2. a man 1.8 m tall walks away from a lamppost whose light is 6 m above the ground. if he walks at a speed…

Explanation:

Step1: Set up similar - triangles

Let $x$ be the distance of the man from the lamppost and $y$ be the length of the shadow. By similar - triangles, we have $\frac{1.8}{6}=\frac{y}{x + y}$. Cross - multiplying gives $1.8(x + y)=6y$, which simplifies to $1.8x+1.8y = 6y$, and further to $1.8x = 4.2y$, or $y=\frac{1.8}{4.2}x=\frac{3}{7}x$.

Step2: Differentiate with respect to time

Differentiate both sides of the equation $y=\frac{3}{7}x$ with respect to time $t$. Using the chain rule, $\frac{dy}{dt}=\frac{3}{7}\frac{dx}{dt}$.

Step3: Substitute the given value of $\frac{dx}{dt}$

We know that $\frac{dx}{dt}=2$ m/s. Substituting this value into the equation $\frac{dy}{dt}=\frac{3}{7}\frac{dx}{dt}$, we get $\frac{dy}{dt}=\frac{3}{7}\times2=\frac{6}{7}\approx0.857$ m/s. The distance of the man from the lamppost ($x = 10$ m) is not needed to find the rate at which the shadow is growing.

Answer:

$\frac{6}{7}$ m/s