Math 21a Old Exam One Fall 2003 Solutions Spring, 2009


 Ella Mills
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1 1 (a) Find the curvature κ(t) of the curve r(t) = cos t, sin t, t at the point corresponding to t = Hint: You ma use the two formulas for the curvature κ(t) = T (t) r (t) = r (t) r (t) r (t) 3 Solution: We compute: r (t) = sin t, cos t, r (t) = cos t, sin t, Therefore and so r (t) r i j k (t) = sin t cos t = sin t, cos t, 1 cos t sin t κ(t) = r (t) r (t) r (t) 3 sin t, cos t, 1 = sin t, cos t, 3 = ( ) 3 = 1 (b) Find the osculating plane of this curve at the point corresponding to t = Solution: The osculating plane is perpendicular to T(t) and N(t), but it is also perpendicular to r (t) and r (t), and those are easier to work with We ve alread computed a normal vector n = r () r (), which is sin(), cos(), 1 =,, 1 This plane passes through the point r() = cos, sin, () = 1,,, so the plane has equation,, 1 ( 1),, = or = or + = (c) Find the normal plane of this curve at the point corresponding to t = Solution: The normal plane is perpendicular to T(), which points in the same direction as r () Thus the normal plane is through the point r() = 1,, and perpendicular to r () = sin(), cos(), =, 1, Thus the plane has equation, 1, ( 1),, = or = (d) Find the curvature κ(t) of the curve r(t) = cos(t), sin(t), 1t at the point corresponding to t = Solution: This is a reparameteriation of the curve in part (a) (we ve replaced t with t) Thus the curve traced out b this vector function is the same and hence it has the same curvature: κ(t) = 1 Consider the plane + 3 = 1 and the line parameteried b = t + 1, = 4t 1, = 3t (a) Find two different unit vectors that are normal to the plane Solution: There are onl two: 1 1 1,, 3 and 1 1 1,, 3 (b) Find a vector a that is parallel to the line
2 Solution: The line is,, = t+1, 4t 1, 3t or r(t) = 1, 1, +t, 4, 3 Thus a vector parallel to the line is a =, 4, 3 (or an scalar multiple of this) (c) Eplain wh the line is not parallel to the plane Solution: The easiest wa to see this is to see that a n, where n is a normal to the plane We ll use n = 1,, 3 for simplicit; we get a n =, 4, 3 1,, 3 = 11, not ero (d) Find the point of intersection of the line and the plane Solution: One wa to do this is to plug in = t + 1, = 4t 1, = 3t into the equation for the plane and then find t We find t =, so (,, ) = (1, 1, ) is the point of intersection (e) Find an equation for a plane that contains both the line and the origin Solution: The line contains the vector a =, 4, 3 and the point (,, ) = (1, 1, ) Thus the plane must contain the vector 1, 1, from the origin to our point We can then compute our plane s normal vector: i j k n =, 4, 3 1, 1, = 4 3 = 3, 3, Since the plane contains the origin, we can write the equation as 3, 3, 6,, = or + = (f) Find the angle between the plane in part (e) and the original plane, + 3 = 1 (Recall that the angle between two planes is defined as the acute angle between their normal vectors) If ou weren t able to find the plane in part (e), then use the plane = instead Solution: We can find the angle θ between the two normal vectors using either the dot product (which will give us cos θ) or the cross product (sin θ) We ll use the dot product since it s easier to compute: 3, 3, 6 1,, 3 cos(θ) = 3, 3, 6 1,, 3 = 1 ( ) ( ) = Before we get too ecited, we should note that the answer is NOT simpl cos ( 1 ), 3 since this is not an acute angle Instead our answer is that the angle between the planes is π cos ( 1 ) 3 3 Let a and b be two vectors such that a b = i + j 3k (a) What is ( 3b) (4a)? Solution: ( 3b) (4a) = 1b a = 1a b = 1,, 3 = 4, 4, 36 (b) Find a (a b) if it eists If it doesn t eist then eplain wh not Solution: This is ero a b is a vector perpendicular to a (and b, not that it matters), so a (a b) = (c) Find a (a b) if it eists If it doesn t eist then eplain wh not Solution: This doesn t eist: a b is a scalar (a number), and we can t cross a vector with a scalar
3 (d) Suppose that not onl does a b = i + j 3k, but also that a b = a b Which of the following statements must be true? (Simpl circle all the statements that must be true no eplanations are necessar) (i) a b (iii) a is perpendicular to b (ii) a and b are parallel (iv) either a or b must be a unit vector Solution: Recall that u v = u v sin(θ), so for a and b we must have sin θ = 1 Thus we have two true statements: (i) a b (iii) a is perpendicular to b 4 Consider the space curve defined b the vector function with t π r(t) = 3 cos(t), t, 3 sin(t) (a) Sketch the curve as neatl as possible (Be sure to label our aes) Solution: This is a heli, starting at (3,, ) and winding its wa around the positive ais to the point (3, 4π, ): (b) Find an equation for a surface that this space curve lies on and describe the surface Solution: Since (3 cos(t)) + (3 sin(t)) = 9, we get + = 9 This is a clinder centered on the ais with radius 3 (c) Find the length of this space curve Solution: The length of a smooth curve from t = a to t = b is b a r (t) dt r (t) = 6 sin(t),, 6 cos(t), so r (t) = 4 = 1 Thus the length of the curve is Here π 1 dt = 4π 1
4 (d) Give parametric equations for the tangent line to this space curve at the point (3, π, ) Solution: The tangent line requested goes through the point (3, π, ) when t = π, so it has direction equal to r (π) = 6 sin(π),, 6 cos(π) =,, 6 Thus one set of parametric equations for the tangent line is (t) = 3, π, + t,, 6 (a) Consider the space curve that is defined as the intersection of the plane = and the clinder + = 1 Eplain wh this space curve can be parameteried b the vector function r(t) = cos t, sin t, sin t with t π Solution: This space curve, when projected verticall into the plane, is simpl the unit circle + = 1, which we parameterie = cos t and = sin t The component can be found because the curve lies in the plane =, so = sin t (b) Now consider the vector function p(t) = r(t) r(t), where r(t) is the vector function given in part (a) Compute the components of the vector function p(t) Solution: Since r(t) = cos t + sin t, the vector function p(t) is cos t p(t) = cos + sin t, sin t cos + sin t, sin t cos + sin t (c) Now consider the space curve formed b the intersection of the same plane = and the unit sphere (the sphere of radius one centered at the origin) Eplain wh this new space curve is parameteried b the vector function p(t) Solution: We onl need to confirm two facts: The space curve p(t) lies on the plane =, and The space curve p(t) lies on the unit sphere The first is obvious (look at the and components of p(t)) The second isn t ver hard: p(t) = r(t) r(t) = 1, so the space curve lies on this sphere 6 (a) Sketch the graph of the function f(, ) = ( 1) 1 Make our sketch as neat as possible, and be sure to label our aes carefull Solution: In the plane, this is simpl a parabola In fact, the = k is alwas the same parabola Thus this is a parabolic clinder as seen here: 3 1
5 (b) Now sketch the graph of the function f(, ) = + 1 Again, make our sketch as neat as possible, and label our aes carefull Solution: If we write = + 1, we can rearrange this to be + = 1, a onesheeted hperboloid But since the square root function outputs nonnegative numbers, we onl get half this hperboloid: 4 7 A moth is fling around a light bulb hanging from the ceiling of a room The light bulb is 1 feet above the floor and the moth stas eactl 1 feet above the floor as it flies around the light bulb The moth flies in a circular path, at constant speed, alwas eactl feet awa from the center of the light bulb (a) Write down a set of equations that describe the circle traced out b the moth in spherical coordinates Consider the origin to be the point on the floor directl below the light bulb Solution: The circle in spherical coordinates is simpl ρ = 14 = 6 and φ = tan 1 ( 1 ) (The angle θ is arbitrar) If it helps, we can write this in rectangular coordinates:,, = 6 sin ( tan 1 ( 1 = cos(θ), sin(θ), 1 Here s a small picture of the situation: )) cos(θ), 6 sin ( tan 1 ( 1 )) ( sin(θ), 6 cos tan 1 ( )) 1 moth 1 φ plane The hpotenuse of this right triangle is ρ = 14 = 6 and tan(φ) = are what produced the spherical coordinates above 1 = 1 These values (b) Write down the equations giving the moth s circular path in clindrical coordinates, and then give a parameteriation of the moth s path in clindrical coordinates
6 Solution: In clindrical coordinates this path is r =, = 1 A parameteriation in clindrical coordinates might be (r, θ, ) = (, t, 1) (simpl θ = t together with the equations involving r and, above) (c) Suppose that the moth s flight path is given b a vector function r(t), with parameter t representing time Given what ou know about the moth s flight answer the following questions: (i) Is it possible to determine r (t) k? If so, what is it? Solution: Yes, r (t) k = The moth alwas flies at the same height, so the tangent vector (velocit) has is horiontal (has no vertical component) Thus this velocit is perpendicular to the vertical vector k, so the dot product is ero (ii) Describe the curve parameteried b r (t) k Solution: We parameterie the curve traced b the moth b r(t) = 1 cos(t), 1 sin(t), (Note we re assuming that the angular speed is 1, but we have no units, so wh not?) Then r (t) = 1 sin(t), 1 cos(t), and so r i j k (t) k = 1 sin(t) 1 cos(t) = 1 cos(t), 1 sin(t), 1 This curve is a circle in the plane 8 Match the following space curves to the appropriate vector functions (Some of the graphs appear on the net page) Solution: (a) r(t) = cos t, sin t, ln t This is Curve III The distinguishing characteristic is that, as the and coordinates revolve in a circle, the coordinate increases quickl from then increases slower and slower Curve III (b) r(t) = t cos t, t sin t, 1 t This is Curve I The ke thing to note is that when and are at radius t, the height is = 1 t So as t decreases to ero, and like on a ver small circle as grows without bound Curve I (c) r(t) = cos t, sin t, sin(3t) This is Curve II As and travel around their customar circle, the coordinate rises and falls through three ccles Curve II
7 (d) r(t) = t cos t, t sin t, t The ke feature to notice is that the and coordinates are revolving with an everincreasing radius that happens to be the same as the coordinate Thus the curve lies on the cone = r, and this is Curve VI Curve VI (e) r(t) = cos t, sin t, t This is our old friend the heli, Curve IV (f) r(t) = cos t, sin t, 1 t This is Curve V, the onl one left The ke here is that the height = 1 t, the inverse of the angle through which the curve has revolved around the ais Thus as t decreases to ero, for eample, grows without bound As t grows larger, however, the height decreases toward ero Curve IV Curve V 9 A planar mirror in space contains the point P (4, 1, ) and is perpendicular to the vector n = 1,, 3 The light ra QP = v = 3, 1, with source Q(7,, 7) hits the mirror plane at the point P (a) Compute the projection u = proj n (v) of v onto n Solution: Recall that proj n (v) = n v n n Since n = 1,, 3 and v = 3, 1,, we get u = proj n (v) = 1,, 3 3, 1, 1,, 3 1,, 3 = 1,, 3 14 (b) Identif u in the figure below and use it to find a vector parallel to the reflected ra Solution: I ve redrawn the figure faintl and drawn in u as more dark Q u P Mirror The two dashed arrows are both QP u, so the vector we want is ( u)+ ( QP u ) = QP u
8 1 What is the distance between the two clinders + = 1 and ( ) + ( ) = 4? Solution: The first clinder is centered along the ais and has radius 1 The second is centered parallel to the ais, but with = and = ; it has radius We ll find the distance between these two center lines and subtract 3 (= 1 + ) The distance between the lines is, and so the distance between the clinders is (One can compute the distance between the lines in the usual wa, but it s easier just to think about it When are the two lines closest?) 11 (a) Parameterie the intersection of the ellipsoid with the plane = ( ) = Solution: When = 3, the and coordinates satisf ( ) ( ) + = 1 We parameterie this circle with = cos t and = sin t, so the full parameteriation is r(t) = cos t, + sin t, 3 (b) Parameterie the ellipsoid itself in the form r(θ, φ) = Solution: In order to use spherical coordinates, we make the equation look like a sphere: Using ρ =, we get or r(θ, φ) = ( ) ( ) ( ) ( ) ( ) + + = 3 = sin(φ) cos(θ) = sin(φ) sin(θ) 3 = cos(φ), sin(φ) cos(θ), + sin(φ) sin(θ), 3 cos(φ) (c) What is the curvature of the curve in part (a) at the point (,, 3) Hint: While ou ma use the curvature formula κ(t) = r (t) r (t) r (t) 3, ou are also allowed to cite a fact which ou know about the curvature Solution: Rather than just plunging into the computation, we ll instead infer from the hint that there s a simpler wa to do this The curve in question is a circle of radius and we recall that the curvature of a circle of radius R is κ = 1 R Thus the curvature of our circle is 1 for all values of t
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