Question

a street light is at the top of a 19 ft tall pole.
a woman 5.25 ft tall walks away from the pole with a speed of 7 ft/sec along a straight path.
how fast is the shadow lengthening when she is 50 ft from the base of the pole?
(round answers to 3 decimal places.)

how fast is the tip of her shadow moving when she is 50 ft from the base of the pole?
(round answers to 3 decimal places.)

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Explanation:

Step1: Define Variables

Let \( x \) be the distance of the woman from the pole, and \( s \) be the length of her shadow. Let \( L \) be the distance of the tip of the shadow from the pole. So \( L = x + s \).

Step2: Similar Triangles

The two triangles (pole - tip of shadow and woman - tip of shadow) are similar. So \(\frac{19}{L}=\frac{5.25}{s}\). Cross - multiplying gives \( 19s = 5.25L \). But \( L=x + s \), so \( 19s=5.25(x + s) \).
Expanding: \( 19s=5.25x+5.25s \).
Subtract \( 5.25s \) from both sides: \( 19s - 5.25s=5.25x \), \( 13.75s = 5.25x \), then \( s=\frac{5.25}{13.75}x=\frac{21}{55}x \).

Step3: Differentiate for Shadow Length Rate

Differentiate \( s=\frac{21}{55}x \) with respect to time \( t \). Using the chain rule, \(\frac{ds}{dt}=\frac{21}{55}\frac{dx}{dt}\). We know \(\frac{dx}{dt} = 7\) ft/sec. So \(\frac{ds}{dt}=\frac{21}{55}\times7=\frac{147}{55}\approx2.673\) ft/sec.

Step4: Differentiate for Tip of Shadow Rate

Differentiate \( L=x + s \) with respect to \( t \): \(\frac{dL}{dt}=\frac{dx}{dt}+\frac{ds}{dt}\). We can also use the similar triangles equation \(\frac{19}{L}=\frac{5.25}{s}\Rightarrow19s = 5.25L\). Differentiate both sides with respect to \( t \): \( 19\frac{ds}{dt}=5.25\frac{dL}{dt} \). But from \( L=x + s \), \(\frac{dL}{dt}=\frac{dx}{dt}+\frac{ds}{dt}\). Substitute \(\frac{ds}{dt}=\frac{5.25}{19}\frac{dL}{dt}\) into \(\frac{dL}{dt}=\frac{dx}{dt}+\frac{5.25}{19}\frac{dL}{dt}\).
\(\frac{dL}{dt}-\frac{5.25}{19}\frac{dL}{dt}=\frac{dx}{dt}\)
\(\frac{19 - 5.25}{19}\frac{dL}{dt}=\frac{dx}{dt}\)
\(\frac{13.75}{19}\frac{dL}{dt}=\frac{dx}{dt}\)
\(\frac{dL}{dt}=\frac{19}{13.75}\times7=\frac{133}{13.75}=\frac{532}{55}\approx9.673\) ft/sec. (Or using \(\frac{dL}{dt}=\frac{dx}{dt}+\frac{ds}{dt}=7 + \frac{147}{55}=\frac{385 + 147}{55}=\frac{532}{55}\approx9.673\))

Answer:

The rate at which the shadow is lengthening is \(\approx\boldsymbol{2.673}\) \(\frac{\text{ft}}{\text{sec}}\).
The rate at which the tip of the shadow is moving is \(\approx\boldsymbol{9.673}\) \(\frac{\text{ft}}{\text{sec}}\).