The critical part which I want to discuss is painted with red below.
A light on top of a lamppost shines$~25~\mathrm{ft}~$above the ground. A main$~6~\mathrm{ft}~~$tall is walking away from the light. What is the length of his shadow when he is$~40~\mathrm{ft}~~$away from the base of the lamppost? If he is walking away at the rate of$~5~{~\mathrm{ft}/\mathrm{sec}}~$, how fast is his shadow increasing at this point?
What I thought for it are as below.
$$\text{slope}={6-25\over 40-0}=-{19\over40}\tag{1}$$
$$y=-{19\over 40}x+25\tag{2}$$
$$0=-{19\over 40}x+25\tag{3}$$
$$0=-19x+25\cdot 40\tag{4}$$
$$19x=1000\tag{5}$$
$$x={1000\over 19}\leftarrow~~x\text{-coordinate where the light ray finishes}\tag{6}$$
$$\therefore~~s={1000\over 19}-40\cdot{19\over 19}\tag{7}$$
$$={1000-760\over 19}={240\over19}~~\leftarrow~~\text{length of the shadow}\tag{8}$$
$${\mathrm{dx}\over\mathrm{dt}}=-5\tag{9}$$
$$x=-5t\tag{10}$$
$$s_{}(t):={1000\over 19}-(-5t)=5t+{1000\over 19}\tag{11}$$
$$\underbrace{{\mathrm{ds}\over\mathrm{dt}}=5~~{\mathrm{ft}\over\mathrm{sec}}}_{\text{My result}}\tag{12}$$
What I can't get is the following equation which is given for the second problem.
$$\color{red}{{\mathrm{ds}\over\mathrm{dt}}=-{30\over 19}~~{\mathrm{ft}\over\mathrm{sec}}}\tag{13}$$
How this equation can be attained?
BTW I want to know any way which can be used to create a link which let display jump to a specific equation(e.g.eqn10).
We are given how $x$ changes with time, and we want to find out how $s$ changes with time. So, we first want a way to relate $x$ and $s$, specifically, a relation that holds regardless of where the man is. One way to go about this is to use similar triangles and notice that $\frac{s}{6}=\frac{x+s}{25}$, which gives us $s=\frac{6}{19}x$. Then, differentiating both sides by time, it follows that the speed of the shadow is $6/19$ times the speed of the guy (upto some sign convention that I probably didn't bother with here), which gives you the answer.