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 person’s 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. Using similar right - triangles, we have $\frac{8}{x + y}=\frac{2}{y}$. Cross - multiplying gives $8y=2(x + y)$.

Step2: Simplify the equation

Expand the right - hand side: $8y = 2x+2y$. Subtract $2y$ from both sides: $6y = 2x$, so $y=\frac{1}{3}x$.

Step3: Differentiate with respect to time

Differentiate both sides of $y=\frac{1}{3}x$ with respect to time $t$. We know that $\frac{dx}{dt}=0.5$ m/s. Using the chain rule, $\frac{dy}{dt}=\frac{1}{3}\frac{dx}{dt}$.

Step4: Calculate the rate of change of the shadow length

Substitute $\frac{dx}{dt}=0.5$ m/s into the equation: $\frac{dy}{dt}=\frac{1}{3}\times0.5=\frac{1}{6}$ m/s.

Answer:

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