Question

A kite 100 ft above the ground moves horizontally at a speed of10 ft/s. At what rate is the angle (inradians) between the string and the horizontal decreasing when 200ft of string have been let out? (Enter your answer as afraction.)

Answer 1

Answer:

$\displaystyle \frac{d\theta}{dt}=-\frac{1}{40}\text{ rad/s}$

Step-by-step explanation:

We can model the situation with the diagram below.

The relationship between the angle between the string and the horizontal can be modeled by:

$\displaystyle \tan\theta =\frac{100}{x}=100x^{-1}$

Differentiate the equation with respect to time t:

$\displaystyle \sec^2(\theta)\frac{d\theta}{dt}=-100x^{-2}\frac{dx}{dt}$

When 200 ft of string have been left out, a = 200.

By the Pythagorean Theorem:

$x=\displaystyle \sqrt{200^2-100^2}=\sqrt{30000}=100\sqrt3$

Then it follows that:

$\displaystyle \sec^2\theta =\left(\frac{200}{100\sqrt{3}}\right)^2=\frac{4}{3}$

And since the kite moves horizontally at a speed of 10 ft/s, dx/dt = 10.

Substitute:

$\displaystyle \left(\frac{4}{3}\right)\frac{d\theta}{dt}=-100(100\sqrt{3})^{-2}(10)$

So:

$\displaystyle \frac{d\theta}{dt}=-\frac{100}{(100^2)(3)}(10)\left(\frac{3}{4}\right)=-\frac{1}{40}\text{ rad/s}$