A man 6 feet tall walks at a rate of 13 feet per second away from a light that is 15 feet above the ground (see figure).When he is 5 feet from the base of the light,at what rate is the length of his shadow changing?

Question

A man 6 feet tall walks at a rate of 13 feet per second away from a light that is 15 feet above the ground (see figure).When he is 5 feet from the base of the light,at what rate is the length of his shadow changing? ​  ​

Quizplus · Accepted Answer

We can use similar triangles to solve this problem. Let x be the distance from the tip of the shadow to the base of the light, and let y be the length of the shadow.

Then, we have two similar right triangles: one with height 15 and base x, and the other with height (15 + 6) = 21 and base (x + 5):

$\frac{y}{x} = \frac{15}{6}$

$\frac{y}{x+5} = \frac{21}{6}$

Solve for y in terms of x:

$y = \frac{5}{2}x$

$y = \frac{7}{2}(x+5)$

Set these equal to each other and solve for x:

$\frac{5}{2}x = \frac{7}{2}(x+5)$

$x = \frac{25}{2}$

Now, we can use the chain rule to find the rate of change of y (the length of the shadow):

$\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$

$\frac{dy}{dx} = \frac{5}{2}$ (from above)

$\frac{dx}{dt} = -13$ (because the man is walking away from the light)

Plug in the values:

$\frac{dy}{dt} = \frac{5}{2} \cdot (-13) = -\frac{65}{2}$

The negative sign indicates that the length of the shadow is decreasing. So the answer is:

Choice: C

A Man 6 Feet Tall Walks at a Rate of 13