Question

7 a person of height 2 meters is walking away from an 8 meter tall street light at a speed of .5 meters per second. at what rate is the persons shadow increasing in length?

Explanation:

Step1: Set up similar - triangles

Let $x$ be the distance of the person from the street - light and $y$ be the length of the person's shadow. By similar triangles, we have $\frac{2}{8}=\frac{y}{x + y}$. Cross - multiplying gives $2(x + y)=8y$, which simplifies to $2x+2y = 8y$, and then $2x = 6y$, or $y=\frac{1}{3}x$.

Step2: Differentiate with respect to time

Differentiate both sides of the equation $y=\frac{1}{3}x$ with respect to time $t$. Using the chain rule, $\frac{dy}{dt}=\frac{1}{3}\frac{dx}{dt}$.

Step3: Substitute the given value

We know that $\frac{dx}{dt}=0.5$ m/s. Substituting this value into the equation $\frac{dy}{dt}=\frac{1}{3}\frac{dx}{dt}$, we get $\frac{dy}{dt}=\frac{1}{3}\times0.5=\frac{1}{6}$ m/s.

Answer:

$\frac{1}{6}$ m/s