1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Related Rates Problem Solving

  1. Aug 25, 2009 #1
    1. The problem statement, all variables and given/known data
    A light is hung 15ft above a straight horizontal path. If a man 6 ft tall is walking away from the light at the rate of 5ft/sec. How fast is his shadow lengthening? And at what rate is the tip of the man's shadow moving?

    y = 6? or 15?
    dx/dt = 6


    2. Relevant equations
    z^2 = x^2 + y^2 ?


    3. The attempt at a solution
    I tried looking for angle of it or whatever that came up on my mind. I really don't know where to start after getting the given.

    My professor didn't show a same problem just like the above. It just shows us some spherical, cone, two cars moving from different location, a helicopter above and a satellite dish below that is rotating. But this one is so different.

    It's not an assignment by the way. I'm just practicing and came up with this problem on the book. The answer at the back for the rate of the tip of the man's shadow moving is 25/3 ft/sec. How the heck did he get that?
     
  2. jcsd
  3. Aug 25, 2009 #2

    Mark44

    Staff: Mentor

    Draw a picture if you haven't already done so.
    Let x(t) = the man's horizontal distance from the light at time t. His height is 6', so don't use a variable for this constant. Let s(t) = the length of the man's shadow at time t.
    You should have right two triangles: a small one whose altitude is 6' and whose base is s(t), and a large one whose altitude is 15' and whose base is x(t) + s(t). These triangles are similar, so you should be able to get an equation for s(t).

    After that, get a new equation with the derivatives (the rates) and substitute the given rates to find an expression for s'(t).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Related Rates Problem Solving
Loading...