Verified Answer :

LammettHash

Surface area of a sphere: \(A=4\pi r^2\)
Volume of a sphere: \(V=\dfrac43\pi r^3\)

Differentiate both with respect to an arbitrary variable for time:
\(\displaystyle\frac{\mathrm dA}{\mathrm dt}=8\pi r\frac{\mathrm dr}{\mathrm dt}\)
\(\displaystyle\frac{\mathrm dV}{\mathrm dt}=4\pi r^2\frac{\mathrm dr}{\mathrm dt}\)

You're given that \(\dfrac{\mathrm dA}{\mathrm dt}=-3\pi\) when \(r=2\), so you can use this in the first equation to solve for \(\dfrac{\mathrm dr}{\mathrm dt}\).

\(-3\pi=8\pi \times2\dfrac{\mathrm dr}{\mathrm dt}\implies\dfrac{\mathrm dr}{\mathrm dt}=-\dfrac3{16}\)

Now use this to find \(\dfrac{\mathrm dV}{\mathrm dt}\).

\(\displaystyle\frac{\mathrm dV}{\mathrm dt}=4\pi \times2^2\times\left(-\dfrac3{16}\right)=-3\pi\)

obasola1

Answer:

9.42cm^3/s

Step-by-step explanation:

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

We know a sphere has a volume and area

V=4/3\(\pi r^3 and A=4\pi r^2\)

dA/dt=dA/drdr/dt............1

also dV/dt=dV/drdr/dt... ........

from equation 1

we can make dr/dt the subject of the equation

dr/dt=dA/dt/(dA/dr)

substitute into equation 2

dV/dt=dV/dr(dA/dt/(dA/dr))

dv/dr=\(4\pi r^2\)=43.142^2=50.24

dA/dt= 3pi

dA/dr=8p12=

3/1650.24

dV/dt=9.42cm^3/s

is the rate at which the volume is decreasing

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