Tom was given this problem: The side s(t) of a square is decreasing at a rate of 2 kilometers per hour. At a certain instant t_(0), the side is 9 kilometers. What is the rate of change of the area A(t) of the square at that instant? Which equation should Tom use to solve the problem? Choose 1 answer: (A) A(t)=4*s(t) (B) A(t)=[s(t)]^(3) (C) A(t)=[s(t)]^(2) (D) [A(t)]^(2)=[s(t)]^(2)+[s(t)]^(2)

Question

Tom was given this problem: The side  s(t) of a square is decreasing at a rate of 2 kilometers per hour. At a certain instant  t_(0), the side is 9 kilometers. What is the rate of change of the area  A(t) of the square at that instant? Which equation should Tom use to solve the problem? Choose 1 answer: (A)  A(t)=4*s(t) (B)  A(t)=[s(t)]^(3) (C)  A(t)=[s(t)]^(2) (D)  [A(t)]^(2)=[s(t)]^(2)+[s(t)]^(2)

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Understand Relationship: Understand the relationship between the side of a square and its area. The area of a square is given by the square of its side length. Therefore, the correct equation relating the side s(t) and the area A(t) of the square is A(t) = [s(t)]^2.Choose Correct Equation: Choose the correct equation from the given options. Option (C) A(t) = [s(t)]^2 is the correct equation because it represents the area of a square as the square of its side length.Use Chain Rule: Use the chain rule to find the rate of change of the area with respect to time. We know that s(t) is decreasing at a rate of 2 km/h, so ds/dt = -2 km/h. To find dA/dt, we differentiate A(t) with respect to t using the chain rule: dA/dt = 2 * s(t) * ds/dt.Substitute Given Values: Substitute the given values into the derivative to find the rate of change of the area at the instant t_0. At the instant t_0, s(t) = 9 km and ds/dt = -2 km/h. Plugging these values into the derivative, we get: dA/dt = 2 * 9 km * (-2 km/h) = -36 km^2/h.

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