Math Problem Statement

A street light is mounted at the top of a 15-ft-tall pole. A man 6 feet tall walks away from the pole with a speed of 4 ft/s along a straight path. How fast (in ft/s) is the tip of his shadow moving when he is 40 feet from the pole?

Solution

To solve the problem of determining how fast the tip of the man's shadow is moving, we will use related rates in calculus. Let's set up the problem with the necessary variables and relationships.

First, define the variables:

  • Let yy be the length of the man's shadow in feet.
  • Let xx be the distance between the man and the pole in feet.
  • Let LL be the distance from the tip of the shadow to the pole, L=x+yL = x + y.

Given:

  • The height of the pole is 15 feet.
  • The height of the man is 6 feet.
  • The man walks away from the pole at a speed of 4 ft/s, so dxdt=4\frac{dx}{dt} = 4 ft/s.

Using similar triangles, we can set up the following proportion based on the heights and lengths of the shadows: 15L=6y\frac{15}{L} = \frac{6}{y}

Solving for LL in terms of yy: 15y=6L    L=156y    L=52y15y = 6L \implies L = \frac{15}{6}y \implies L = \frac{5}{2}y

Since L=x+yL = x + y: x+y=52y    x=52yy    x=32yx + y = \frac{5}{2}y \implies x = \frac{5}{2}y - y \implies x = \frac{3}{2}y

Differentiate both sides with respect to time tt: ddt(x)=ddt(32y)\frac{d}{dt}(x) = \frac{d}{dt}\left(\frac{3}{2}y\right) dxdt=32dydt\frac{dx}{dt} = \frac{3}{2} \frac{dy}{dt}

We know dxdt=4\frac{dx}{dt} = 4 ft/s, so: 4=32dydt4 = \frac{3}{2} \frac{dy}{dt}

Solving for dydt\frac{dy}{dt}: dydt=423=83 ft/s\frac{dy}{dt} = \frac{4 \cdot 2}{3} = \frac{8}{3} \text{ ft/s}

Thus, the speed at which the tip of the shadow is moving is: dLdt=dxdt+dydt=4+83=123+83=203 ft/s\frac{dL}{dt} = \frac{dx}{dt} + \frac{dy}{dt} = 4 + \frac{8}{3} = \frac{12}{3} + \frac{8}{3} = \frac{20}{3} \text{ ft/s}

So, the tip of the shadow is moving at a rate of 203\frac{20}{3} ft/s when the man is 40 feet from the pole.

Would you like further details or have any questions?

Here are 8 related questions you might consider:

  1. What is the length of the man's shadow when he is 40 feet away from the pole?
  2. How would the speed of the shadow's tip change if the man walked at a different speed?
  3. How is the rate of change of the shadow's length affected by the height of the pole?
  4. What happens to the rate of change of the shadow's length if the man's height is different?
  5. How would you set up a similar problem for a different height of the pole?
  6. How does the concept of similar triangles help in solving this problem?
  7. What if the man were walking towards the pole instead? How would that affect the rates?
  8. Can you explain why we use the differentiation approach to solve this problem?

Tip: Always double-check your related rates setup by ensuring the units of measurement are consistent throughout the problem.

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Math Problem Analysis

Mathematical Concepts

Related Rates
Similar Triangles

Formulas

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Theorems

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Suitable Grade Level

Advanced High School