A spherical balloon is being inflated at a rate of 10 cm3/min . How fast is the radius increasing when the radius is 5 cm? ( V = (4)/(3)pi r3 )

Mathematics

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Answer

110π cm/min\frac{1}{10\pi} \text{ cm/min}

Step-by-step solution

4 steps

  1. Identify the given information and the formula.

    We are given the rate of change of volume dVdt=10 cm3/min\frac{dV}{dt} = 10 \text{ cm}^3/\text{min} and the current radius r=5 cmr = 5 \text{ cm}. We need to find the rate of change of the radius drdt\frac{dr}{dt}.
    The formula for the volume of a sphere is V=43πr3V = \frac{4}{3}\pi r^3.

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

    Using the chain rule, we differentiate both sides of the volume equation with respect to tt:

    dVdt=ddt(43πr3)\frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right) dVdt=43π⋅3r2drdt\frac{dV}{dt} = \frac{4}{3}\pi \cdot 3r^2 \frac{dr}{dt} dVdt=4πr2drdt\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}

  3. Substitute the given values into the differentiated equation.

    We have dVdt=10\frac{dV}{dt} = 10 and r=5r = 5.

    10=4π(5)2drdt10 = 4\pi (5)^2 \frac{dr}{dt} 10=4π(25)drdt10 = 4\pi (25) \frac{dr}{dt} 10=100πdrdt10 = 100\pi \frac{dr}{dt}

  4. Solve for drdt\frac{dr}{dt}.

    drdt=10100π\frac{dr}{dt} = \frac{10}{100\pi} drdt=110π cm/min\frac{dr}{dt} = \frac{1}{10\pi} \text{ cm/min}

    The rate at which the radius is increasing when the radius is 5 cm is 110π\frac{1}{10\pi} cm/min.

    110π cm/min\frac{1}{10\pi} \text{ cm/min}
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