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