A ladder 5.1m long is placed on a level ground against a vertical wall. The foot of the ladder is pushed towards the wall at 1.5m/s. At what rate is the top of the ladder rising when its bottom is 2.4m from the wall?
We have a right angled triangle, with sides x,y, and r.
r is the hypotenuse, with a length of 5.1
x is given as 2.4
We solve for y and find it is 4.5
$\displaystyle x^2 + y^2 = r^2$
Differentiate.
$\displaystyle 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$ (Because r is a constant.)
$\displaystyle 2(2.4)(1.5) + 2(4.5) \left( \frac{dy}{dt} \right) = 0$
Solve for $\displaystyle \frac{dy}{dt}$