A 6 ft man walks at a rate of 3ft//s away from a street light that is 21 feet above the ground. (a) At what rate is the length of his shadow changing when he is 8 ft from the light? (b) At what rate is the tip of his shadow moving when (i) 10ft from the light?

This is a classic related rates problem that involves understanding the relationship between the man, his shadow, and the light source.

### Step-by-Step Solution:

Let’s define the variables:
- Let \( x \) be the distance from the man to the base of the streetlight (in feet).
- Let \( s \) be the length of the man's shadow (in feet).
- The height of the streetlight is 21 feet, and the height of the man is 6 feet.

### Part (a): Rate of change of the shadow length when he is 8 feet from the light

We are given:
- \( \frac{dx}{dt} = 3 \text{ ft/s} \) (the speed of the man walking away from the light).

To relate \( x \) and \( s \), we use similar triangles. The ratio of the height of the man to the height of the light equals the ratio of the length of his shadow to his distance from the light:

\[
\frac{6}{21} = \frac{s}{x+s}
\]

Simplifying this:

\[
\frac{2}{7} = \frac{s}{x+s}
\]

Cross-multiply:

\[
2(x + s) = 7s
\]

Expanding this:

\[
2x + 2s = 7s
\]

Simplifying for \( s \):

\[
2x = 5s \quad \Rightarrow \quad s = \frac{2x}{5}
\]

Now differentiate both sides of this equation with respect to time \( t \):

\[
\frac{ds}{dt} = \frac{2}{5} \frac{dx}{dt}
\]

Substitute \( \frac{dx}{dt} = 3 \text{ ft/s} \):

\[
\frac{ds}{dt} = \frac{2}{5} \times 3 = 1.2 \text{ ft/s}
\]

Thus, the length of the man's shadow is increasing at a rate of **1.2 ft/s** when he is 8 feet from the light.

### Part (b): Rate at which the tip of the shadow is moving when \( x = 10 \) ft

The tip of the shadow is located at the total distance from the light pole, which is \( x + s \). To find how fast the tip of the shadow is moving, we need to compute \( \frac{d}{dt}(x + s) \):

\[
\frac{d}{dt}(x + s) = \frac{dx}{dt} + \frac{ds}{dt}
\]

We already know that \( \frac{dx}{dt} = 3 \text{ ft/s} \), and we previously derived that \( \frac{ds}{dt} = \frac{2}{5} \frac{dx}{dt} \), so:

\[
\frac{d}{dt}(x + s) = \frac{dx}{dt} + \frac{2}{5} \frac{dx}{dt}
\]

Substitute \( \frac{dx}{dt} = 3 \text{ ft/s} \):

\[
\frac{d}{dt}(x + s) = 3 + \frac{2}{5} \times 3 = 3 + 1.2 = 4.2 \text{ ft/s}
\]

Thus, the tip of the shadow is moving at a rate of **4.2 ft/s** when the man is 10 feet from the light.

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### Summary:
- (a) The length of the man's shadow is increasing at **1.2 ft/s** when he is 8 feet from the light.
- (b) The tip of his shadow is moving at **4.2 ft/s** when he is 10 feet from the light.

Would you like a deeper explanation of any step, or do you have other questions on related rates?

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Here are 5 follow-up questions you might consider:
1. How would the rate of change of the shadow length vary if the man walked faster or slower?
2. How would the problem change if the height of the streetlight were different?
3. What happens to the rate of change of the shadow when the man is very far from the light?
4. Can you calculate how long the shadow would be if the man was 15 feet from the light?
5. What other real-life situations can be modeled with similar triangles and related rates?

**Tip:** In related rates problems, always identify which quantities are changing and use geometry to relate them.

A 6 ft man walks at a rate of 3 ft/s away from a street light that is 21 feet above the ground. (a) At what rate is the length of his shadow changing when he is 8 ft from the light? (b) At what rate is the tip of his shadow moving when he is 10 ft from the light?

Solve a classic related rates problem where a 6 ft man walks at 3 ft/s away from a street light 21 ft tall. Learn how to calculate the rate of shadow length change when the man is 8 ft from the light, and the rate of the tip of his shadow when he is 10 ft from the light. Suitable for Grades 10-12.

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