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The radius of a sphere is increasing at 0.5 cm/s. When r = 12 cm, find the rates at which the surface area and the volume are increasing.

Mathematics

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Here's the solution to the first rate problem:

1. The radius of a sphere is increasing at 0.5 cm/s0.5 \text{ cm/s}. When r=12 cmr = 12 \text{ cm}, find the rates at which the surface area and the volume are increasing.

Given:
drdt=0.5 cm/s\frac{dr}{dt} = 0.5 \text{ cm/s}
r=12 cmr = 12 \text{ cm}

Formulas for a sphere:
Surface Area: S=4πr2S = 4\pi r^2
Volume: V=43πr3V = \frac{4}{3}\pi r^3

Part a· 2 steps

Rate of increase of Surface Area (dSdt\frac{dS}{dt})

  1. Differentiate the surface area formula with respect to time tt.

    S=4πr2S = 4\pi r^2 dSdt=ddt(4πr2)\frac{dS}{dt} = \frac{d}{dt}(4\pi r^2)

    Using the chain rule, we get:

    dSdt=4π(2r)drdt\frac{dS}{dt} = 4\pi (2r) \frac{dr}{dt} dSdt=8πrdrdt\frac{dS}{dt} = 8\pi r \frac{dr}{dt}

  2. Substitute the given values into the derivative.

    dSdt=8π(12 cm)(0.5 cm/s)\frac{dS}{dt} = 8\pi (12 \text{ cm}) (0.5 \text{ cm/s}) dSdt=8π(6 cm2/s)\frac{dS}{dt} = 8\pi (6 \text{ cm}^2/\text{s}) dSdt=48\picm\textsuperscript2/s\frac{dS}{dt} = \boxed{\textbf{48\pi cm\textsuperscript{2}/s}}

Part b· 2 steps

Rate of increase of Volume (dVdt\frac{dV}{dt})

  1. Differentiate the volume formula with respect to time tt.

    V=43πr3V = \frac{4}{3}\pi r^3 dVdt=ddt(43πr3)\frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)

    Using the chain rule, we get:

    dVdt=43π(3r2)drdt\frac{dV}{dt} = \frac{4}{3}\pi (3r^2) \frac{dr}{dt} dVdt=4πr2drdt\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}

  2. Substitute the given values into the derivative.

    dVdt=4π(12 cm)2(0.5 cm/s)\frac{dV}{dt} = 4\pi (12 \text{ cm})^2 (0.5 \text{ cm/s}) dVdt=4π(144 cm2)(0.5 cm/s)\frac{dV}{dt} = 4\pi (144 \text{ cm}^2) (0.5 \text{ cm/s}) dVdt=4π(72 cm3/s)\frac{dV}{dt} = 4\pi (72 \text{ cm}^3/\text{s}) dVdt=288\picm\textsuperscript3/s\frac{dV}{dt} = \boxed{\textbf{288\pi cm\textsuperscript{3}/s}}
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