(a) Find cylindrical coordinates for the point whose rectangular coordinates are (x, y, z) = ( 4, 8, 2). Solution: r = p x 2 + y 2 =

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1 MATH 03 Exam Solutions February 16, 004 S. F. Ellermeyer Name Instructions. This exam contains seven problems, but only six of them will be graded. You maychooseanysixtodo. PleasewriteDON TGRADEontheonethatyoudon twantme to grade. In writing your solution to each problem, include sufficient detail and use correct notation. (For instance, don t forget to write = when you mean to say that two things are equal.) Your method of solving the problem should be clear to the reader (me). If I have to struggle to understand what you have written, then you might not get full credit for your solution even if you get a correct answer. 1. Make sure to show all details of your computations here. You may express angles in terms of radians or degrees (whichever you choose) and you may approximate angles by rounding to two decimal places. (a) Find cylindrical coordinates for the point whose rectangular coordinates are (x, y, z) = ( 4, 8, ). Solution: r = p x + y = q ( 4) +(8) = 80 = 4. Also, tan (θ) =y/x = and θ is in the second quadrant of the xy plane, so θ = arctan ( ) + π Cylindrical coordinates of this point are thus ³ (r, θ,z)= 4, arctan ( ) + π 116.6,. Let us check our answer: and x = r cos (θ) =4 cos (arctan ( ) + π) = 4 cos (arctan ( )) = 4 µ 1 = 4 y = r sin (θ) =4 sin (arctan ( ) + π) = 4 sin (arctan ( )) = 4 µ =8. 1

2 (b) Find the rectangular coordinates for the point that has cylindrical coordinates are (r, θ,z)=, π,. 4 Solution: Note that this point is in the room below us. ³ π x = r cos (θ) =cos = 4 ³ π y = r sin (θ) =sin = 4 so the rectangular coordinates are à (x, y, z) =,!,. (c) Find spherical coordinates for the point whose rectangular coordinates are (x, y, z) = ( 4, 8, ). Solution: ρ = p x + y + z q = ( 4) +(8) +( ) = 84 =. Also, so In addition, cos (φ) = z ρ = Ã! φ = arccos = cos (θ) = = = x ρ sin (φ) 4 ³ sin arccos ³ ³ sin arccos = ³ 10 = ³

3 and θ is in the second quadrant of the xy plane, so Ã! θ =arccos Spherical coordinates for this point are thus à (ρ, θ, φ) = Ã! Ã!!, arccos, arccos ³, 116.6, Let us check our work: and and x = ρ sin (φ)cos(θ) = à Ã!! à Ã!! sin arccos cos arccos = Ã!Ã! 10 = 4. y = ρ sin (φ)sin(θ) = à Ã!! à Ã!! sin arccos sin arccos = Ã!à 10! =8 z = ρ cos (φ) = à Ã!! cos arccos = Ã! =. (d) Find the rectangular coordinates for the point that has spherical coordinates are (ρ, θ, φ) =, π, 3π

4 Solution: Note that this point is in the room below us. x = ρ sin (φ)cos(θ) µ 3π ³ π =sin cos 4 4 Ã!Ã! = and = y = ρ sin (φ)sin(θ) µ 3π ³ π =sin sin 4 4 Ã!Ã! = and = µ Ã! 3π z = ρ cos (φ) =cos = 4 so the rectangular coordinates of this point are à (x, y, z) =,!,. =. Here are six parametric descriptions of curves labelled 1 6 and six pictures of curves labelled A F. Match each equation with the curve that it describes. 4

5 1) x =cos(t),y=sin(t),z= t ) x = t cos (t),y= t sin (t),z= t 3) x = t cos (t),y= t sin (t),z= t 4) x =cos(t),y=sin(t),z=1 ) x = t cos (t),y= t sin (t),z=1 6) x =sin(t),y= t, z = t Equations 1 match curve F.

6 Equations match curve E. Equations 3 match curve D. Equations 4 match curve A. Equations match curve B. Equations 6 match curve C. 3. Sketch the plane curve defined by the vector function r (t) =t 3 i + t j. Makeyoursketchonthegridprovidedbelow. Itmightbehelpfultomakeatableof values so that you can get a good idea of what the curve looks like. Find the general equation for the tangent vector, r 0 (t), and use this to find the tangent vector to the curve r (t) at the point corresponding to the parameter value t =1.(In other words, find r 0 (1).) Illustrate this tangent vector in your picture. Solution: r 0 (t) =3t i +tj so r 0 (1) = 3i +j. 6

7 4. Find the curvature function, κ (x), forthecurvef (x) =e x.thendeterminethepoint on the graph of f at which this graph has maximum curvature. (Show all details.) Solution: For the curve f (x) =e x,wehave so the curvature is κ (x) = f 0 (x) =e x f 00 (x) =e x f 00 (x) 1+(f 0 (x)) 3/ = e x (1 + e x ) 3/. We would like to find at what value of x the curvature is maximum: Since ³ (1 + e x ) 3/ e x e x 3 (1 + κ 0 ex ) 1/ (e x ) (x) = (1 + e x ) 3 (1 + e x ) 3/ e x e ³3(1+e x x ) 1/ (e x ) = (1 + e x ) 3 = ex (1 + e x ) 1/ (1 e x ) (1 + e x ) 3 = ex (1 e x ) (1 + e x ) / we see that κ 0 (x) =0when 1 e x =0. Solving this equation for x gives us e x =1/ or (e x ) =1/ or e x = / or Ã! x =ln. 7

8 Since this x gives us the only critical point of κ and since it is clear that the curvature must be maximum at some point for the graph of f (x) =e x,weconcludethatthe curvatureismaximumatthepoint à Ã!! ln, ( , ).. The position function of a moving particle is r (t) =t i +tj+ t 16t k. Find the time, t, at which the speed of the particle is at its minimum. details.) Solution: position=r (t) = t, t, t 16t velocity=r 0 (t) =ht,, t 16i (Show all speed= r 0 (t) q = 4t ++(t 16) = 8t 64t +81 To find when the speed is minimum, we compute d ³ 8t 64t +81 = dt 8t 3 8t 64t +81 This derivative is equal to zero when t =4. Also, this derivative is negative when t<4 and positive when t > 4. Thus t =4is where the minimum speed occurs. 8

9 6. Here are six parametric descriptions of surfaces labelled 1 6 and six pictures of surfaces labelled A F. Match each equation with the curve that it describes. 1) r (u, v) =hcos (u), sin (u),vi ) r (u, v) =hu, v, v i 3) r (u, v) =hv cos (u),vsin (u),v i 4) r (u, v) =h1+u, 1+v, u + vi ) r (u, v) =hv cos (u),vsin (u),vi 9

10 6) r (u, v) =hv cos (u),vsin (u), 1i Equation 1 match surface F. Equation match surface A. Equation 3 match surface C. Equation 4 match surface B. Equation match surface D. Equation 6 match surface E. 10