Calculus III. Exam 2
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1 Calculus III Math 143 Spring 011 Professor Ben Richert Exam Solutions Problem 1. (0pts) Computational mishmash. For this problem (and only this problem), you are not required to supply any English explanation. You don t need simplify either. (a 5pts) Consider the point (3, 5) in Cartesian coordinates. Give the polar coordinates for this point. ( ) Solution. (r,θ) = 3 +5,arctan(5/3) (b 5pts) Consider the curve described by the equation x y +y 3 =. Find a Polar equation for this curve. Let me reiterate that you do not need to simplify. Solution. (r cos θ)rsinθ+r 3 sin 3 θ = (c 5pts) Give a vector which is parallel to 1,,3 and twice as long as,4,1. Solution ,,3 = ,,3. (d 5pts) Write an integral which computes the surface area obtained by rotating the polar curve r = e sin(θ) for 0 θ π/6 about the x-axis. Do not solve the integral. Hint: first convert to a parametric curve. 0.8 r e Sin Θ 0 Θ Π Solution. We know x = rcosθ = e sin(θ) cosθ so x (t) = e sin(θ) cos θ e sin(θ) sinθ and y = rsinθ = e sin(θ) sinθ so y (t) = e sin(θ) sinθcosθ +e sin(θ) cosθ. So using the formula π/6 0 b a πy (x (t)) +(y (t)) dt yields πe sin(θ) sinθ (e sin(θ) cos θ e sin(θ) sinθ) +(e sin(θ) sinθcosθ +e sin(θ) cosθ) dθ.
2 Problem. (10pts) Consider the graph of the parametric curve whose component functions x(t) and y(t) have graphs as follows: 1.4 x t y t Give all t such that the tangent line to the parametric curve x = x(t), y = y(t) is horizontal. Solution. We know that the tangent is horizontal exactly when dy/dx = y (t)/x (t) = 0 and this occurs when y (t) = 0, that is, when the graph of y (t) has a horizontal tangent (note that x (t) is never zero since its tangent is never horizontal, thus it truly is enough to consider y (t) = 0). By inspection this occurs exactly when t = 1 and t =. Problem 3. (10pts) You like to talk math with a friend named Euler, but he has a terrible cell phone connection which cuts out frequently. Do exactly one of the following two problems. (a 10pts) Euler claims to have discovered a new rule for cross products and dot products, not given in the text. He begins tosay I have discovered that v ( u w) =... but then the connection cuts out. Explain (in a few sentences) how you know that he can t possibly be correct, no matter what he was about to say. Solution. The problem is that u w is a scalar, so v ( u w) would be the cross of a vector and a scalar, which isn t defined. No matter what Euler was about to say, it is nonsense. (b 10pts) Euler is very excited about vectors and says, My two favorite vectors are 1,,3 and, static,4, and the best part is that they are orthogonal. Unfortunately, you didn t catch all of his second vector because of the bad phone connection (static momentarily drowned out his voice), but you can still figure it out. What are his two favorite vectors?
3 Solution. Suppose that the second vector is, t, 4. Then because the two vectors are perpendicular, we have that 0 = 1,,3,t,4 = +t+, or solving for t, that t = 7. Thus Euler s two favorite vectors are 1,,3 and, 7,4 Problem 4. (10pts) You move a sled (full of chalk, that precious national resource) in a straight line from the point (0,0) to the point (10,10) (in meters) by applying the force given by the vector F = 40,60 (in Newtons). How much work is done moving the sled? Solution. Here we apply force F = 40,60 to affect movement along the displacement vector D = 10,10. According to the text, work = F D = 40,60 10,10 = = 1000 Joules. Another way to think about it is as follows. First compute the scalar projection of F onto the direction vector 10,10 to determine the magnitude of the force which affects movement in the 10, 10 direction, and then multiply by distance traveled. That is, compute comp 10,10 40,60 = 40,60 10, = Newtons and then multiply by = 00 meters. One gets the same answer, of course. Problem 5. (15pts) Sketch the polar curve r = 1/+sin(θ), 0 θ π, and set up an integral which computes the area enclosed by one of the smaller loops (you do not need to compute this integral). Solution. We begin with the graph in the θr plane, as follows: r 1 1 Π 7Π Π 3Π Π Θ 1 and note that 1/+sinθ is zero if sinθ = 1/. Now sinθ = 1/ at θ = 7π/6, 11π/6, 19π/6, and 3π/6 (thinking about the unit circle). So it follows that sinθ = 1/ at θ = 7π/, 11π/, 19π/, and 3π/. Thus the curve traces the first loop while 0 θ 7π/,
4 7Π another smaller loop (in the negative direction) while 7π/ θ 11π/ 7Π and behaves similarly for 11π/ θ π, yielding
5 7Π From this we can also setup the integral to compute area. A small loop is traced over the range 7π/ θ 11π/, so the β (f(θ)) formula for area, dθ gives α 11π/ (1/+sin(θ)) area = dθ. 7π/ Problem 6. (10pts) Write the equation of a plane which contains the line 1,,3 + t,4,6 and is parallel to the line 1,,1 +t 1,0,1. Solution. The equation of the plane containing (a, b, c) and with perpendicular A, B, C is A(x a)+b(y b)+c(z c) = 0. So we need a point and a normal. We ll use the point (1,,3) since that is on the line (set t = 0 in the first equation above) and the line is to be contained in our plane. For the perpendicular, we take i j k,4,6 1,0,1 = 4 6 = 4 0, ( 6),0 4 = 4,4, This vector is perpendicular to the plane because the cross product is perpendicular to both of its component vectors and in this case each component vector is parallel to the plane (and they themselves are not parallel, else the cross product would have been zero). So, the plane in question is 4(x 1)+4(y ) 4(z 3) = 0.
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