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When the growth of a spherical cell depends on the flow of nutrients through the surface, it is reasonable to assume that the growth rate, dV/dtd V / d t, is proportional to the surface area, S. Assume that for a particular cell dV/dt=13Sd V / d t=\dfrac{1}{3} \cdot S. At what rate is its radius rr increasing?

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In this exercise, let us find the rate at which the radius rr of a spherical cell is increasing. Given that the growth of the spherical cell depends on the transfer of nutrients through its surface. We assume that the growth rate, dV/dtdV/dt is proportional to the surface area such that: dV/dt=13SdV/dt = \frac{1}{3}\cdot S.
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