A street light is mounted at the top of a 15 foot-tall pole. A man 6 ft tall walks away from the pole with the speed of 5 ft/s along a straight path. How fast is the tip of his shadow moving when he is 40 ft from the pole?

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A street light is mounted at the top of a 15 foot-tall pole. A man 6 ft tall walks away from the pole with the speed of 5 ft/s along a straight path. How fast is the tip of his shadow moving when he is 40 ft from the pole?

OCW Scholar - Single Variable Calculus
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I'll say that x is the distance from the pole and s is the distance from the man to the tip of his shadow. The actual location of the shadow with respect to the pole would be x+s. The speed of the man would be dx/dt=5. |dw:1353905474708:dw| Using similar triangles...\[\frac{15}{x+s}=\frac{6}{s}\]\[15s=6(x+s)\]\[15s=6x+6s\]\[9s=6x\]\[s=\frac{6}{9}x=\frac{2}{3}x\]Differentiate with respect to time in order to find the rate of change of the shadow relative to the man.\[\frac{ds}{dt}=\frac{2}{3} \frac{dx}{dt}\]\[\frac{ds}{dt}=\frac{2}{3}(5)=\frac{10}{3}ft/s\]Now you could just add the rate of the man to the rate of the shadow... 10/3 +5=25/3 ft/s or you can use a little more calculus. From above, the location of the shadow is x+s. So after looking at the similar triangles, but keeping x and s together gives:\[x+s=\frac{15}{6}s=\frac{5}{2}s\] Taking the derivative here would give:\[\frac{d(x+s)}{dt}=\frac{5}{2} \frac{ds}{dt}\]Using our rate of change of the shadow from above (ds/dt=10/3):\[\frac{ds(x+s)}{dt}=\frac{5}{2} \frac{10}{3}=\frac{25}{3} ft/s\]Just like before.
In that last line of equations... ignore the first s in ds(x+s). It was really meant to to be the derivative of x+s together with respect to time.
thanks :)_

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