A man standing 5 feet away from the base of a lamppost casts a 3 feet long shadow. If he is 6 feet tall and walks away from the lamppost at a speed of 400 feet per minute, at what rate will his shadow lengthen ? In the figure, x represents the distance of the man from the bottom of the lamppost at time t, y represents the length of the man’s shadow at time t, and H represents the height of the lamppost.

We are given the following information:

- The man is 6 feet tall.
- The initial distance between the man and the lamppost is 5 feet.
- The length of the man's shadow is 3 feet when he is standing 5 feet from the lamppost.
- The man walks away from the lamppost at a speed of 400 feet per minute.
- \( x \) represents the distance of the man from the lamppost at time \( t \).
- \( y \) represents the length of the man's shadow at time \( t \).
- \( H \) represents the height of the lamppost.

We want to find the rate at which the man's shadow lengthens (\( \frac{dy}{dt} \)) when he walks away from the lamppost at 400 feet per minute (\( \frac{dx}{dt} = 400 \) feet per minute).

### Step 1: Use Similar Triangles
From the setup, we can use the principle of similar triangles. The triangles formed by the man and his shadow, and by the lamppost and the ground, are similar.

For similar triangles, we have the ratio:

\[
\frac{H}{x + y} = \frac{6}{y}
\]

Here:
- \( H \) is the height of the lamppost,
- \( x \) is the distance of the man from the lamppost,
- \( y \) is the length of the man's shadow,
- \( 6 \) is the height of the man.

### Step 2: Express \( y \) in terms of \( x \)
Rearranging the equation:

\[
H \cdot y = 6 \cdot (x + y)
\]

Expanding the right-hand side:

\[
H \cdot y = 6x + 6y
\]

Now, solve for \( y \):

\[
H \cdot y - 6y = 6x
\]

\[
y(H - 6) = 6x
\]

\[
y = \frac{6x}{H - 6}
\]

### Step 3: Differentiate with respect to time
Now, differentiate both sides of the equation \( y = \frac{6x}{H - 6} \) with respect to time \( t \).

\[
\frac{dy}{dt} = \frac{6}{H - 6} \cdot \frac{dx}{dt}
\]

We are given \( \frac{dx}{dt} = 400 \) feet per minute. So, substitute that in:

\[
\frac{dy}{dt} = \frac{6 \cdot 400}{H - 6}
\]

\[
\frac{dy}{dt} = \frac{2400}{H - 6}
\]

### Step 4: Find \( H \)
We can use the initial conditions to find \( H \). When the man is 5 feet away from the lamppost (\( x = 5 \)) and his shadow is 3 feet long (\( y = 3 \)), we can substitute these values into the similar triangles equation:

\[
\frac{H}{5 + 3} = \frac{6}{3}
\]

\[
\frac{H}{8} = 2
\]

\[
H = 16
\]

### Step 5: Substitute \( H = 16 \) into the equation
Now, substitute \( H = 16 \) into the rate equation:

\[
\frac{dy}{dt} = \frac{2400}{16 - 6}
\]

\[
\frac{dy}{dt} = \frac{2400}{10} = 240 \text{ feet per minute}
\]

### Conclusion
The man's shadow is lengthening at a rate of 240 feet per minute.

Would you like more details or have any questions?

Here are 5 related questions:
1. What is the relationship between the height of the man and the length of his shadow?
2. How do similar triangles help in solving related rates problems?
3. How would the result change if the man walked away from the lamppost at a different speed?
4. How can this problem be solved graphically using functions of time?
5. What if the man was shorter or taller, how would that affect the shadow's rate of change?

**Tip:** In related rates problems, always look for geometric relationships like similar triangles to set up equations involving multiple changing quantities.

A man standing 5 feet away from the base of a lamppost casts a 3 feet long shadow. If he is 6 feet tall and walks away from the lamppost at a speed of 400 feet per minute, at what rate will his shadow lengthen? In the figure, x represents the distance of the man from the bottom of the lamppost at time t, y represents the length of the man’s shadow at time t, and H represents the height of the lamppost.

Learn how to solve a related rates problem where a man standing 5 feet away from a lamppost casts a 3 feet shadow. Discover how the rate of the shadow lengthens as the man walks away at 400 feet per minute. This problem involves using similar triangles and derivatives. Suitable for high school students in Grades 11-12.

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