This problem involves related rates. We can use similar triangles to set up an equation and then take derivatives to find the rate we want.

1. Set up the similar triangles: The height of the light pole to the length of its shadow is proportional to the height of the woman to the length of her shadow. This gives us the equation: 20/(50+x) = 6/x, where x is the length of the woman's shadow.

2. Solve for x: Cross-multiplying gives us 20x = 300 + 6x. Simplifying gives us 14x = 300, so x = 300/14 = 21.43 feet.

3. Differentiate both sides with respect to time (t): Differentiating gives us dx/dt = -20/(50+x)^2 * ds/dt, where s is the distance from the woman to the pole and ds/dt is the rate at which she is walking away from the pole.

4. Plug in the given values: We know that ds/dt = 7 ft/sec and x = 21.43 feet. Plugging these into the equation gives us dx/dt = -20/(50+21.43)^2 * 7 = -0.12 ft/sec.

So, the distance from the base of the pole to the tip of the shadow is decreasing at a rate of 0.12 feet per second when the woman is 50 feet from the base of the pole.

Question

This problem involves related rates. We can use similar triangles to set up an equation and then take derivatives to find the rate we want.

1. Set up the similar triangles: The height of the light pole to the length of its shadow is proportional to the height of the woman to the length of her shadow. This gives us the equation: 20/(50+x) = 6/x, where x is the length of the woman's shadow.

2. Solve for x: Cross-multiplying gives us 20x = 300 + 6x. Simplifying gives us 14x = 300, so x = 300/14 = 21.43 feet.

3. Differentiate both sides with respect to time (t): Differentiating gives us dx/dt = -20/(50+x)^2 * ds/dt, where s is the distance from the woman to the pole and ds/dt is the rate at which she is walking away from the pole.

4. Plug in the given values: We know that ds/dt = 7 ft/sec and x = 21.43 feet. Plugging these into the equation gives us dx/dt = -20/(50+21.43)^2 * 7 = -0.12 ft/sec.

So, the distance from the base of the pole to the tip of the shadow is decreasing at a rate of 0.12 feet per second when the woman is 50 feet from the base of the pole.

Knowee AI · Accepted Answer

This problem involves related rates. We can use similar triangles to set up an equation and then take derivatives to find the rate we want.

1. Set up the similar triangles: The height of the light pole to the length of its shadow is proportional to the height of the woman to the length of her shadow. This gives us the equation: 20/(50+x) = 6/x, where x is the length of the woman's shadow.

2. Solve for x: Cross-multiplying gives us 20x = 300 + 6x. Simplifying gives us 14x = 300, so x = 300/14 = 21.43 feet.

3. Differentiate both sides with respect to time (t): Differentiating gives us dx/dt = -20/(50+x)^2 * ds/dt, where s is the distance from the woman to the pole and ds/dt is the rate at which she is walking away from the pole.

4. Plug in the given values: We know that ds/dt = 7 ft/sec and x = 21.43 feet. Plugging these into the equation gives us dx/dt = -20/(50+21.43)^2 * 7 = -0.12 ft/sec.

So, the distance from the base of the pole to the tip of the shadow is decreasing at a rate of 0.12 feet per second when the woman is 50 feet from the base of the pole.

) A street light is at the top of a 20 foot tall pole. A 6 foot tall woman walks away from the pole with a speed of 7 ft/sec along a straight path. At what rate is the distance from the base of pole to the tip of the shadow changing when she is 50 ft from the base of the pole?

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