Question

A 66​-foot-tall woman walks at 55 ​ft/s toward a street light that is 2424 ft above the ground. What is the rate of change of the length of her shadow when she is 19 ft19 ft from the street​ light? At what rate is the tip of her shadow​ moving? Let L be the length of the​ woman's shadow and let x be the​ woman's distance from the street light. Write an equation that relates L and x.

Answer 1

Answer:

a. $\frac{dx}{dt}=20ft/s$

b. $\frac{d(x+L)}{dt}==25ft/s$

Explanation:

Using the triangle theorem both triangle the woman makes between the light so the rate of change of length can use geometry first

$tan(\beta)=\frac{24ft}{L+x}=\frac{6ft}{x}$

Solve to find the rate relation

$x=\frac{24}{6}*L$

$x=4*L$

Now the rate of the change rate

$\frac{dx}{dt}=4*\frac{dL}{dt}$

$\frac{dx}{dt}=4*5ft/s=20ft/s$

Finally the rate of her shadow moving

$\frac{d(x+L)}{dt}=\frac{dx}{dt}+\frac{dL}{dt}$

$\frac{d(x+L)}{dt}=20ft/s+5ft/s=25ft/s$