Respuesta :

LammettHash
Surface area of a sphere: [tex]A=4\pi r^2[/tex]
Volume of a sphere: [tex]V=\dfrac43\pi r^3[/tex]

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

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

[tex]-3\pi=8\pi \times2\dfrac{\mathrm dr}{\mathrm dt}\implies\dfrac{\mathrm dr}{\mathrm dt}=-\dfrac3{16}[/tex]

Now use this to find [tex]\dfrac{\mathrm dV}{\mathrm dt}[/tex].

[tex]\displaystyle\frac{\mathrm dV}{\mathrm dt}=4\pi \times2^2\times\left(-\dfrac3{16}\right)=-3\pi[/tex]
obasola1

Answer:

9.42cm^3/s

Step-by-step explanation:

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 ?

We know a sphere has a volume and area

V=4/3[tex]\pi r^3 and A=4\pi r^2[/tex]

dA/dt=dA/dr*dr/dt............1

also dV/dt=dV/dr*dr/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=[tex]4\pi r^2[/tex]=4*3.14*2^2=50.24

dA/dt= 3*pi

dA/dr=8*p1*2=

3/16*50.24

dV/dt=9.42cm^3/s

is the rate at which the volume is decreasing

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