You need to find \(\displaystyle \frac{d\theta}{dt}\)
You know it is 100ft off the ground and moving horizontally with a velocity 8ft/sec (lets say eastward). You also know you want to find \(\displaystyle \frac{d\theta}{dt}\) when the string is 200ft long. It should be obvious that we have a triangle here.
At any given time, one leg of the triangle is 100ft (off the ground). The other leg, the easward component, is variable, and is a function of the length of the string.
Code:
/ |
/ |
s / | 100ft
/ |
/theta____|
x
Lets call the bottom leg of the triangle x. Well, \(\displaystyle x=\sqrt{s^2-100^2}\)
We know we want to get a relation with \(\displaystyle \theta\), s, 100 and x. Lets use the tangent function:
\(\displaystyle tan\theta=\frac{100}{x}\)
Take derivative implicitly:
\(\displaystyle sec^2\theta\frac{d\theta}{dt}=\frac{-100}{x^2}\frac{dx}{dt}\)
Substitute back in for x and \(\displaystyle \frac{dx}{dt}\):
\(\displaystyle sec^2\theta\frac{d\theta}{dt}=\frac{100}{s^2-100^2}(8\frac{ft}{s})\)
edit: To get rid of the secant, use the triangle again.
Now we have an equation with theta and s. Can you take it from here?
