Answer to Question #265670 in Calculus for ozan.ture

Question #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?


1
Expert's answer
2021-11-16T11:23:27-0500

Solution:


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

from graph,

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

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

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

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


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