Answer to Question #265670 in Calculus for ozan.ture

Question

 A spotlight on the ground is shining on a wall 24m

 away. If a woman 2m

 tall walks from the spotlight toward the building at a speed of 1.2m/s,

 how fast is the length of her shadow on the building decreasing when she is 2m

 from the building?

Answer 1

Solution:

Given $\dfrac{d x}{d t}=1.2 \ m/s$

from graph,

$\begin{aligned} &\frac{y}{24}=\frac{2}{x} \\ &\Rightarrow y=\frac{48}{x} \end{aligned}$

differentative w.r.to $\mathrm{t}$ ,

$\begin{aligned} &\Rightarrow \frac{d y}{d t}=48 \frac{d}{d t}\left(\frac{1}{x}\right) \\ &\Rightarrow \frac{d y}{d t}=48\left[-\frac{1}{x^{2}}\right] \frac{d x}{d t} \\ &\Rightarrow \frac{d y}{d t}=48\left[-\frac{1}{(24-2)^{2}}\right](1.2) \ [Using \ (i)] \\& \Rightarrow \frac{d y}{d t}=-0.119 \end{aligned}$

Thus, the shadow on building is decreasing at a rate of 0.119 m/s.