 (b) Give a geometric interpretation of this problem. Solution. (a) This is a constrained optimization problem with the objective function being f(,, z) = ( 2) 2 + ( 1) 2 + z 2 and the constraint function being g(,, z) = z 2 1. B Theorem in the CLP III tet, an local minimum or maimum (,, z) must obe the Lagrange multiplier equations f = 2( 2) = 2λ = λg f = 2( 1) = 2λ = λg f z = 2z = 2λz = λg z z 2 = 1 (E1) (E2) (E) (E4) for some real number λ. B equation (E), 2z(1 λ) =, which is obeed if and onl if at least one of z =, λ = 1 is obeed. If z = and λ 1, the remaining equations reduce to 2 = λ 1 = λ = 1 (E1) (E2) (E4) Substituting = 2 1, from (E1), and =, from (E2), into (E) gives 1 λ 1 λ 4 (1 λ) (1 λ) = 1 (1 2 λ)2 = 1 λ = ± and hence (,, z) = ± 1 (2, 1, ) To aid in the evaluation of f(,, z) at these points note that, at these points, 2 = λ = 2λ 1 λ, 1 = λ = λ 1 λ = f(,, z) = 4λ2 (1 λ) 2 + λ 2 (1 λ) 2 = λ2 (1 λ) 2 = λ2 = ( 1 ) 2 If λ = 1, the remaining equations reduce to 2 = 1 = z 2 = 1 (E1) (E2) (E4) Since 2 and 1, neither (E1) nor (E2) has an solution. So we have the following candidates for the locations of the min and ma

6 1 point (2, 1, ) 1 (2, 1, ) ( ) 2 ( ) 2 value of f min ma So the minimum is ( 1 ) 2 = 6 2. (b) The function f(,, z) = ( 2) 2 + ( 1) 2 + z 2 is the square of the distance from the point (,, z) to the point (2, 1, ). So the minimum of f subject to the constraint z 2 = 1 is the square of the distance from (2, 1, ) to the point on the sphere z 2 = 1 that is nearest (2, 1, ). 4. (a) Find the minimum of the function h(, ) = on the closed bounded domain defined b (b) Eplain wh Question 4 gives another wa of solving Question. Solution. (a) Since h = 4, 2 is never zero, h has no critical points and the minimum of h on the disk must be taken on the boundar, = 1, of the disk. To find the minimum on the boundar, we parametrize b = cos θ, = sin θ and find the minimum of Since H(θ) = 4 cos θ 2 sin θ + 6 = H (θ) = 4 sin θ 2 cos θ = = cos θ = 2 sin θ = 2 So 1 = = = 2 = = ± 1, = ± 2 At these two points h = = 6 1 = 6 1 = 6 2 The minimum is 6 2. (b) In Question 4 we had to find the minimum of f(,, z) = ( 2) 2 + ( 1) 2 + z 2 subject to the constraint z 2 = 1. But when the constraint is satisfied, f = ( 2) 2 + ( 1) 2 + z 2 = ( 2) 2 + ( 1) = = h(, ) and (, ) runs over

7 . This question is about the integral ln ( ) d d (a) Sketch the domain of integration. (b) Evaluate the integral b transforming to polar coordinates. Solution. (a) On the domain of integration runs from to 1. In inequalities, 1. For each fied in that range, runs from to 4 2. In inequalities, that is 4 2. Note that the inequalities 4 2, are equivalent to ,. Note that the line = and the circle intersect when = 4, i.e. = ±1. Here is a sketch. 2 ` 2 4 p?,1q? π{6 (b) In polar coordinates, the circle = 4 is r = 2 and the line =, i.e. = 1, is tan θ = 1 or θ = π. As d d = r dr dθ, the domain of integration is 6 { (r cos θ, r sin θ) θ π /6, r 2 } and ln ( ) d d = 2 dr π/6 = π du ln(u) 12 1 = π [ u ln(u) u 12 = π [ ] ln() 4 12 dθ r ln(1 + r 2 ) = π 6 ] 1 2 dr r ln(1 + r 2 ) with u = 1 + r 2, du = 2r dr 7

8 6. Evaluate 2 e d d Solution. The antiderivative of the function e 2 cannot be epressed in terms of elementar functions. So the inside integral 2 2 e2 d cannot be evaluated using standard calculus 2 techniques. The trick for dealing with this integral is to reverse the order of integration. On the domain of integration runs from 1 to. In inequalities, 1. For each fied in that range, runs from 2 to 2. In inequalities, 2 2. The domain of integration, namel { (, ) 1, 2 2 } is sketched in the figure on the left below. p 1, 2q 2 2 p 1, 2q 2 {2 Looking at the figure on the right above, we see that we can also epress the domain of integration as { (, ) 2, /2 } So the integral e 2 d d = e 2 2 /2 e 2 = 1 2 [ 1 2 ] d d d = e2 = 1 [ e 4 1 ] 4 2 8

9 7. Let a > be a fied positive real number. Consider the solid inside both the clinder = a and the sphere z 2 = a 2. Compute its volume. Hint: sin (θ) = 1 12 cos(θ) 4 cos(θ) + C Solution. We ll use clindrical coordinates. In clindrical coordinates the sphere z 2 = a 2 becomes r 2 + z 2 = a 2 and the circular clinder = a (or equivalentl ( a/2) = a 2 /4) becomes r 2 = ar cos θ or r = a cos θ. Here is a sketch of the top view of the solid. 2 ` 2 a pa{2,q The solid is { (r cos θ, r sin θ, z) π/2 θ π /2, r a cos θ, a 2 r 2 z a 2 r 2 } B smmetr, the volume of the specified solid is four times the volume of the solid { (r cos θ, r sin θ, z) θ π/2, r a cos θ, z a 2 r 2 } Since dv = r dr dθ dz, the volume of the solid is 4 π/2 dθ a cos θ dr a 2 r 2 dz r = 4 = 4 = 4 = 4a = 4a = 4a = 4a π/2 π/2 π/2 π/2 a cos θ dθ dr r a 2 r 2 dθ ( a 2 r 2) /2 a cos θ dθ [a ( a 2 a 2 cos 2 θ ) ] /2 dθ [ 1 sin θ ] [θ 112 cos(θ) + 4 cos θ ] π/2 [ π ] 4 [ π 2 2 ] 9

10 8. (a) Sketch the surface given b the equation z = 1 2. (b) Let E be the solid bounded b the plane =, the clinder z = 1 2, and the plane = z. Set up the integral as an iterated integral. E f(,, z) dv Solution. (a) Each constant cross section of z = 1 2 is an upside down parabola. So the surface is a bunch of upside down parabolas stacked side b side. The figure on the left below is a sketch of the part of the surface with and z (both of which conditions will be required in part (b)). z z z (b) The figure on the right above is a sketch of the plane = z. It intersects the surface z = 1 2 in the solid blue sloped parabolic curve in the figure below. z z 1 2 p1,, q z , z Observe that, on the curve z = 1 2, z =, we have = 1 2. So that when one looks at the solid E from high on the z ais, one sees { (, ) 1 2 } The = 1 2 boundar of that region is the dashed blue line in the plane in the figure above. So E = { (,, z) 1 1, 1 2, z 1 2 } and the integral E f(,, z) dv = d d dz f(,, z) 1

11 9. (a) Find the volume of the solid inside the surface defined b the equation ρ = 8 sin(ϕ) in spherical coordinates. Hint: ou do not need a sketch to answer this question; and sin 4 (ϕ) = 1 2( 12ϕ 8 sin(2ϕ) + sin(4ϕ) ) + C (b) Sketch this solid or describe what it looks like. Hint: it is a solid of revolution. Solution. (a) Recall that, in spherical coordinates, ϕ runs from (that s the positive z ais) to π (that s the negative z ais), θ runs from to 2π (θ is the regular polar or clindrical coordinate) and dv = ρ 2 sin ϕ dρ dθ dϕ. So π 2π 8 sin ϕ Volume = dϕ dθ dρ ρ 2 sin ϕ = π dϕ = 2(8 ) π = 2(8 ) π = 2(8 ) π 2π π [ 1 dθ (8 sin ϕ) dϕ sin 4 ϕ 2 12 π 2 = 128π2 sin ϕ ( 12ϕ 8 sin(2ϕ) + sin(4ϕ) ) ] π (b) Fi an ϕ between and π. If ρ = 8 sin ϕ, then as θ runs from to 2π, (,, z) = ( ρ sin ϕ cos θ, ρ sin ϕ sin θ, ρ cos ϕ ) = ( 8 sin 2 ϕ cos θ, 8 sin 2 ϕ sin θ, 8 sin ϕ cos ϕ ) = ( R cos θ, R sin θ, Z ) with R = 8 sin 2 ϕ, Z = 8 sin ϕ cos ϕ sweeps out a circle of radius R = 8 sin 2 ϕ contained in the plane z = Z = 8 sin ϕ cos ϕ and centred on (,, Z = 8 sin ϕ cos ϕ ). So the surface is a bunch of circles stacked one on top of the other. It is a surface of revolution. We can sketch it b first sketching the θ = section of the surface (that s the part of the surface in the right half of the z plane) and then rotate the result about the z ais. The θ = part of the surface is { (,, z) = 8 sin 2 ϕ, =, z = 8 sin ϕ cos ϕ, ϕ π } = { (,, z) = 4 4 cos(2ϕ), =, z = 4 sin(2ϕ), ϕ π } 11

12 It s a circle of radius 4, contained in the z plane (i.e. = ) and centred on (4,, )! The figure on the left below is a sketch of the top half of the circle. When we rotate the circle about the z ais we get a torus (a donut) but with the hole in the centre shrunk to a point. The figure on the right below is a sketch of the part of the torus in the first octant. z z 12