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A drone is flying at a level altitude
of 20 meters straight towards an
observer on the ground. If the
drone is flying at 8 meters per
second, what is the rate of change
between it and the observer when
it is 25 meters away from the
doserver ? Remember the sides of a right triangle are related by a ² + b² = c²

Respuesta :

[tex]\begin{gathered} h^2=\text{ \lparen20\rparen}^2\text{ + \lparen25\rparen}^2 \\ h\text{ = }\sqrt[2]{400+625} \\ h=\sqrt[2]{1025} \\ h\text{ }\approx\text{ 32m} \end{gathered}[/tex]

Now we have to find the derivative:

[tex]\begin{gathered} \left(2h\right)dh/dt\frac{}{}\text{ = 2x \lparen}\frac{dx}{dt}) \\ \frac{dh}{dt}\text{ = 2 \lparen8m/s\rparen / 2 \lparen32\rparen} \\ \frac{dh}{dt}\text{ = }\frac{16}{64} \\ \frac{dh}{dt}\text{ = }\frac{1}{4} \\ ^{=\text{ 0.25m/s}} \end{gathered}[/tex]

The rate of change is 0.25m/s

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