Hello, janedoe!
A conical tank is 12 ft across the top and 17 ft deep.
If water is flowing into the tank at a rate of 8 ft³/min,
find the rate of change of the depth when the water is 2 ft deep. Code:
: 6 :
- *-------+-------*
: \ | /
: \ | r /
: *----+----*
17 \ | /
: \ h| /
: \ | /
: \|/
- *
The volume of the water is: .$\displaystyle V \:=\:\frac{\pi}{3}r^2h$ .[1]
From the similar right triangles: .$\displaystyle \frac{r}{h} \:=\:\frac{6}{17} \quad\Rightarrow\quad r \:=\:\frac{6}{17}\,h$
Substitute into [1]: .$\displaystyle V \;=\;\frac{\pi}{3}\left(\frac{6}{17}\,h\right)\!\! ^2h \quad\Rightarrow\quad V \:=\:\frac{12\pi}{289}\,h^3$
Differentiate with respect to time: .$\displaystyle \frac{dV}{dt} \:=\:\frac{36\pi}{289}\,h^2\,\frac{dh}{dt}$
Since $\displaystyle \frac{dV}{dt} = 8,\;h = 2$. we have: .$\displaystyle 8 \:=\:\frac{36\pi}{289}(2^2)\frac{dh}{dt}$
. . Therefore: .$\displaystyle \frac{dh}{dt} \:=\:\frac{289}{18\pi}$ ft/min.
And thank you, thank you for sharing your work and your thoughts!
You put in a lot of work before asking for help.
It is an extremely rare occurence . . .We truly appreciate it!