# An inverted cone has a diameter of 42 in and a height of 15in. If the water flowing o

• Mar 19th 2010, 09:40 PM
yoman360
An 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 $\pi$ $in^3$/sec , how fast is the depth of the water dropping when the height is 5in?
• Mar 19th 2010, 11:08 PM
sa-ri-ga-ma
Quote:

Originally Posted by yoman360
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 $\pi$ $in^3$/sec , how fast is the depth of the water dropping when the height is 5in?

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 PM
yoman360
Quote:

Originally Posted by sa-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.

I followed your steps and this is what I get:
$r=\frac{21}{15} h$

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

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

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

plug in h and dv/dt

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

solve for dh/dt i get

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

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

what did i do incorrectly?
• Mar 20th 2010, 03:18 PM
yoman360
Quote:

Originally Posted by yoman360
I followed your steps and this is what I get:

plug in h $dv/dt=\pi\frac{49}{25}(15)^2* dh/dt$

Never mind. h=5 and H= 15
I figured it out I was suppose to plug in 5 here instead if 15 then solving for dh/dt i got 5/7 in/sec

Thanks for the help (Rofl)