Question: A $25$ foot ladder is resting against a wall so that the bottom of the ladder is $7$ feet from the wall. The bottom of the ladder starts slipping away from the wall at a rate of $1$ foot per second. How many feet per second is the top of the ladder sliding down the wall when it is $15$ feet above the ground?

I got $2w\frac{dw}{dt} + 2h\frac{dh}{dt}=0.$ I know $w$ and $h$ and the derivative of $w$ and that’s about it. Solving for $\frac{dh}{dt}$ gave me $-\frac{3}{4}$, although this is incorrect. What did I do wrong?

If you do decide to solve it, please state how you got the answer as well! Thank you!

  • $\begingroup$ There are several related questions in the handy list at right that are effectively duplicates of your question. Have a look at them to see how to solve this problem. $\endgroup$ – amd Aug 31 '18 at 1:03
  • $\begingroup$ the answer is $\frac{-4}{3}$ $\endgroup$ – Anonymous Aug 31 '18 at 1:20

From $2w\frac{dw}{dt} + 2h\frac{dh}{dt} =0$,

$\begin{array}\\ \frac{dh}{dt} &=-\dfrac{w\frac{dw}{dt}}{h}\\ &=-\dfrac{\sqrt{25^2-15^2}}{15}\\ &=-\dfrac{5\sqrt{5^2-3^2}}{15}\\ &=-\dfrac{4}{3}\\ \end{array} $

  • $\begingroup$ unfortunately this is still incorrect... i think you mean +4/3 ? $\endgroup$ – mathperson1234 Aug 31 '18 at 0:28
  • $\begingroup$ Wouldn't the phrasing of the problem be asking for a positive number? $\endgroup$ – random Aug 31 '18 at 0:29
  • 1
    $\begingroup$ That's a directed velocity - to the left for negative. $\endgroup$ – marty cohen Aug 31 '18 at 0:31
  • $\begingroup$ hm.... i'm not quite sure that's what they mean.. after all, the answer is still incorrect... $\endgroup$ – mathperson1234 Aug 31 '18 at 0:34
  • $\begingroup$ @mathperson1234: Are you saying both $4/3$ and $-4/3$ are incorrect? $\endgroup$ – Clayton Aug 31 '18 at 0:36

assuming $v_x$ constant:

$$ x = l\cos\theta \rightarrow \dot{x} = -l(\sin\theta)\dot{\theta} \quad ...(i)\\ y = l \sin\theta \rightarrow \dot{y} = (l\cos\theta) \dot{\theta} = x \dot{\theta} ...(ii) $$ $$ (i) / (ii) \Rightarrow \dot{y} = -\frac{x}{l}\left(\frac{l}{\sqrt{l^2-x^2}}\right)\dot{x} = \frac{-\sqrt{25^2-15^2}}{15}\cdot 1= -\frac{20}{15}=-\frac{4}{3} $$


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