a 5.5 - foot - tall woman walks at 6 ft/s toward a street light that is 22 ft above the ground. what is the…

Explanation:

Step1: Set up similar - triangles

Let $x$ be the distance of the woman from the street - light and $y$ be the length of her shadow. By similar triangles, we have $\frac{5.5}{22}=\frac{y}{x + y}$. Cross - multiplying gives $5.5(x + y)=22y$, which simplifies to $5.5x+5.5y = 22y$, and then $5.5x=16.5y$, or $y=\frac{1}{3}x$.

Step2: Differentiate with respect to time

Differentiate both sides of $y=\frac{1}{3}x$ with respect to time $t$. Using the chain rule, $\frac{dy}{dt}=\frac{1}{3}\frac{dx}{dt}$. We know that $\frac{dx}{dt}=- 6$ ft/s (negative because $x$ is decreasing as the woman walks towards the street - light).

Step3: Find the rate of change of the shadow length

Substitute $\frac{dx}{dt}=-6$ into $\frac{dy}{dt}=\frac{1}{3}\frac{dx}{dt}$. So $\frac{dy}{dt}=\frac{1}{3}\times(-6)=- 2$ ft/s.

The tip of the shadow is at a distance $L=x + y$. Since $y=\frac{1}{3}x$, then $L=x+\frac{1}{3}x=\frac{4}{3}x$. Differentiating $L$ with respect to $t$ gives $\frac{dL}{dt}=\frac{4}{3}\frac{dx}{dt}$. Substituting $\frac{dx}{dt}=-6$ ft/s, we get $\frac{dL}{dt}=\frac{4}{3}\times(-6)=-8$ ft/s.

Answer:

The rate of change of the length of the woman's shadow is $2$ ft/s. The rate at which the tip of her shadow is moving is $8$ ft/s.