An inverted cone has a diameter of 42 in and a height of 15in. If the water flowing out of the vertex of the container at a rate of 35$\displaystyle \pi$ $\displaystyle in^3$/sec , how fast is the depth of the water dropping when the height is 5in?

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- Mar 19th 2010, 09:40 PMyoman360An inverted cone has a diameter of 42 in and a height of 15in. If the water flowing o
An inverted cone has a diameter of 42 in and a height of 15in. If the water flowing out of the vertex of the container at a rate of 35$\displaystyle \pi$ $\displaystyle in^3$/sec , how fast is the depth of the water dropping when the height is 5in?

- Mar 19th 2010, 11:08 PMsa-ri-ga-ma
Volume of the cone V = 1/3*π*r^2*h.....(1)

You want to find dh/dt. So you have to find r in terms of h.

r and h are proportional tp R and H, where R = 21" and H = 15".

So R/H = r/h.

So r = (R/H)*h.

Substitute this value in the eq.(1) and find dV/dt.

dV/dt is given. Find dh/dt. - Mar 20th 2010, 03:14 PMyoman360
I followed your steps and this is what I get:

$\displaystyle r=\frac{21}{15} h$

$\displaystyle v=\frac{1}{3}\pi(\frac{21}{15}h)^2*h$

simplify so $\displaystyle v=\pi\frac{49}{75}h^3$

$\displaystyle dv/dt=\pi\frac{49}{25}h^2* dh/dt$

plug in h and dv/dt

$\displaystyle 35\pi=\pi\frac{49}{25}(15)^2* dh/dt$

solve for dh/dt i get

$\displaystyle dh/dt=\frac{5}{63} in/sec$

the problem is that the answer key says the answer is $\displaystyle \frac{5}{7}$ in/sec

what did i do incorrectly? - Mar 20th 2010, 03:18 PMyoman360