   Need help with a related rates calculus problem
  I need help with a calculus problem.  This is a homework question, and the answer is 7.1 inches per minute.  However, I do not understand how to arrive at this solution (we don't get graded for homework anyway, it's for our own benefit).  I'm not in university, only highschool (non AP) calculus, and to give you an idea of the level we have not yet started integrals/antiderivatives, but I know all of the calculus up until that point.  Here is the problem and my approach:"On a morning when the sun will pass directly overhead, the shadow of an 80 ft tall building on level ground is 60 ft long.  At the moment in question, the angle theta that the sun makes with the ground is increasing at a rate of .27 radians/min.  At what rate is the shadow decreasing?  Express your answer in inches per minute."Here is a picture (this is copied as closely as possible from the book, i.e I know the angle is in the correct place):www.thedevelopersalliance.com/~shadow12345/calculus.JPGFirst off, theta is equal to tan-1 (80/60) = .927 radians.I have tried two different approaches to solving this problem.  Here is my first setup (and I forgot to mention, my calculator is in fact in radians, as it should be):I labelled the shadow length as S.S = 80 / tan(.927)ds/dt' = (-80 * sec(.927)^2 * .27) / (tan(.927)^2) = -33.764 ft/min = -405.16 inches/minI got my approach from using the quotient rule and the chain rule and the derivative of tangent.My second approach was this (remember S was originally 60 in the instant given in the problem):tan(.927) = 80 / S(sec(.927)^2 * .27) = (-80 / S^2) * ds/dtds/dt = (sec(.927)^2 * .27) * (60^2 / -80) = -33.72ft/min = -404.68 inches / minI have absolutely no clue how they get 7.1 inches/minute.[edited by - Shadow12345 on February 9, 2004 8:37:17 PM][edited by - Shadow12345 on February 10, 2004 12:21:18 PM]
  that looks simple... at time t0   , you have [y0 = 80, x0 = 60, a0 = atan(y0 / x0)]at time t0+dt, you have [y1 = 80, x1 = ??, a1 = a0 + 0.27/60 * dt]dt = 1.0f;a0 = 0.927295f;a1 = 0.931795f;x1 = 80 / tan(a1) = 59.4394f;(x1 - x0) / dt = 0.560612f feet = 6.71 inches / sec = 402 inches/minI find the same as you, which is slightly less than 7.1 inch / mindecreasing dt brings the same result. 6.7.... ;quote:Original post by shadow12345  I need help with a calculus problem.  This is a homework question, and the answer is 7.1 inches per second....Here is the problem and my approach:Hey shadow12345, THANK YOU for following my policy on homework by disclosing this information and your detailed approach! This helps me to maintain the forum!Now, that said, I concur with oliii. The book answer appears to be wrong. These things happen sometimes. quote:Original post by shadow12345  First off, theta is equal to tan-1 (80/60) = .927 radians.Your figure looks more like an alpha than a theta? Anyway, doesn''t matter.quote:Original post by shadow12345  I labelled the shadow length as S.S = 80 / tan(.927)ds/dt'' = (-80 * sec(.927)^2 * .27) / (tan(.927)^2) = -33.764 ft/min = -405.16 inches/minThis is kind of a bizarre derivation, but the ds/dt equation is correct. If you substitute exactly 80/60 for the tan in the denominator, and use exactly the result of atan(80/60) instead of ".927" when calculating sec, then your result will be -33.75 feet/minute exactly, which is exactly -6.75 feet/second.quote:Original post by shadow12345  My second approach was this (remember S was originally 60 in the instant given in the problem):tan(.927) = 80 / S(sec(.927)^2 * .27) = (-80 / S^2) * ds/dtds/dt = (sec(.927)^2 * .27) * (60^2 / -80) = -33.72ft/min = -404.68 inches / minAgain, there''s some weirdness about your notation in your derivation, e.g., the way you wrote the right-hand-side looks like ds/dt is in the numerator. But the result is fine. The ONLY reason you get a different answer from the first one is due to different rounding error.The way I did this was to solve tan(theta) = 80/s for s, so:s = 80/tan = 80 * cot(theta)Then, take the derivative of that. I just find the derivative is less "dirty" if you are able to solve for the one unknown first.   Graham RhodesSenior ScientistApplied Research Associates, Inc. ;  I think I might have confused you guys a little bit, because I realized something I did wrong.  The final answer is supposed to be 7.1 inches per minute, but in the first paragraph I wrote inches per second (I do not know why I did this).  First, we know that book answer is wrong because it isn't negative!  The shadow length is decreasing as the angle increases.-33.75 feet/minute seems to be the collective answer (grhodes you said -33.75 feet/minute exactly, and oliii you said 402inches/minute which is pretty much equivalent, and I came up with something similar).  I think that the book meant to put inches per second, not inches per minute, as the final answer.  Subsequently if the book answer was requested as inches/second it would be 6.75 (which is what you guys also collectively suggested) and the only discrepancies would be because I didn't round correctly in all steps.Okay, whew, thank you guys a lot.  I guess after all I did have a pretty good idea of how to set it up, which is significant because we get graded mostly on the setup, and only about 10-20% on actually getting the right answer.quote:Hey shadow12345, THANK YOU for following my policy on homework by disclosing this information and your detailed approach! This helps me to maintain the forum!Yes you are very welcome and thanks for helping.  [edited by - Shadow12345 on February 10, 2004 12:34:01 PM]