Question

a 5 - foot - tall woman walks at 8 ft/s toward a street light that is 30 ft above the ground. what is the rate of change of the length what rate is the tip of her shadow moving?
the rate of change of the length of the womans shadow is $-\frac{8}{5}$ ft/s when she is 18 ft from the street light.
(type an exact answer in simplified form.)
the tip of her shadow moves at a rate of $\square$ when she is 18 ft from the street light.
(type an exact answer in simplified form.)

Explanation:

Step1: Set up the relationship

Let \(x\) be the distance of the woman from the street - light and \(y\) be the length of her shadow.
By similar triangles, \(\frac{5}{30}=\frac{y}{x + y}\).
Cross - multiply: \(5(x + y)=30y\).
Expand: \(5x+5y = 30y\).
Rearrange: \(5x=25y\), so \(y=\frac{1}{5}x\).

Step2: Differentiate with respect to time

We know that \(\frac{dx}{dt}=- 8\) (negative because \(x\) is decreasing).
Differentiate \(y=\frac{1}{5}x\) with respect to \(t\) using the chain rule \(\frac{dy}{dt}=\frac{1}{5}\frac{dx}{dt}\).

Step3: Find the rate of the tip of the shadow

Let \(L=x + y\) be the distance from the base of the street - light to the tip of the shadow.
Since \(y=\frac{1}{5}x\), then \(L=x+\frac{1}{5}x=\frac{6}{5}x\).
Differentiate \(L\) with respect to \(t\): \(\frac{dL}{dt}=\frac{6}{5}\frac{dx}{dt}\).

Step4: Substitute the value of \(\frac{dx}{dt}\)

Substitute \(\frac{dx}{dt}=-8\) into \(\frac{dL}{dt}=\frac{6}{5}\frac{dx}{dt}\).
\(\frac{dL}{dt}=\frac{6}{5}\times(-8)=-\frac{48}{5}\)

Answer:

The tip of her shadow moves at a rate of \(-\frac{48}{5}\text{ ft/s}\)