A -meter ladder is sliding down a vertical wall so the distance between the top of the ladder and the ground is decreasing at meters per minute. At a certain instant, the bottom of the ladder is meters from the wall. What is the rate of change of the area formed by the ladder at that instant (in square meters per minute)?

Question

A    -meter ladder is sliding down a vertical wall so the distance between the top of the ladder and the ground is decreasing at     meters per minute. At a certain instant, the bottom of the ladder is     meters from the wall. What is the rate of change of the area formed by the ladder at that instant (in square meters per minute)?

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Define Variables: Let's denote the length of the ladder as L meters, the rate at which the top of the ladder is sliding down as T meters per minute, and the distance of the bottom of the ladder from the wall at a certain instant as B meters. We are asked to find the rate of change of the area A formed by the ladder, the wall, and the ground at that instant.  The area A of the right-angled triangle formed by the ladder, the wall, and the ground can be expressed as: A = (1/2) * height * base Since the ladder is sliding down, the height is changing with time, and so is the base. We need to find dA/dt, the rate of change of the area with respect to time.  First, we need to express the height and base in terms of the length of the ladder and the distance of the bottom from the wall using the Pythagorean theorem: L^2 = height^2 + base^2Pythagorean Theorem: At the instant we are considering, the height of the ladder is decreasing at T meters per minute, which is the rate of change of the height with respect to time (dh/dt = -T). We need to find the rate of change of the base with respect to time (db/dt).  Differentiating both sides of the Pythagorean theorem with respect to time, we get: 0 = 2 * height * (dh/dt) + 2 * base * (db/dt) Since dh/dt = -T, we can solve for db/dt: base * (db/dt) = -height * (dh/dt) db/dt = -(height/base) * TRate of Change Calculation: Now we can find the rate of change of the area A with respect to time (dA/dt) by differentiating the area formula with respect to time: dA/dt = (1/2) * (height * (db/dt) + base * (dh/dt))  Substituting the values of dh/dt and db/dt from the previous steps, we get: dA/dt = (1/2) * (height * (-(height/base) * T) + base * (-T)) dA/dt = (1/2) * (-height^2/base * T - base * T)Simplify Expression: We can simplify the expression for dA/dt by factoring out -T/2: dA/dt = -T/2 * (height^2/base + base)  Since we know that L^2 = height^2 + base^2, we can express height^2/base as (L^2 - base^2)/base: dA/dt = -T/2 * ((L^2 - base^2)/base + base) dA/dt = -T/2 * (L^2/base - base + base) dA/dt = -T/2 * (L^2/base)Final Calculation: Finally, we substitute the given values for L, T, and B (where B is the base at the instant we are considering) into the expression for dA/dt: dA/dt = -T/2 * (L^2/B)  This gives us the rate of change of the area formed by the ladder at that instant in square meters per minute.

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