## MarcLeclair 2 years ago 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?

1. EulerGroupie

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.

2. EulerGroupie

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.

3. MarcLeclair

thanks :)_