# The surface area of a sphere is decreasing at the constant rate of 3*pi sq. cm/sec . ? At what rate is the volume of the sphere decreasing at the instant its radius is 2 cm ?

Surface area of a sphere: [latex]A=4\pi r^2[/latex] Volume of a sphere: [latex]V=\dfrac43\pi r^3[/latex] Differentiate both with respect to an arbitrary variable for time: [latex]\displaystyle\frac{\mathrm dA}{\mathrm dt}=8\pi r\frac{\mathrm dr}{\mathrm dt}[/latex] [latex]\displaystyle\frac{\mathrm dV}{\mathrm dt}=4\pi r^2\frac{\mathrm dr}{\mathrm dt}[/latex] You're given that [latex]\dfrac{\mathrm dA}{\mathrm dt}=-3\pi[/latex] when [latex]r=2[/latex], so you can use this in the first equation to solve for [latex]\dfrac{\mathrm dr}{\mathrm dt}[/latex]. [latex]-3\pi=8\pi \times2\dfrac{\mathrm dr}{\mathrm dt}\implies\dfrac{\mathrm dr}{\mathrm dt}=-\dfrac3{16}[/latex] Now use this to find [latex]\dfrac{\mathrm dV}{\mathrm dt}[/latex]. [latex]\displaystyle\frac{\mathrm dV}{\mathrm dt}=4\pi \times2^2\times\left(-\dfrac3{16}\right)=-3\pi[/latex]