# Bases are located on the field 90 feet away from one another Jimmy is running at a speed of 10 ft/sec from second to third base. When Jimmy is halfway to third base, how quickly is the distance between him and home plate decreasing?

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Use the same picture, but relabel, so that B is second base, A is third and C is home plate. Also note the difference in speeds of the runners.

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We can use the Pythagorean theorem to find the distance between Jimmy and home plate at any given point. Let x be the distance from Jimmy to the second base. Then, according to the Pythagorean theorem:

(x^2 + 90^2 = (90 + 10t)^2)

Differentiating both sides with respect to time t, we get:

(2x \frac{dx}{dt} = 2(90 + 10t) \cdot 10)

At the halfway point to third base, (x = 45) feet. Substituting this value into the equation and solving for (\frac{dx}{dt}), we find:

(\frac{dx}{dt} = \frac{(90 + 10t) \cdot 10}{x})

Plugging in (x = 45) and (t = 90/10 = 9) (since Jimmy takes 9 seconds to run from second to third base), we get:

(\frac{dx}{dt} = \frac{(90 + 10(9)) \cdot 10}{45} = \frac{180}{3} = 60 \text{ ft/sec})

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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