This is a problem of related rates in calculus. We can solve it step by step as follows:

1. First, let's denote the following:
- x: the distance between the man and the pole
- y: the length of the shadow

2. From the problem, we know that the man is moving away from the pole at a rate of 7 ft/s, so dx/dt = 7 ft/s.

3. We need to find dy/dt, the rate at which the tip of his shadow is moving, when x = 30 ft.

4. We can set up a relationship between x and y using similar triangles. The height of the pole and the length of the shadow form one triangle, and the height of the man and the distance he has walked form another. So, we have 15/y = 6/(x-30).

5. Solve this equation for y to get y = 15x/(x-30).

6. Now, differentiate both sides with respect to time (t) to get dy/dt = 15(dx/dt)/(x-30) - 15x(dx/dt)/((x-30)^2).

7. Substitute dx/dt = 7 ft/s and x = 30 ft into the equation to get dy/dt = 15*7/30 - 15*30*7/(30-30)^2.

8. The denominator of the second term becomes zero, which makes the whole term undefined. So, we only consider the first term, which gives dy/dt = 3.5 ft/s.

So, the tip of the man's shadow is moving at a speed of 3.5 ft/s when he is 30 feet from the pole.

Question

This is a problem of related rates in calculus. We can solve it step by step as follows:

1. First, let's denote the following:
   - x: the distance between the man and the pole
   - y: the length of the shadow

2. From the problem, we know that the man is moving away from the pole at a rate of 7 ft/s, so dx/dt = 7 ft/s.

3. We need to find dy/dt, the rate at which the tip of his shadow is moving, when x = 30 ft.

4. We can set up a relationship between x and y using similar triangles. The height of the pole and the length of the shadow form one triangle, and the height of the man and the distance he has walked form another. So, we have 15/y = 6/(x-30).

5. Solve this equation for y to get y = 15x/(x-30).

6. Now, differentiate both sides with respect to time (t) to get dy/dt = 15(dx/dt)/(x-30) - 15x(dx/dt)/((x-30)^2).

7. Substitute dx/dt = 7 ft/s and x = 30 ft into the equation to get dy/dt = 15*7/30 - 15*30*7/(30-30)^2.

8. The denominator of the second term becomes zero, which makes the whole term undefined. So, we only consider the first term, which gives dy/dt = 3.5 ft/s.

So, the tip of the man's shadow is moving at a speed of 3.5 ft/s when he is 30 feet from the pole.

Knowee AI · Accepted Answer

This is a problem of related rates in calculus. We can solve it step by step as follows:

1. First, let's denote the following:
   - x: the distance between the man and the pole
   - y: the length of the shadow

2. From the problem, we know that the man is moving away from the pole at a rate of 7 ft/s, so dx/dt = 7 ft/s.

3. We need to find dy/dt, the rate at which the tip of his shadow is moving, when x = 30 ft.

4. We can set up a relationship between x and y using similar triangles. The height of the pole and the length of the shadow form one triangle, and the height of the man and the distance he has walked form another. So, we have 15/y = 6/(x-30).

5. Solve this equation for y to get y = 15x/(x-30).

6. Now, differentiate both sides with respect to time (t) to get dy/dt = 15(dx/dt)/(x-30) - 15x(dx/dt)/((x-30)^2).

7. Substitute dx/dt = 7 ft/s and x = 30 ft into the equation to get dy/dt = 15*7/30 - 15*30*7/(30-30)^2.

8. The denominator of the second term becomes zero, which makes the whole term undefined. So, we only consider the first term, which gives dy/dt = 3.5 ft/s.

So, the tip of the man's shadow is moving at a speed of 3.5 ft/s when he is 30 feet from the pole.

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 7 ft/s along a straight path. How fast (in ft/s) is the tip of his shadow moving when he is 30 feet from the pole?

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