A 25-meter ladder is sliding down a vertical wall so the distance between the bottom of the ladder and the wall is increasing at 3.5 meters per minute. At a certain instant, the top of the ladder is 7 meters from the ground. What is the rate of change of the distance between the top of the ladder and the ground at that instant (in meters per minute)? Choose 1 answer: (A) -3.5 (B) -12 (C) -7 (D) -24

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Given Information: We have a 25-meter ladder sliding down a wall. We're given the rate at which the bottom moves away from the wall (3.5 m/min) and the height of the top of the ladder above the ground (7 m) at a certain instant.Pythagorean Theorem: We can use the Pythagorean theorem to find the distance of the bottom of the ladder from the wall at that instant. Let's call this distance 'x'. So, we have $ x^2 + 7^2 = 25^2 $.Calculating Distance: Calculating 'x', we get $ x^2 = 25^2 - 7^2 = 625 - 49 = 576 $. So, $ x = \sqrt{576} = 24 $ meters.Rate of Change: Now, we need to find the rate of change of the height of the ladder with respect to time, which is $ \frac{dy}{dt} $, where 'y' is the height of the ladder above the ground.Implicit Differentiation: Using implicit differentiation on the Pythagorean theorem $ x^2 + y^2 = 25^2 $, we get $ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 $.Substitute Values: We know $ \frac{dx}{dt} = 3.5 $ m/min (the rate at which the bottom moves away from the wall) and at the instant we're considering, $ y = 7 $ m.Solving Equation: Plugging in the values, we get $ 2(24)(3.5) + 2(7) \frac{dy}{dt} = 0 $.Final Rate Calculation: Solving for $ \frac{dy}{dt} $, we get $ \frac{dy}{dt} = -\frac{2(24)(3.5)}{2(7)} $.Simplifying, $ \frac{dy}{dt} = -\frac{168}{14} = -12 $ meters per minute.

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