Math Problem Statement

A 7.00-ft-tall man walks at 4.00 ft/s toward a streetlight that is 15.0 ft above the ground. At what rate is the end of the man's shadow moving when he is 10.0 ft from the base of the light? Use the direction in which the distance from the streetlight increases as the positive direction.

Solution

This is a related rates problem where a man walks toward a streetlight, and the rate at which the shadow moves is to be determined.

Let's break it down:

Problem summary:

  • A 7-foot-tall man is walking toward a streetlight at a speed of 4 ft/s.
  • The streetlight is 15 feet tall.
  • The man is 10 feet away from the base of the streetlight.
  • We are to find the rate at which the end of his shadow is moving.

Step-by-step solution:

Step 1: Setting up a relationship

Let:

  • xx be the distance between the man and the streetlight.
  • ss be the length of the man's shadow.

By similar triangles, we can set up the following proportion:

15s+x=7s\frac{15}{s + x} = \frac{7}{s}

Multiplying both sides by s(s+x)s(s + x), we get:

15s=7(s+x)15s = 7(s + x)

Expanding:

15s=7s+7x15s = 7s + 7x

Now solve for ss:

15s7s=7x15s - 7s = 7x 8s=7x8s = 7x s=78xs = \frac{7}{8}x

Step 2: Differentiating with respect to time

Now we differentiate both sides of the equation s=78xs = \frac{7}{8}x with respect to time tt to find the rate of change of ss (i.e., dsdt\frac{ds}{dt}):

dsdt=78dxdt\frac{ds}{dt} = \frac{7}{8} \frac{dx}{dt}

Given:

  • dxdt=4\frac{dx}{dt} = -4 ft/s (negative because the man is moving toward the streetlight).

Substitute dxdt\frac{dx}{dt}:

dsdt=78(4)\frac{ds}{dt} = \frac{7}{8}(-4) dsdt=3.5 ft/s\frac{ds}{dt} = -3.5 \text{ ft/s}

Step 3: Finding the rate at which the shadow's tip moves

The rate at which the tip of the shadow moves is the rate at which the man is moving dxdt\frac{dx}{dt} plus the rate at which the shadow is changing dsdt\frac{ds}{dt}:

Rate of tip of shadow=dxdt+dsdt\text{Rate of tip of shadow} = \frac{dx}{dt} + \frac{ds}{dt} Rate of tip of shadow=4+(3.5)\text{Rate of tip of shadow} = -4 + (-3.5) Rate of tip of shadow=7.5 ft/s\text{Rate of tip of shadow} = -7.5 \text{ ft/s}

Final Answer:

The tip of the man's shadow is moving at a rate of 7.5 ft/s toward the light.

Would you like further clarification or details on any step?


Here are 5 related questions you might explore:

  1. What happens to the speed of the shadow as the man gets closer to the light?
  2. How does the rate change if the height of the streetlight changes?
  3. Can you derive a formula for the shadow's speed at any given distance?
  4. How would the problem change if the man were moving away from the streetlight?
  5. What would happen if the man were walking at a different speed?

Tip: Always ensure that units are consistent throughout related rates problems to avoid calculation errors.

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Math Problem Analysis

Mathematical Concepts

Related Rates
Geometry
Similar Triangles
Differentiation

Formulas

\frac{15}{s + x} = \frac{7}{s}
s = \frac{7}{8}x
\frac{ds}{dt} = \frac{7}{8} \frac{dx}{dt}

Theorems

Similar Triangles
Differentiation in Calculus

Suitable Grade Level

High School Calculus or Early College Level