Answer to Question #195395 in Calculus for Angela Artita

Question #195395

A woman standing on a cliff is watching a motorboat through a telescope as the boat approaches the shoreline directly below her. If the telescope is 25 meters above the water level and if the boat is approaching the cliff at 20 m/s, at what rate is the acute angle made by the telescope with the vertical changing when the boat is 250 meters from the shore? 


with a full solution please and please include the given.


1
Expert's answer
2021-05-19T18:22:21-0400



The picture is a geometric depiction of the given information.

We introduce variables "x" and "\\theta" as shown.

We are given that "dx\/dt=-20 \\ m\/s." The sign is negative because "x" is decreasing with time.

From right triangle


"\\tan \\theta=\\dfrac{x}{25}"

Differentiate both sides with respect to "t"


"\\dfrac{d}{dt}\\big(\\tan \\theta\\big)=\\dfrac{d}{dt}\\big(\\dfrac{x}{25}\\big)"


Use the Chain Rule


"\\dfrac{1}{\\cos ^2\\theta }\\cdot\\dfrac{d \\theta}{dt}=\\dfrac{1}{25}\\cdot\\dfrac{dx}{dt}"

"(1+\\tan ^2 \\theta)\\cdot\\dfrac{d \\theta}{dt}=\\dfrac{1}{25}\\cdot\\dfrac{dx}{dt}"

Substitute

"\\bigg(1+\\big(\\dfrac{x}{25}\\big)^2\\bigg)\\cdot\\dfrac{d \\theta}{dt}=\\dfrac{1}{25}\\cdot\\dfrac{dx}{dt}"

Solve for "\\dfrac{d \\theta}{dt}"


"(x^2+625)\\cdot\\dfrac{d \\theta}{dt}=25\\cdot\\dfrac{dx}{dt}"


"\\dfrac{d \\theta}{dt}=\\dfrac{25}{x^2+625}\\cdot\\dfrac{d x}{dt}"

When "x=250\\ m"


"\\dfrac{d \\theta}{dt}=\\dfrac{25}{(250)^2+625}\\cdot(-20\\ m\/s)"

"\\dfrac{d \\theta}{dt}=-\\dfrac{4}{505} \\ rad\/s"

"\\dfrac{d \\theta}{dt}=-0.008 \\ rad\/s"


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