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Natali [406]
3 years ago
11

A 66​-foot-tall woman walks at 55 ​ft/s toward a street light that is 2424 ft above the ground. What is the rate of change of th

e length of her shadow when she is 19 ft19 ft from the street​ light? At what rate is the tip of her shadow​ moving? Let L be the length of the​ woman's shadow and let x be the​ woman's distance from the street light. Write an equation that relates L and x.
Physics
1 answer:
I am Lyosha [343]3 years ago
5 0

Answer:

a. \frac{dx}{dt}=20ft/s

b. \frac{d(x+L)}{dt}==25ft/s

Explanation:

Using the triangle theorem both triangle the woman makes between the light so the rate of change of length can use geometry first

tan(\beta)=\frac{24ft}{L+x}=\frac{6ft}{x}

Solve to find the rate relation

x=\frac{24}{6}*L

x=4*L

Now the rate of the change rate

\frac{dx}{dt}=4*\frac{dL}{dt}

\frac{dx}{dt}=4*5ft/s=20ft/s

Finally the rate of her shadow moving

\frac{d(x+L)}{dt}=\frac{dx}{dt}+\frac{dL}{dt}

\frac{d(x+L)}{dt}=20ft/s+5ft/s=25ft/s

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