Given that the volume \(V\) of a sphere with radius \(r\) is given by the formula:
\[V = \frac{4}{3}\pi r^3\]
Differentiating both sides with respect to time \(t\), we get:
\[\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}\]
Given \(\frac{dV}{dt} = -4 \text{ cm}^3\), and the radius \(r = 6 \text{ cm}\), we can solve for \(\frac{dr}{dt}\):
\[-4 = 4\pi (6)^2\frac{dr}{dt}\]
\[-4 = 144\pi \frac{dr}{dt}\]
\[\frac{dr}{dt} = \frac{-4}{144\pi}\]
\[\frac{dr}{dt} = \frac{-1}{36\pi}\]
To find the rate at which the surface area \(A\) is diminishing, we use the formula for the surface area of a sphere:
\[A = 4\pi r^2\]
Differentiating both sides with respect to time \(t\), we get:
\[\frac{dA}{dt} = 8\pi r\frac{dr}{dt}\]
Given the radius \(r = 6 \text{ cm}\) and \(\frac{dr}{dt} = \frac{-1}{36\pi}\), we can calculate the rate at which the surface area is diminishing:
\[\frac{dA}{dt} = 8\pi (6)\left(\frac{-1}{36\pi}\right)\]
\[\frac{dA}{dt} = \frac{-8}{6}\]
\[\frac{dA}{dt} = \frac{-4}{3} \text{ cm}^2/\text{s}\]
Therefore, when the radius of the balloon is 6 cm, the surface area is diminishing at a rate of \(\frac{4}{3} \text{ cm}^2/\text{s}\).
A spherical balloon is decreasing in volume at a rate of 4cm^3 .find the rate at which the surface area is diminishing when the radius is 6 cm
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1 answer