# Calculus - Ladder Optimization Velocity

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!

• 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. – amd Aug 31 '18 at 1:03
• the answer is $\frac{-4}{3}$ – Anonymous Aug 31 '18 at 1:20

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

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

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

assuming $v_x$ constant:

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