"A street light is mounted at the top of a 15-ft-tall pole. A man 6ft tall walks away from the pole with a speed of 5ft/s along a straight path. How fast is the tip of his shadow moving when he is 40 ft from the pole?"(adsbygoogle = window.adsbygoogle || []).push({});

This is how I did it:

[tex]\frac{15}{6} = \frac{x+y}{y}[/tex]

[tex]0 = \frac{y(\frac{dy}{dx} + \frac{dy}{dt}) - (x + y)\frac{dy}{dt}}{y^2}[/tex]

[tex]y\frac{dx}{dt} - x\frac{dy}{dt} = 0[/tex]

[tex]\frac{dy}{dt} = \frac{y}{x} \frac {dx}{dt}[/tex]

[tex]\frac{dy}{dt} = \frac{6}{9}(5)[/tex]

[tex]\frac{dy}{dt} = \frac{30}{9} = \frac{10}{3} ft/s[/tex]

then, since the guy is also moving at 5ft/s, add 5 to dy/dt, and the answer is:

[tex]

\frac{25}{3} ft/s

[/tex]

This is the right answer. However, my solution manual does it another way. They find y = (2/3)x, and then say that the tip of the shadow moves at a rate of:

[tex]\frac{d}{dt}(x + y) = \frac{d}{dt}(\frac {5}{3} x) = \frac {5}{3} (5) = \frac {25}{3} ft/s.

[/tex]

Are both ways correct? I was trying to explain my solution to someone, but now I'm confused as to whether my solution is correct.

According to my way, why don't you simply stop at

[tex]

\frac{10}{3} ft/s

[/tex]

I'm thinking that this is the rate of the shadow moving relative to the man. But to find the rate of the shadow relative to the street light (or the earth), you must add the speed of the man. Is my reasoning correct??

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# Homework Help: Related rates solution

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