V = (4/3) PI * R^3

dV/dt = 4 PI R^2 * dR/dt

4.9 = 4 * pi * (1.5)^2 * dr/dt

4.9 / [ 4 * pi * 1/.5^2] = dr/dt

Kendra M.

asked • 12dA spherical balloon is being inflated by a compressor that is pumping air at a rate of At 4.9ft^3 /min . what rate is the radius of the balloon increasing when the radius is 1.5 ft Round to three decimal places.

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V = (4/3) PI * R^3

dV/dt = 4 PI R^2 * dR/dt

4.9 = 4 * pi * (1.5)^2 * dr/dt

4.9 / [ 4 * pi * 1/.5^2] = dr/dt

Let the rate that the volume increases/decreases be represented by:

*dV/dt* = 4.9

Your goal is to solve for *dr/dt* (the rate at which the radius is increasing/decreasing).

Start with the general formula for the volume of a sphere:

*V *= (4/3)*π r *^{3}

Differentiate both sides with respect to *t *(time).

*dV/dt = 4π r *^{2}*(dr/dt)*

Plug in the information provided in the question, *dV/dt* = 4.9 and *r *=* *1.5.

**4.9 = 4π (1.5)**

Solve for the expression *dr/dt*, which represents the rate the radius is changing.

Ask me if you are unsure about anything in this explanation.

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