Problem 6 - The Calculus with Analytic Geometry

Question

A light is hung \(15 \mathrm{ft}\) above a straight horizontal path. If a man \(6 \mathrm{ft}\) tall is walking away from the light at the rate of \(5 \mathrm{ft} / \mathrm{sec}\), how fast is his shadow lengthening?

Step-by-Step Solution

Answer

His shadow is lengthening at a rate of \( \frac{10}{3} \) ft/sec.

Step 1 - Understand the Problem

A light is positioned 15 feet above the ground, and a man who is 6 feet tall is walking away from the light at a rate of 5 feet per second. We need to determine how fast the length of the man's shadow is increasing.

Step 2 - Set Up the Variables

Let \( x \) be the distance from the man to the point directly below the light, and let \( s \) be the length of his shadow. We know \( \frac{dx}{dt} = 5 \) ft/sec. We aim to find \( \frac{ds}{dt} \).

Step 3 - Use Similar Triangles

Notice that we can form two similar triangles: one with the light, the top of the man's head, and the tip of his shadow, and the other with the light and the ground. The relationship between the heights and base lengths of the triangles is given by: \[ \frac{15}{x+s} = \frac{6}{s} \]

Step 4 - Simplify the Equation

Cross-multiply to simplify the equation: \[ 15s = 6(x+s) \] Simplify further to get: \[ 15s = 6x + 6s \] \[ 9s = 6x \] \[ s = \frac{2}{3}x \]

Step 5 - Differentiate Both Sides

Differentiate both sides with respect to \( t \): \[ \frac{d}{dt}(s) = \frac{2}{3} \frac{d}{dt}(x) \] Substitute \( \frac{dx}{dt} = 5 \) ft/sec: \[ \frac{ds}{dt} = \frac{2}{3} \times 5 \] So: \[ \frac{ds}{dt} = \frac{10}{3} \text{ ft/sec} \]

Key Concepts

Similar Triangles

Understanding similar triangles is crucial in solving this shadow problem. Take a look at the light, the man, and their shadows. The light, the top of the man's head, and the endpoint of his shadow form one triangle. Similarly, the light touching the ground and the man's feet forms another. Both these triangles are similar because their angles are equal. This similarity helps us set up a proportion between the lengths of corresponding sides. Here, we relate the height of the light and the total length (man + shadow) to the man's height and the shadow's length: \[ \frac{15}{x+s} = \frac{6}{s} \]

Differentiation

Differentiation in calculus helps us understand how things change over time. In this problem, we are asked to find how fast the man's shadow lengthens. When we set up our similar triangles' equation, we got: \[ s = \frac{2}{3}x \]To find how the shadow length (\