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A man 6 feet tall walks at a rate of 5 feet per second away from a light that is 15 feet above the ground. When he is 10 feet from the base of the light,

(a) at what rate is the tip of his shadow moving?
(b) at what rate is the length of his shadow changing?

My Attempt

Given:

$a=6\ \mathbb{ft}$; $b'=5\ \frac{\mathbb{ft}}{\mathbb{s}}$; $h=15\ \mathbb{ft}$; $d=10\ \mathbb{ft}$

I used the triangle proportionality theorem and got my $b=4\ \mathbb{ft}$, and then by Pythagoras Theorem I got my $c=\sqrt{52}$. I do not know how to show you the triangle I drew, but I drew a triangle with in a larger triangle. I am sort of confused on what to do in the problem, and how solve both parts. I assumed that I would have to use Pythagoras Theorem:

$$a^2+b^2=c^2$$ Then take the derivative of that $$a(a')+b(b')=c(c')$$

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  • $\begingroup$ Instead of Pythagorean Theorem, use similar triangles. $\endgroup$
    – B. Goddard
    Nov 12, 2016 at 16:10

1 Answer 1

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If $(0,0) $ is the base of the light and $(0,15) $ is the light, if $k$ is the distance between the man feet and the base of the light, and if the man moves to the right walking on the $x $-axis with the highest point of his head at height $y=6$, the line connecting the light and the highest point of his head is $y=-\frac {9}{k} x +15$. Therefore, the intersection with the $x $-axis (which is the shadow tip) is in $(\frac {5}{3} k, 0)$.

Since $k $ increases at a constant rate of $dk/dt=5$ feet per second, then the shadow tip moves at a constant rate of $\displaystyle \frac {5}{3} 5=25/3 \,\,$ feet per second. Also, since the dimension of the shadow is $\displaystyle \frac {5}{3} k -k= \frac {2}{3} k \,\, $, the shadow length moves at a rate of $\displaystyle \frac {2}{3} 5=10/3 \,\,$ feet per second.

Note that the information that he is $10$ feet from the base of the light is not necessary to the problem, since the rates are constant.

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