man 6 feet tall walks at a rate of feet per second away from a light that is 15 feet above the ground (see figure). When he is feet from the base of the light, at what rate is the Length of his shadow changing?

Question

Quizplus · Accepted Answer

Let's call the man's height $h$ and the distance between the man and the light post $x$. We want to find $\frac{d\ell}{dt}$ when $x=20$. We can begin by using similar triangles to express $\ell$ in terms of $h$ and $x$:
$$\frac{\ell}{h}=\frac{x+\ell}{h+15}$$
Cross-multiplying and solving for $\ell$ gives
$$\ell = \frac{hx}{h+15}$$
Now we can implicitly differentiate with respect to time $t$:
$$\frac{d\ell}{dt} = \frac{h\frac{dx}{dt}(h+15)-hx\frac{dh}{dt}}{(h+15)^2}$$
We are given that $h=6$ ft and $\frac{dh}{dt} = -\frac{4}{3}$ ft/sec (since the man is walking away from the light, his shadow is getting longer, so the height of his head relative to the ground is decreasing). We are also given that $\frac{dx}{dt} = \frac{8}{15}$ ft/sec. Plugging in these values and $x=20$ gives
$$\frac{d\ell}{dt} = \boxed{\frac{26}{3}	ext{ ft/sec}}$$

Man 6 Feet Tall Walks at a Rate of Feet 52ft/sec\frac { 5 } { 2 } \mathrm { ft } / \mathrm { sec }25​ft/sec

Man 6 Feet Tall Walks at a Rate of Feet $\frac { 5 } { 2 } \mathrm { ft } / \mathrm { sec }$