A ladder 5 metres long is leaning against the side of a building.

A ladder 5 metres long is leaning against the side of a building. If the foot of the ladder is pulled away from the building at a constant rate of 1.5 metres per second, how fast is the angle formed by the ladder and the ground changing the moment when the top of the ladder is 3 metres above the ground?

I basically started with the equation $cos(p) = \frac{x}{5}$ and then took the derivative of this, subbed in the values of x, x', and z from the diagram, and my final answer was -5/6 rad/s

Is this approach correct? If so, is the answer correct? My friend got an entirely different answer but I am not sure what exactly he did and whether or not it is correct.

Thanks.

• You should find out what he did. It would help you either figure out your mistake or figure out his, which is helpful either way. – Teresa Lisbon Dec 1 '17 at 5:00
• Share with us how do you obtain $-\frac56$ so that I can point out your mistake if any? what do you get after you take the derivative? how do you substitute in those value? you are not showing us sufficient details to let us know your mistake if any. – Siong Thye Goh Dec 1 '17 at 5:57
• @Craig D Siong Thye Goh is right. Please edit your question to explain what exactly you did . – Abhinav Dhawan Dec 1 '17 at 6:37

Your method appears correct. We start from here $$\cos(p) = \frac{x}{5}$$ Then taking the derivative and using chain rule, we get $$-\sin(p)\frac{dp}{dt}=\frac{1}{5}\frac{dx}{dt}$$ Now plug in and solve to get $$\frac{dp}{dt} = -\frac{1}{5\sin(p)}\frac{dx}{dt} = -\frac{1}{5\frac{3}{5}}\frac{3}{2}=-\frac{1}{2}$$. You must have gone wrong somewhere to get $-\frac{5}{6}$. Hopefully, this lets you find your mistake.