Answer in Calculus for ozan.ture #265670

 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?

Solution:

Given “dfrac{d x}{d t}=1.2 m/s”

from graph,

“begin{aligned}nn&frac{y}{24}=frac{2}{x} \nn&Rightarrow y=frac{48}{x}nnend{aligned}”

differentative w.r.to “mathrm{t}” ,

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

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