Taima was given this problem: The radius r(t) of the base of a cone is decreasing at a rate of 2 centimeters per minute. The height of the cone is fixed at 12 centimeters. At a certain instant t_(0), the radius is 13 centimeters. What is the rate of change of the surface area S(t) of the cone at that instant? Which equation should Taima use to solve the problem? Choose 1 answer: (A) S(t)=pi[r(t)]^(2)+2pi*r(t)*12 (B) S(t)=pi[r(t)]^(2)*12 (C) S(t)=pi[r(t)]^(2)+pi*r(t)sqrt([r(t)]^(2)+12^(2)) (D) S(t)=(pi[r(t)]^(2)*12)/(3)

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Taima was given this problem: The radius  r(t) of the base of a cone is decreasing at a rate of 2 centimeters per minute. The height of the cone is fixed at 12 centimeters. At a certain instant  t_(0), the radius is 13 centimeters. What is the rate of change of the surface area  S(t) of the cone at that instant? Which equation should Taima use to solve the problem? Choose 1 answer: (A)  S(t)=pi[r(t)]^(2)+2pi*r(t)*12 (B)  S(t)=pi[r(t)]^(2)*12 (C)  S(t)=pi[r(t)]^(2)+pi*r(t)sqrt([r(t)]^(2)+12^(2)) (D)  S(t)=(pi[r(t)]^(2)*12)/(3)

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Surface Area Formula: To find the rate of change of the surface area of the cone, we need to use the formula for the surface area of a cone, which includes both the base area and the lateral surface area. The base area is given by πr^2, where r is the radius, and the lateral surface area is given by πrl, where l is the slant height of the cone. Since the height is fixed, we can use the Pythagorean theorem to find the slant height l in terms of r, which is l = sqrt(r^2 + h^2), where h is the height of the cone. The correct formula for the surface area of a cone is therefore S(t) = πr^2 + πrl. Substituting l with sqrt(r^2 + h^2), we get S(t) = πr^2 + πr*sqrt(r^2 + h^2). This corresponds to option (C).Differentiating Base Area: Now that we have identified the correct formula, we can differentiate S(t) with respect to time t to find the rate of change of the surface area. We will use the chain rule for differentiation since r is a function of t.Differentiating Lateral Area: First, let's differentiate the base area πr^2 with respect to t. The derivative of πr^2 with respect to r is 2πr, and then we multiply by dr/dt (the rate of change of the radius with respect to time) to get the rate of change of the base area with respect to time. Since dr/dt is given as -2 cm/min, the rate of change of the base area is 2πr * (-2) = -4πr.Calculating Values: Next, we differentiate the lateral surface area πr*sqrt(r^2 + h^2) with respect to t. Using the product rule and chain rule, we get the derivative with respect to r as π*sqrt(r^2 + h^2) + πr*(1/2)*(1/sqrt(r^2 + h^2))*(2r). Then we multiply by dr/dt to get the rate of change of the lateral surface area with respect to time. Substituting dr/dt = -2 and h = 12, we get the rate of change of the lateral surface area as π*(-2)*sqrt(13^2 + 12^2) + π*13*(1/2)*(1/sqrt(13^2 + 12^2))*(2*13)*(-2).Simplifying Expression: Now we need to calculate the actual values. Substituting r = 13 cm and h = 12 cm into the expression we derived for the rate of change of the lateral surface area, we get: π*(-2)*sqrt(13^2 + 12^2) + π*13*(1/2)*(1/sqrt(13^2 + 12^2))*(2*13)*(-2) = -2π*sqrt(169 + 144) + π*13*(1/2)*(1/sqrt(169 + 144))*(2*13)*(-2).Total Rate of Change: Simplifying the expression, we get: -2π*sqrt(313) + π*13*(1/2)*(1/sqrt(313))*(2*13)*(-2) = -2π*sqrt(313) - π*13^2/sqrt(313).Adding the rate of change of the base area to the rate of change of the lateral surface area gives us the total rate of change of the surface area: Total rate of change = -4π*13 - 2π*sqrt(313) - π*13^2/sqrt(313).

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