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

a.
`(dx)/(dt)=20ft/s`

b.
`(d(x+L))/(dt)==25ft/s`

Step-by-step 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)=(24ft)/(L+x)=(6ft)/(x)`

Solve to find the rate relation

`x=(24)/(6)*L`

`x=4*L`

Now the rate of the change rate

`(dx)/(dt)=4*(dL)/(dt)`

`(dx)/(dt)=4*5ft/s=20ft/s`

Finally the rate of her shadow moving

`(d(x+L))/(dt)=(dx)/(dt)+(dL)/(dt)`

`(d(x+L))/(dt)=20ft/s+5ft/s=25ft/s`