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The volume of a sphere is increasing at a constant rate of 5409 cubic meters per second. At the instant when the volume of the sphere is 32763276 cubic meters, what is the rate of change of the surface area of the sphere?

User Tuomur
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Final answer:

To answer the student's question, we must calculate the rate of change of a sphere's surface area using differentiation and related rates, given the constant rate of volume increase.

Step-by-step explanation:

The question is asking about the rate at which the surface area of a sphere changes given that the volume of the sphere is increasing at a constant rate. To find this rate, we need to use calculus, specifically the concept of related rates, and the formulae for the volume and surface area of a sphere. The formula for the volume V of a sphere is given by V = (4/3)πr^3 and the formula for the surface area S is S = 4πr^2. Since the volume is changing at a rate of 5409 cubic meters per second, we denote this by dV/dt = 5409 m^3/sec. To find the rate of change of the surface area (dS/dt), we need to find the relationship between the surface area and the volume. First, express the radius r in terms of the volume, r = (3V/4π)^(1/3), and then differentiate the surface area with respect to time using the chain rule.

User RedRocket
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