The foot of a ladder 50 ft. long rests on horizontal ground, and the top of the ladder rests against the side of a pyramid which makes an angle of 120° with the ground. If the foot of the ladder is drawn directly away from the base of the pyramid at the uniform rate of 2 ft. per second, how fast will the top of the ladder slide down the side of the pyramid ?
I tried the following: First found the position of B as $$ \left(-\frac{y}{\sqrt{3}},\ y\right) $$
Now the distance OB is $$\frac{2y}{\sqrt{3}}$$. Differentiating with respect to time we get, $$\frac{d\left|BO\right|}{dt}=\frac{2}{\sqrt{3}}\cdot\frac{dy}{dt}$$
Now using the fact that the length of the ladder remains constant, $$ \left(x+\frac{y}{\sqrt{3}}\right)^{2}+y^{2}=2500$$
Meaning, $$x^{2}+\frac{4y^{2}}{3}+\frac{2xy}{\sqrt{3}}=2500$$
After differentiation,
$$\left(2x+\frac{2y}{\sqrt{3}}\right)\cdot\frac{dx}{dt}+\left(\frac{2x}{\sqrt{3}}+\frac{8y}{3}\right)\cdot\frac{dy}{dt}=0$$
Giving,
$$\frac{dy}{dt}=\ \frac{-\left(x+\frac{y}{\sqrt{3}}\right)\cdot\frac{dx}{dt}}{\left(\frac{4y}{3}+\frac{x}{\sqrt{3}}\right)}$$
Now as $$\frac{dx}{dt\ }\ =\ 2\ \frac{ft}{s}$$,
$$ \frac{dy}{dt}=\ \frac{-2\left(x+\frac{y}{\sqrt{3}}\right)}{\left(\frac{4y}{3}+\frac{x}{\sqrt{3}}\right)}$$.
From here we can get rate of sliding of the top of the ladder. But my answer seems to be wrong.
Here the side BO is the side of the pyramid inclined to OA assumed to be the x axis, and OK is assumed to be the y axis.
Am I missing something here?
The answer given is : $$\frac{2x+4y}{2x+y} ft. per second$$ , where x is the distance of the top of ladder, and y is the distance of foot of the ladder, from the base of the pyramid.
PS-- The variables in the answer given are interchanged w.r.t the figure I have drawn.
I got the answer. The question really is ambiguous,
as we got this: $$\frac{dy}{dt}=\ \frac{-\left(x+\frac{y}{\sqrt{3}}\right)\cdot\frac{dx}{dt}}{\left(\frac{4y}{3}+\frac{x}{\sqrt{3}}\right)}$$
I figured by the comment posted by Dan(thanks! Dan) we are being asked about the distance to the top of the ladder from the base of the pyramid, meaning we are to find in terms of distance along the line OB.
So let's convert all the y's in the above equation to OB, we get,
$$ y\ =\ \frac{\sqrt{3}}{2}\cdot OB$$
meaning,
$$\frac{dy}{dt}=\ \frac{-\left(x+\frac{OB}{2}\right)\cdot\frac{dx}{dt}}{\left(\frac{2OB}{\sqrt{3}}+\frac{x}{\sqrt{3}}\right)}$$
but now as, $$\frac{d\left|OB\right|}{dt}\ =\ \frac{2}{\sqrt{3}}\cdot\frac{dy}{dt}$$
we have ,
$$\frac{d\left|OB\right|}{dt}\ =\ \frac{2}{\sqrt{3}}\cdot\ \frac{-\left(2x+OB\right)\cdot\frac{dx}{dt}}{\left(2OB+x\right)\cdot\frac{2}{\sqrt{3}}}$$
finally we put $$\frac{dx}{dt}\ =\ 2\ $$
and we replace OB by variable y for neatness(and to get the form given in the answer...):
$$\frac{d\left|OB\right|}{dt}\ =\ \ \left|\frac{-\left(2x+y\right)\cdot2}{\left(2y+x\right)}\right|$$
But as the variables in the answer are so weirdly interchanged, we also do the same to get the final answer as:
$$\frac{d\left|OB\right|}{dt}\ =\ \ \frac{4y+2x}{2x+y}$$