Single Variable Calculus, Chapter 3, 3.8, Section 3.8, Problem 16

At what rate is the length of his shadow on the building decreasing when he is 4m from the building?

Illustration







Given:

$\qquad$ height of the man = $2m$

$\qquad$ distance of the spotlight to the wall = $12m$

$\qquad$ horizontal velocity of the man = $1.6 m/s$

Required: the rate how fast is the length of the man's shadow on the building decreasing when he is $4m$ from the building.

Solution:

By applying similar triangles we have,


$\begin{equation} \begin{aligned} \frac{y}{12} =& \frac{2}{12 - x} \\ \\ y =& \frac{24}{12 - x} = 24(12 - x)^{-1} \end{aligned} \end{equation}$


By taking the derivative with respect to time,


$\begin{equation} \begin{aligned} \frac{dy}{dt} =& \frac{dy}{dx} \left( \frac{dx}{dt} \right) = 24 \cdot \frac{dy}{dx} (12 - x)^{-1} \frac{dx}{dt} \\ \\ \frac{dy}{dt} =& 24(-1)(12 - x)^{-2} \cdot (-1) \frac{dx}{dt} \\ \\ \frac{dy}{dt} =& 24(12 - x)^{-2} \frac{dx}{dt} \end{aligned} \end{equation}$


We know that $\displaystyle \frac{dx}{dt} = 1.6$ and $x = 4m$ so


$\begin{equation} \begin{aligned} \frac{dy}{dt} =& 24(12 - 4)^{-2} (1.6) \\ \\ \frac{dy}{dt} =& 0.6 m/s \end{aligned} \end{equation}$


The length of the shadow is decreasing at a rate of $0.6 m/s$ when the man is $4m$ from the building.