My Math Forum water is poured into an inverted circular cone of base radiu

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 October 31st, 2009, 03:12 AM #1 Newbie   Joined: Oct 2009 Posts: 7 Thanks: 0 water is poured into an inverted circular cone of base radiu water is poured into an inverted circular cone of base radius 5 cm and height 15 cm, at the rate of 10 cm/sec. calculate a) the rate of increase of the height of water level b) the rate of increase of the surface area of the water when the water level is 4cm height please solve it their answers are Ans (a)45/8pi (b)5 square centi meter
 October 31st, 2009, 04:54 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,599 Thanks: 941 Math Focus: Elementary mathematics and beyond a) Volume, $V$, of a cone is $V\,=\,\frac{1}{3}\pi r^2h$, where $r$ is radius and $h$ is height. From the information given $r\,=\,\frac{1}{3}h$ and $\frac{dV}{dt}\,=\,10cm^3$ Volume, in terms of height, is $V\,=\,\frac{1}{27}\pi h^3$. Implicit differentiation with respect to time: $\frac{dV}{dt}\,=\,\frac{1}{9}\pi h^2\,\frac{dh}{dt}\,\rightarrow\,10\,=\,\frac{16}{ 9}\pi\frac{dh}{dt}\,\rightarrow\,\frac{dh}{dt}\,=\ ,\frac{45}{8\pi}cm$. b) 1. Rewrite the volume equation in terms of radius: $V\,=\,\pi r^3$. 2. Find $\frac{dr}{dt}$--> $\frac{dV}{dt}\,=\,3\pi r^2\,\frac{dr}{dt}\,\rightarrow\,10\,=\,\frac{16}{ 3}\pi\frac{dr}{dt}\,\rightarrow\,\frac{dr}{dt}\,=\ ,\frac{30}{16\pi}$ 3. Find $\frac{dS}{dt}$, where $S$ is surface area. $S\,=\,\pi r^2\,\rightarrow\,\frac{dS}{dt}\,=\,2\pi r\frac{dr}{dt}=5cm^2$
 October 31st, 2009, 05:31 AM #3 Newbie   Joined: Oct 2009 Posts: 7 Thanks: 0 Re: water is poured into an inverted circular cone of base radiu Bro Thanks To Solve my problem, now I fully understood the problem Thanks Again

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