Here's the question: The base x of a right triangle increases at a rate of 5cm/sec, while the height remains constant at h=10. How fast is the angle θ changing when x=10cm?
What I've done so far is the following: I know that $\frac{dx}{dt}=5$ and that I'm supposed to find dθ when x=10.
- From the triangle drawn, I can determine that $tanθ=\frac{10}{x}$
- The next thing to do would be take the derivative of both sides, and this is where I mess up. The left side should obviously be $sec^2(θ)\frac{dθ}{dt}$, and I know I should substitute x=10 and $\frac{dx}{dt}=5$, but I don't know what to do with the h or how to take the derivative of $10x^{-1}$ for this and solve it.