Math Problem Statement

A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks away from the pole with a speed of 5 ft/s along a straight path. How fast is the tip of his shadow moving when he is 35 ft from the pole?

Solution

To solve this problem, let's set up a relationship using similar triangles and related rates.

Step-by-Step Solution

  1. Define the Variables:

    • Let xx be the distance between the man and the base of the pole, changing over time. The man is moving away from the pole at a speed of 5 ft/s, so dxdt=5 ft/s\frac{dx}{dt} = 5 \text{ ft/s}.
    • Let yy be the length of the shadow of the man.
  2. Determine the Rate We Need to Find:

    • We are asked to find how fast the tip of the shadow is moving. The tip of the shadow is at a distance x+yx + y from the base of the pole, so we need to determine d(x+y)dt\frac{d(x + y)}{dt}.
  3. Set Up a Relationship Using Similar Triangles:

    • The height of the pole (15 ft) and the height of the man (6 ft) form similar triangles with the distances from the pole to the tip of the shadow.
    • Therefore, we can write: 15x+y=6y\frac{15}{x + y} = \frac{6}{y}
  4. Solve for yy:

    • Cross-multiply to clear the fractions: 15y=6(x+y)15y = 6(x + y)
    • Distribute 66: 15y=6x+6y15y = 6x + 6y
    • Rearrange to solve for yy: 9y=6x9y = 6x y=23xy = \frac{2}{3}x
  5. Differentiate Both Sides with Respect to tt:

    • Since y=23xy = \frac{2}{3}x, differentiate with respect to tt: dydt=23dxdt\frac{dy}{dt} = \frac{2}{3} \frac{dx}{dt}
    • Substitute dxdt=5 ft/s\frac{dx}{dt} = 5 \text{ ft/s}: dydt=23×5=103 ft/s\frac{dy}{dt} = \frac{2}{3} \times 5 = \frac{10}{3} \text{ ft/s}
  6. Find the Rate at Which the Tip of the Shadow Moves:

    • The speed of the tip of the shadow is ddt(x+y)\frac{d}{dt}(x + y).
    • Differentiate x+yx + y with respect to tt: ddt(x+y)=dxdt+dydt\frac{d}{dt}(x + y) = \frac{dx}{dt} + \frac{dy}{dt}
    • Substitute the values of dxdt=5\frac{dx}{dt} = 5 and dydt=103\frac{dy}{dt} = \frac{10}{3}: ddt(x+y)=5+103=15+103=253 ft/s\frac{d}{dt}(x + y) = 5 + \frac{10}{3} = \frac{15 + 10}{3} = \frac{25}{3} \text{ ft/s}

Answer:

The tip of the shadow is moving at a rate of 253\frac{25}{3} ft/s, or approximately 8.338.33 ft/s, when the man is 35 ft from the pole.


Would you like further details, or do you have any questions?

Here are five related questions for further practice:

  1. How would the answer change if the pole were 20 ft tall instead of 15 ft?
  2. What if the man was walking at a different speed, say 6 ft/s? How would that affect the rate of the shadow's tip movement?
  3. If the man started closer to the pole, would the rate of the tip of the shadow change initially?
  4. How would this problem change if the man’s height were different, say 5 ft?
  5. What if the man was moving toward the pole instead of away? How would this affect the rate of the shadow's movement?

Tip: In related rates problems involving similar triangles, look for proportions to set up relationships, as they can simplify calculations significantly.

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Math Problem Analysis

Mathematical Concepts

Related Rates
Similar Triangles

Formulas

Similar triangles ratio: height of pole / (distance from pole to tip of shadow) = height of man / length of shadow
Rate of tip of shadow movement: d/dt(x + y) = dx/dt + dy/dt

Theorems

Similarity of triangles in related rates problems

Suitable Grade Level

Grades 11-12