the rate of change of volume of a sphere with respect to its surface area when radiis is 4cm is

winner · January 14th, 2011, 07:18 AM

Gas is being pumped into a spherical balloon at the rate of 100 cubic centimeter per min. When the radius is 10 cm, the surface area changed at a rate of ???

ANS - 20 square centimeter per minute

Thanx for reading

greg1313 · January 14th, 2011, 08:08 AM

$\frac{dV}{dt}\,=\,100$ rate of change of volume with respect to time.

$\frac{dr}{dt}$ rate of change of radius with respect to time.

$\frac{ds}{dt}$ rate of change of surface area with respect to time (what we want to compute).

Formula for volume of sphere: $\frac{4}{3}\pi r^3$

Formula for surface area of sphere: $4\pi r^2$

With radius at 10 cm., calculate dr/dt using derivative of formula for volume:

$\frac{dV}{dt}\,=\,100\,=\,4\pi r^2\frac{dr}{dt}\,=\,4\pi (100)\frac{dr}{dt}\,\Rightarrow\,\frac{dr}{dt}\,=\ ,\frac{1}{4\pi}$

With radius at 10 cm., calculate ds/dt using derivative of formula for surface area:

$\frac{ds}{dt}\,=\,8\pi r\frac{dr}{dt}\,=\,8\pi (10)\,\cdot\,\frac{1}{4\pi}\,=\,20$

sivela · January 14th, 2011, 08:18 AM

This should be in calculus not in geometry

$\text{\frac{dv}{dt}=100\frac{cm^3}{min}}$

$\text{r= 10cm and 4{\pi}r^2 = S.A}$

$\text{v= \frac{4}{3}\pi{r^3}}$

$\text{ \frac{dv}{dt}= \frac{dr}{dt}{4\pi}{r^2}=100\frac{cm^3}{min}}$

$\text{\frac{100}{4{\pi}r^2}=\frac{dr}{dt}}$

$\text{\frac{1}{4\pi}(\frac{cm}{min})=\frac{dr}{dt} }$

$\text{\frac{1}{4\pi}(\frac{cm}{min})(8{\pi}r)=\fra c{d.S.A}{dt}}$

substituting 10cm for r we have

$\text{\frac{1}{4\pi}(\frac{cm}{min})(8{\pi}*10cm)= \frac{d.S.A}{dt}=20\frac{cm^2}{min}}$

skipjack · January 14th, 2011, 11:30 AM

Quote:

Originally Posted by sivela

$\text{\frac{1}{4\pi}=\frac{dr}{dt}}$

Since you're giving units, that should be $\text{\frac{1}{4\pi} \frac{cm}{min}= \frac{dr}{dt}}$ (and similarly for your next line).

$\frac{ds}{dr}$ should be $\frac{ds}{dt}$ in greg1313's last line.

January 14th, 2011, 07:18 AM	#1
winner Newbie Joined: Jan 2011 Posts: 17 Thanks: 0	Surface area and volume Gas is being pumped into a spherical balloon at the rate of 100 cubic centimeter per min. When the radius is 10 cm, the surface area changed at a rate of ??? ANS - 20 square centimeter per minute Thanx for reading
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January 14th, 2011, 08:08 AM	#2
greg1313 Global Moderator Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,932 Thanks: 1127 Math Focus: Elementary mathematics and beyond	Re: Surface area and volume $\frac{dV}{dt}\,=\,100$ rate of change of volume with respect to time. $\frac{dr}{dt}$ rate of change of radius with respect to time. $\frac{ds}{dt}$ rate of change of surface area with respect to time (what we want to compute). Formula for volume of sphere: $\frac{4}{3}\pi r^3$ Formula for surface area of sphere: $4\pi r^2$ With radius at 10 cm., calculate dr/dt using derivative of formula for volume: $\frac{dV}{dt}\,=\,100\,=\,4\pi r^2\frac{dr}{dt}\,=\,4\pi (100)\frac{dr}{dt}\,\Rightarrow\,\frac{dr}{dt}\,=\ ,\frac{1}{4\pi}$ With radius at 10 cm., calculate ds/dt using derivative of formula for surface area: $\frac{ds}{dt}\,=\,8\pi r\frac{dr}{dt}\,=\,8\pi (10)\,\cdot\,\frac{1}{4\pi}\,=\,20$
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January 14th, 2011, 08:18 AM	#3
sivela Senior Member Joined: Jan 2009 Posts: 344 Thanks: 3	Re: Surface area and volume This should be in calculus not in geometry $\text{\frac{dv}{dt}=100\frac{cm^3}{min}}$ $\text{r= 10cm and 4{\pi}r^2 = S.A}$ $\text{v= \frac{4}{3}\pi{r^3}}$ $\text{ \frac{dv}{dt}= \frac{dr}{dt}{4\pi}{r^2}=100\frac{cm^3}{min}}$ $\text{\frac{100}{4{\pi}r^2}=\frac{dr}{dt}}$ $\text{\frac{1}{4\pi}(\frac{cm}{min})=\frac{dr}{dt} }$ $\text{\frac{1}{4\pi}(\frac{cm}{min})(8{\pi}r)=\fra c{d.S.A}{dt}}$ substituting 10cm for r we have $\text{\frac{1}{4\pi}(\frac{cm}{min})(8{\pi}*10cm)= \frac{d.S.A}{dt}=20\frac{cm^2}{min}}$
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