Question

A man of 5 feet height is walking away from a light fixed at a height of 15 feet at the rate of K miles/hour. If the rate of increase of his shadow is $\frac{11}{5}$ feet/sec, then K = (Take 1 mile = 5280 feet)

• A. 2
• B. 3
• C. 4
• D. 5

Hint:

Related rates problems involving shadows almost always use similar triangles. Set up the proportion, create a simple algebraic relationship between the variables, and then differentiate with respect to time. Always be careful with units.

Answer 1

The Correct Option is B

Let H be the height of the light post, so $H=15$ ft.
Let h be the height of the man, so $h=5$ ft.
Let $x$ be the distance of the man from the base of the light post.
Let $s$ be the length of the man's shadow.
By similar triangles (the large triangle formed by the light post and the tip of the shadow, and the small triangle formed by the man and the tip of his shadow):
$\frac{\text{Height of Post}}{\text{Base of large triangle}} = \frac{\text{Height of man}}{\text{Base of small triangle}}$.
$\frac{H}{x+s} = \frac{h}{s}$.
$\frac{15}{x+s} = \frac{5}{s}$.
$15s = 5(x+s) \implies 15s = 5x+5s \implies 10s = 5x \implies x=2s$.
Now, we differentiate this relation with respect to time $t$:
$\frac{dx}{dt} = 2\frac{ds}{dt}$.
We are given the rates:
$\frac{dx}{dt}$ is the speed of the man, given as K miles/hour.
$\frac{ds}{dt}$ is the rate of change of the shadow's length, given as $\frac{11}{5}$ feet/sec.
We must convert the units to be consistent. Let's use feet per second.
$\frac{dx}{dt} = K \frac{\text{miles}}{\text{hour}} = K \frac{5280 \text{ feet}}{3600 \text{ sec}} = K \frac{528}{360} \frac{\text{ft}}{\text{sec}} = K \frac{22}{15} \frac{\text{ft}}{\text{sec}}$.
Substitute the rates into the differentiated equation:
$K \frac{22}{15} = 2 \left(\frac{11}{5}\right)$.
$K \frac{22}{15} = \frac{22}{5}$.
Solve for K:
$K = \frac{22}{5} \times \frac{15}{22} = \frac{15}{5} = 3$.
So, the speed of the man is K = 3 miles/hour.