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Having trouble with Related Rates, Please help

  1. Mar 2, 2013 #1
    A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.0 ft/s, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 8 ft from the wall?

    So, I'm having a difficult time figuring out related rates. I have grasped derivatives just fine, but developing the diagram and labeling seem to be my problems.
    I’ve set up my graph like this:
    photo(5) by d.smith292, on Flickr

    θ is my changing rate in this problem. I've set my equation up like this :

    x = 8ft
    x' = 1ft/sec

    cosθ = x/10

    When I differentiate the problem I get

    -sinθ = 1/10(x')

    When I input this into my calculator I get some crazy long decimal answer, like -.1001674212 Where am I going wrong. I've probably got the entire thing wrong. I'm having trouble with this section. I appreciate the help.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Mar 2, 2013 #2


    User Avatar
    Homework Helper

    On the left hand side you should have
    \frac{\mathrm{d}}{\mathrm{d}t} \cos\theta
    = (- \sin\theta) \frac{\mathrm{d}\theta}{\mathrm{d}t}
  4. Mar 2, 2013 #3


    Staff: Mentor

    No, x is variable that depends on (is a function of) time. At a particular moment x is 8, but at other times it has different values.
    Not quite.
    You are differentiating both sides with respect to t, so you should get
    -sin(θ) dθ/dt = (1/10) dx/dt
  5. Mar 3, 2013 #4
    OK, so I see that I wasn't taking the derivative of θ with respect to time (t). Here is where I'm at now and stuck again.

    Cos θ = x/10 → (Now I take the derivative with respect to time)

    (-sin θ)dθ/dt = (1/10)dx/dt

    Now since the rate of change for x = 1 ft/sec I get

    (-sin θ)dθ/dt = 1/10

    From here I don't know if θ should equal 8/10 or 6/10. Either way I continue to get some crazy answer.

    And am I even on the right track here?
  6. Mar 3, 2013 #5


    User Avatar
    Homework Helper

    It's neither. It should be
    [itex]\cos \theta = \frac{8}{10}[/itex].

    Solve for θ, and then plug that, and dx/dt = 1, into
    [itex]-\sin \theta \frac{d\theta}{dt} = \frac{1}{10} \frac{dx}{dt}[/itex]

    and solve for dθ/dt.
  7. Mar 5, 2013 #6
    So, I had everything right up until I solved for theta. I've rewritten the problem with detailed steps from the beginning so that I can remember this for future reference. Thank you.
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