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An extensible ladder rests vertically against the side of a 10 foot wall. You start dragging the bottom of the ladder along the floor away from the wall at a rate of 0.5 ft/sec. At the moment when the wall makes a 60 degree angle with the ladder, how fast is the distance between bottom and top of the ladder changing?
I'm unsure how to make my equation. I think I'm supposed to start with a trigonometry function, but I'm not sure how to start.
First, make a diagram of the situation:
A
|\
| \
| \
| \
O----B
Here $OA=10$ and $OB=k$ is increasing at 0.5 per second. The length of the ladder $AB=L$ in terms of $k$ is, by the Pythagorean theorem: $$L=\sqrt{k^2+100}$$ Taking derivatives: $$\frac{dL}{dk}=\frac{2k}{2\sqrt{k^2+100}}=\frac k{\sqrt{k^2+100}}$$ We also know that $$\frac{dk}{dt}=0.5$$ Multiplying the two derivatives together: $$\frac{dL}{dt}=\frac{dL}{dk}\cdot\frac{dk}{dt}=\frac k{2\sqrt{k^2+100}}$$ When the angle betwen ladder and wall is 60 degrees, $k=10\tan60^\circ=10\sqrt3$ and hence $$\frac{dL}{dt}=\frac{10\sqrt3}{2\sqrt{(10\sqrt3)^2+100}}=\frac{10\sqrt3}{2\cdot20}=\frac{\sqrt3}4$$ Hence the distance between the ladder's ends is increasing at around 0.433 feet per second at the given instant.
