Shadow Moving Along a WallDate: 08/16/99 at 09:20:53 From: Kris Subject: Inverse trig functions B - | - - | - S - | - - | C - L ------------------ - | - - | - - | - - | - The Problem: At the front of a circular arena there is a light L. A boy starting from B runs at the rate of 3 m/s toward the center C. At what rate will his shadow be moving along the rear wall when he is halfway from B to C? This problem was presented in my assignment and is the only one I can't seem to get right. I think the rate might be 8 m/s? (My friend used vector calculus, but we can't use this because our teacher said it wouldn't be relevant for this course.) I drew a Triangle LCS Angle CLS = Angle CSL and from this I derived that Angle CLS + Angle CSL = Angle LCS Then found out the height From C towards B within the Triangle LCS is equal to R (the radius) - 3t, since distance = vt. then Angle BLC = Tan^-1(R-3t/R) Letting Angle BLC = theta Angle LCS = 2theta then 2theta = 2Tan^-1(R-3t/R) d(theta) -6 -------- = ------------ dt R^2-(R-3t)^2 Then I found t = d/v = (R/2)(1/3) = R/6 I then substituted t into the above equation, and in the end the R's did not cancel out. Nobody in my class can get it either. Have I gone about this wrong? Thanks heaps, Kris Date: 08/17/99 at 08:24:49 From: Doctor Jerry Subject: Re: Inverse trig functions Hi Kris, Let's put in some coordinates. Let L and S have coordinates (-R,0) and (x,y). At time t the boy is at (0,R-3t). We can figure out equations for x and y. The line through L and the boy is: y-0 = ((R-3t)/R)(x+R) The point (x,y) satisfies x^2+y^2 = R^2 Let's not try, however, to solve these equations simultaneously just yet. Let f(t) be the angle between the positive x-axis and the line from C to S. Letting the arc subtended by f(t) on the circle be s, we know that s = R*f(t) We want to calculate ds/dt, right? We see that ds/dt = R*f'(t) so that's what we'll work on. Now, tan(f(t)) = y/x So, differentiating, sec^2(f(t))*f'(t) = [x*y' - x'*y]/x^2 When the boy is halfway to the center, we'll figure out x', y', x, and y. I'll outline this. If you solve y-0 = ((R-3t)/R)(x+R) and x^2+y^2 = R^2 simultaneously, when R-3t = R/2, you'll find x = 3R/5 and y = 4R/5 Next, differentiate the equations y-0 = ((R-3t)/R)(x+R) and x^2+y^2 = R^2 Replace x with 3R/5 and y with 4R/5, then solve for x' and y'. I think you'll get x' = 21/5 and y' = -21/10 Put everything together, and I think you'll find that R*f'(t) = -231/50 m/s. - Doctor Jerry, The Math Forum http://mathforum.org/dr.math/ |
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