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A tourist at scenic Point Loma, California uses a telescope to track a boat approaching the shore. If the boat moves at a rate of5 meters per second, and the lens of the telescope is 30 meters above water level, how fast is the angle of depression of thetelescope (0) changing when the boat is 200 meters from shore? Round any intermediate calculations to no less than sixdecimal places, and round your final answer to four decimal places.

A Tourist At Scenic Point Loma California Uses A Telescope To Track A Boat Approaching The Shore If The Boat Moves At A Rate Of5 Meters Per Second And The Lens class=

Sagot :

Lest first we hte sine theorem to relate the given measures:

[tex]\frac{\sin (\theta)}{30}=\frac{\sin(90^{\circ})}{\sqrt[]{x^2+30^2}}[/tex]

x represents the distance from the boat to the shore.

[tex]\frac{\sin (\theta)}{30}=\frac{1}{\sqrt[]{x^2+30^2}}[/tex][tex]\frac{\sin(\theta)}{1}=\frac{30}{\sqrt[]{x^2+30^2}}[/tex][tex]\sin (\theta)=\frac{30}{\sqrt[]{x^2+30^2}}[/tex][tex]\theta=\sin ^{-1}(\frac{30}{\sqrt[]{x^2+30^2}})[/tex]

Then we must calculate the derivative in order to know the rate of change at a certain point.

[tex]\frac{d}{dx}(\sin ^{-1}(\frac{30}{\sqrt[]{x^2+30^2}}))=-\frac{30x}{\sqrt[]{\frac{x^2}{x^2+900}}\cdot(x^2+900)^{\frac{3}{2}}}[/tex]

To find how fast is the angle of depression of the telescope is changing when the boat is 200 meters from shore, replace by 200 on the derivative:

[tex]-\frac{30\cdot200}{\sqrt[]{\frac{200^2}{200^2^{}+900}}\cdot(200^2+900)^{\frac{3}{2}}}=-0.0007\text{ rad/s}[/tex]

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View image TymierS768757
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