Math Problem Statement

A street light is at the top of a 16ft pole. A 5ft tall girl walks along a straight path away from the pole with a speed of 5ft/sec. At what rate is the tip of her shadow moving away from the light (ie. away from the top of the pole) when the girl is 29ft away from the pole?

Solution

Let's solve this problem step-by-step using related rates.

Step 1: Label the variables

  • Let xx be the distance of the girl from the base of the pole (in feet).
  • Let yy be the length of the girl's shadow (in feet).
  • The height of the pole is 16 feet, and the height of the girl is 5 feet.

Step 2: Set up the geometric relationship

At any given time, the triangle formed by the top of the pole, the tip of the shadow, and the base of the pole is similar to the triangle formed by the top of the girl's head, the tip of her shadow, and the base of the pole.

Using the properties of similar triangles, we have the following proportion:

height of poledistance from pole to tip of shadow=height of girllength of shadow\frac{\text{height of pole}}{\text{distance from pole to tip of shadow}} = \frac{\text{height of girl}}{\text{length of shadow}}

Substituting the known values:

16x+y=5y\frac{16}{x + y} = \frac{5}{y}

Step 3: Solve the equation for yy

Now, let's solve this proportion for yy:

16y=5(x+y)16y = 5(x + y)

Expanding:

16y=5x+5y16y = 5x + 5y

Subtract 5y5y from both sides:

11y=5x11y = 5x

Solve for yy:

y=511xy = \frac{5}{11}x

Step 4: Differentiate with respect to time

We need to find how fast the tip of the shadow is moving, which is the rate of change of x+yx + y with respect to time. First, differentiate both sides of the equation y=511xy = \frac{5}{11}x with respect to time tt:

dydt=511dxdt\frac{dy}{dt} = \frac{5}{11} \frac{dx}{dt}

Given that the girl is walking at a speed of 5 ft/sec, we have:

dxdt=5 ft/sec\frac{dx}{dt} = 5 \text{ ft/sec}

So,

dydt=511×5=2511 ft/sec\frac{dy}{dt} = \frac{5}{11} \times 5 = \frac{25}{11} \text{ ft/sec}

Step 5: Find the rate at which the tip of the shadow is moving

The rate at which the tip of the shadow is moving is given by:

ddt(x+y)=dxdt+dydt\frac{d}{dt}(x + y) = \frac{dx}{dt} + \frac{dy}{dt}

Substitute the known values:

ddt(x+y)=5+2511\frac{d}{dt}(x + y) = 5 + \frac{25}{11}

To simplify:

5=5511,5511+2511=80115 = \frac{55}{11}, \quad \frac{55}{11} + \frac{25}{11} = \frac{80}{11}

Thus, the rate at which the tip of the shadow is moving is:

8011 ft/sec\frac{80}{11} \text{ ft/sec}

Step 6: Evaluate when x=29x = 29

At the moment when the girl is 29 feet from the pole (x=29x = 29), we substitute this value into the equation for yy to find the length of the shadow:

y=511×29=1451113.18 feety = \frac{5}{11} \times 29 = \frac{145}{11} \approx 13.18 \text{ feet}

So, the tip of the shadow is moving at a rate of 8011\frac{80}{11} ft/sec when the girl is 29 feet away from the pole.

Final Answer:

The tip of the shadow is moving at 8011\frac{80}{11} feet per second, or approximately 7.27 ft/sec when the girl is 29 feet away from the pole.


Would you like more details on any part of the solution? Here are some related questions that might interest you:

  1. How would the rate of the tip of the shadow change if the girl were walking faster or slower?
  2. What happens if the girl walks towards the pole instead of away from it?
  3. How can we model this situation using other geometric principles?
  4. How would the rate change if the pole were taller or shorter?
  5. What is the exact point where the girl's shadow reaches its maximum length in relation to her position?

Tip: It's helpful to remember that when two triangles are similar, their corresponding sides are proportional—this is a key concept in solving related rates problems involving shadows!

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Math Problem Analysis

Mathematical Concepts

Related Rates
Geometry
Similar Triangles
Rate of Change

Formulas

\frac{16}{x + y} = \frac{5}{y}
y = \frac{5}{11}x
\frac{d}{dt}(x + y) = \frac{dx}{dt} + \frac{dy}{dt}

Theorems

Similar Triangles

Suitable Grade Level

Grades 11-12