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 50 ft from the pole?

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

1. First, let's denote the distance from the man to the pole as x, and the length of the shadow as y. We know that the speed at which the man is moving away from the pole (dx/dt) is 5 ft/s.

2. We can set up a relationship between x, y, and the heights using similar triangles. The height of the pole is to the length of the entire shadow (x + y) as the height of the man is to the length of the shadow (y). So, we have 15/(x + y) = 6/y.

3. We can cross-multiply to get 15y = 6x + 6y, and then simplify to get 9y = 6x.

4. Now, we can differentiate both sides with respect to time (t) to get 9(dy/dt) = 6(dx/dt).

5. We know that dx/dt = 5 ft/s, so we can substitute that in to get 9(dy/dt) = 30.

6. Solving for dy/dt, we get dy/dt = 30/9 = 10/3 ft/s.

So, the tip of the man's shadow is moving at a speed of 10/3 ft/s when he is 50 ft from the pole.

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

1. First, let's denote the distance from the man to the pole as x, and the length of the shadow as y. We know that the rate of change of x (dx/dt) is 5 ft/s.

2. The situation forms a similar triangle problem, where the height of the pole is to the length of the shadow (15 ft to y) as the height of the man is to the distance between the man and the pole (6 ft to x). So we can write the equation as 15/y = 6/x.

3. We can rearrange the equation to y = 15x/6 = 2.5x.

4. Now we can differentiate both sides with respect to time (t) to find dy/dt, the rate of change of the shadow length. Using the chain rule, we get dy/dt = 2.5 dx/dt.

5. We know that dx/dt = 5 ft/s, so we can substitute this into the equation to find dy/dt = 2.5 * 5 = 12.5 ft/s.

6. However, this is the rate at which the end of the shadow is moving away from the man. The question asks for the rate at which the tip of the shadow is moving, which is the sum of the man's speed and the shadow's speed. So the tip of the shadow is moving at 5 ft/s + 12.5 ft/s = 17.5 ft/s when the man is 50 ft from the pole.

No 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 distance from the man to the pole as x, and the length of the shadow as y. We know that the rate of change of x (dx/dt) is 5 ft/s.

2. The situation forms a similar triangle, so we can set up the following equation: 15/(x+y) = 6/x.

3. Differentiate both sides with respect to time t to get the related rates. We get: -15(x+y)'/(x+y)^2 = -6x'/x^2.

4. Simplify the equation to get: (x+y)' = (x+y)(x'/x).

5. Substitute the known values into the equation: (x+y)' = (50+y)(5/50).

6. Solve for (x+y)', which is the speed of the tip of the shadow. We get: (x+y)' = 5 + 0.1y'.

7. Since we want to find y' when x = 50, we substitute x = 50 into the original equation to solve for y. We get: 15/(50+y) = 6/50, which gives y = 35.

8. Substitute y = 35 into the equation from step 6 to solve for y'. We get: y' = 5 + 0.1*35 = 8.5 ft/s.

So, the tip of the man's shadow is moving at a speed of 8.5 ft/s when he is 50 ft from the pole.

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

1. First, we need to set up a relationship between the quantities given. We can use similar triangles to do this. The height of the light pole is to the length of the shadow as the height of the man is to the distance between the man and the shadow tip. This gives us the equation: 15/(x + s) = 6/x, where x is the distance of the man from the pole and s is the length of the shadow.

2. We need to find ds/dt, the rate at which the tip of the shadow is moving. To do this, we first need to differentiate both sides of the equation with respect to time (t). This gives us: -15/(x + s)^2 * (dx/dt + ds/dt) = -6/x^2 * dx/dt.

3. We know that dx/dt, the speed of the man, is 5 ft/s. We can substitute this into the equation to get: -15/(x + s)^2 * (5 + ds/dt) = -30/x^2.

4. We are asked to find ds/dt when x = 50 ft. We can substitute this into the equation to get: -15/(50 + s)^2 * (5 + ds/dt) = -30/2500.

5. Solving this equation for ds/dt gives us: ds/dt = (30/2500 * (50 + s)^2 / 5) - 5.

6. However, we still don't know the value of s when x = 50 ft. We can find this by substituting x = 50 ft into the original equation to get: 15/(50 + s) = 6/50. Solving this equation for s gives us: s = 50 * 15/6 - 50 = 125 ft.

7. Finally, we can substitute s = 125 ft into the equation for ds/dt to get: ds/dt = (30/2500 * (50 + 125)^2 / 5) - 5 = 7 ft/s.

So, the tip of the man's shadow is moving at 7 ft/s when he is 50 ft from the pole.

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

1. First, we need to set up a relationship between the variables. We can use similar triangles to do this. The height of the light pole is to the length of the shadow as the height of the man is to the distance between the man and the pole. This gives us the equation: 15/(x + s) = 6/x, where x is the distance of the man from the pole and s is the length of the shadow.

2. Next, we differentiate both sides of the equation with respect to time (t). This gives us: -15/(x + s)^2 * (dx/dt + ds/dt) = -6/x^2 * dx/dt.

3. We know that dx/dt = 5 ft/s (the speed of the man) and we want to find ds/dt (the speed of the tip of the shadow). We can rearrange the equation to solve for ds/dt: ds/dt = ((-15/(x + s)^2) + (6/x^2)) * dx/dt.

4. Finally, we substitute the given values into the equation: x = 50 ft, dx/dt = 5 ft/s, and s = 45 ft (from the similar triangles, when x = 50 ft, s = 45 ft). This gives us: ds/dt = ((-15/(95)^2) + (6/50^2)) * 5 = 3.95 ft/s.

So, the tip of the man's shadow is moving at a speed of approximately 3.95 ft/s when he is 50 ft from the pole.