Solution Check for a Related Rates Problem Involving Shadow Length and Position

Question

I am working on the following problem:

A girl 5 ft tall is running at the rate of 12 ft/s and passes under a street light 20 ft above the ground. Find how rapidly the tip of her shadow is moving when she is (a) 20 ft past the street light, and (b) 50 ft past the street light.

And the follow up:

Find how rapidly the length of the girl's shadow is increasing at each of the stated moments.

$ACE$ and $DEF$ are similar triangles.

Let $s=AF$ and $x=EF$. Then

$$\frac{{s+x}}{{20}}=\frac{{x}}{{5}}$$

then

$$s + x = 4x$$

then

$$s = 3x$$

Let's differentiate both sides:

$$\frac{{ds}}{{dt}} = 3\frac{{dx}}{{dt}}$$

Since $$\frac{{ds}}{{dt}} = 12$$, then $$\frac{{dx}}{{dt}} = 4$$

This answers the follow-up question. So, regardless of the girl's position, the length of her shadow is increasing at a constant rate of $4$ ft/sec.

Now, to answer the original problem we need to find: $$\frac{{d(s+x)}}{{dt}} $$

So,

$$\frac{{d(s+x)}}{{dt}}=4\frac{{dx}}{{dt}}$$

Since $$\frac{{dx}}{{dt}}=4$$, then

$$\frac{{d(s+x)}}{{dt}}=16$$

So, the tip of the girl's shadow is moving at a constant rate of $16$ ft/sec, regardless of her position.

Please, verify my solution.

Answer 1

Use your notations, we know the running velocity of the girl: $\frac{ds}{dt}=12$. We want to find:

(1) The velocity of the shadow, namely, $\frac{dAE}{dt}=~?$

(2) The changing rate for the length of the shadown, namely, $\frac{dx}{dt}=~?$

Solution:

Since $\triangle CBD\sim\triangle CAE$, we have

$$\frac{CB}{CA}=\frac{BD}{AE}\Longrightarrow \frac{15}{20}=\frac{s}{AE}\Longrightarrow AE=\frac43s$$

hence, for part (1), we get

$$\frac{dAE}{dt}=\frac43\frac{ds}{dt}=\frac43\cdot12=16$$

Next, for part (2), since $x=AE-s$, we get

$$\frac{dx}{dt}=\frac{dAE}{dt}-\frac{ds}{dt}=16-12=4$$