# Related rates & derivatives/integration

A ladder 10 feet long leans against a vertical wall. If the bottom if the ladder slides away from the base of the wall at a speed of 2ft/sec, how fast is the angle between the ladder and the wall changing when the bottom of the ladder is 6ft away from the base of the wall?

Draw a labelled picture. Let $P$ be the point where the ladder meets the wall. Let $C$ be the point where the wall meets the ground, and let $F$ be the point where the ladder meets the ground. The points $P$ and $F$ are changing.

At any time $t$, let $x=x(t)$ be the length of $CF$. So $x$ is is the distance from the foot of the ladder to the wall. We are told that $\frac{dx}{dt}=2$.

Let $\theta=\angle CPF$. Then $\theta$ is the angle between the ladder and the wall. We are being asked how fast the angle is changing, so we are being asked about $\frac{d\theta}{dt}$. So we are told one "rate" (the rate of change of $x$) and are asked about another rate (the rate of change of $\theta$).

To solve the problem, we need a relationship between $x$ and $\theta$. This is easy. The geometry tells us that $$x=10\cos\theta.$$ We could solve for $\theta$ in terms of $x$. I prefer to differentiate immediately. Using the Chain Rule, we find that $$\frac{dx}{dt}=-10\sin \theta \frac{d\theta}{dt}.$$ Now at the instant when $x=6$, we have, by the Pythagorean Theorem, $CP=8$, and therefore $\sin\theta=\frac{8}{10}$. So at that instant, $$2=-(10)\left(\frac{8}{10}\right)\frac{d\theta}{dt},$$ and therefore $-\frac{d\theta}{dt}=\frac{2}{8}$ (radians per second).

The angle is decreasing at $1/4$ of a radian per second.

• Shouldn't it be $10cos\theta$ when you differentiate? Commented Jul 16, 2014 at 17:29
• @IrresponsibleNewb: Thank you for spotting it! Fixed, I think. Commented Jul 16, 2014 at 17:45
• No problem, perhaps you could reward me by looking at my new question on related rates! :p Commented Jul 16, 2014 at 17:48