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July 10th, 2012, 03:45 AM   #1
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Rate problem

Quote:
 A snowball melts in volume @ 1 cm³/min. At what rate is the diameter decreasing when the diameter is 10 cm?
Volume of a sphere: $\frac{4}{3}\pi r^{3}$
Volume of a sphere in terms of diameter: $\frac{\pi d^{3}}{6}$

Rate of volume: $\frac{d}{d(time)}volume= -1 cm^{3}/min.$

Rate of diameter: $\frac{d}{d(time)}10 cm= ?$

I'm not sure what to do at this point

 July 10th, 2012, 05:14 AM #2 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: Rate problem [color=#000000]$V(t)=\frac{4}{3}\pi r(t)^3\Leftrightarrow V(t)=\frac{4}{3}\pi \frac{d(t)^3}{8}=\frac{\pi d(t)^3}{6}$ $V'(t)=\frac{\pi d(t)^2d#39;(t)}{2}$ $V'(t)=-1cm^3/min$ when d=10cm at the time instant $t_0$ then $-1=\frac{\pi 10^2 d'(t_0)}{2}\Leftrightarrow d#39;(t_0)=-\frac{2}{100\pi}cm/min.$[/color]
 July 10th, 2012, 05:28 AM #3 Senior Member   Joined: Jan 2010 Posts: 205 Thanks: 0 Re: Rate problem ZardoZ, how were you able to just put in (time) into the equation of the volume? I don't see how you can get the function Volume(time)
July 10th, 2012, 11:02 AM   #4
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Re: Rate problem

Hello, daigo!

Quote:
 $\text{A snowball melts in volume at }1\text{ cm}^3\text{/min.} \;\;\text{At what rate is the diameter decreasing when the diameter is 10 cm?}$

I would put everything in terms of the radius, and switch to the diameter at the end.

$\text{W\!e have: }\:V \:=\:\frac{4}{3}\pi r^3$

$\text{Then: }\:\frac{dV}{dt} \;=\;4\pi r^2\,\frac{dr}{dt}\;\;[1]$

$\text{W\!e are told: }\:\frac{dV}{dt} \,=\,-1,\;\;r \,=\,5$

$\text{Then [1] becomes: }\:-1 \:=\:4\pi(5^2)\,\frac{dr}{dt} \;\;\;\Rightarrow\;\;\;\frac{dr}{dt} \:=\:-\,\frac{1}{100\pi}\text{ cm/min}$

$\text{The rate of change of the diameter }(D)\text{ is: }\:\frac{d\!D}{dt} \;=\;-\,\frac{1}{50\pi}\text{ cm/min}$

 July 10th, 2012, 11:11 AM #5 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Rate problem You correctly found the volume V as a function of the diameter d is: $V=\frac{\pi}{6}d^3$ Differentiate with respect to time t: $\frac{dV}{dt}=\frac{\pi}{6}$$3d^2\cdot\frac{dd}{dt }$$=\frac{\pi}{2}d^2\cdot\frac{dd}{dt}$ We are given: $\frac{dV}{dt}=-1\text{ \frac{cm^3}{min}}$ hence: $-1\text{ \frac{cm^3}{min}}=\frac{\pi}{6}$$3d^2\cdot\frac{dd }{dt}$$=\frac{\pi}{2}d^2\cdot\frac{dd}{dt}$ Thus, we find: $\frac{dd}{dt}=-\frac{2}{\pi d^2}\:\text{\frac{cm}{min}}$ And so: $\frac{dd}{dt}\|_{d=10}=-\frac{2}{\pi\cdot10^2}\:\text{\frac{cm}{min}}=-\frac{1}{50\pi}\:\text{\frac{cm}{min}}$

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