Question

A conical tank has height 3 m and radius 2 m at the top. Water flows in at a rate of 0.7m^3/min. How fast is the water level rising when it is 0.8 m?
I tried to solve it but got wrong answer

Answer 1

i honestly dont know the right answer, but im just going to give it a go, so bear with me? ...and sorry if it leads nowhere man

Answer 2

Sounds good lol I compare mine to yours

Answer 3

so volume of a cone = pi x r^2 x h x 1/3
\[V = \pi \times r^2 \times h \ \frac{ 1 }{ 3 }\]
and what you want is a rate of change of Volume with regards to height
so if V = y
and h = x
y = pi x r^2 x (x) x 1/3
so rate of change = gradient = derivative of y
so
y' = pi x r^2 x 1/3 ???

so gradient is constant, and is (4/3) pi...

now if your volume is increasing at a rate of 0.7m^3/minute...

...nah dude, im getting lost, think im just muddying the waters here, sorry, will keep thinking about it and get back to u if i get a clearer picture in my head

Answer 4

yea I am stuck but thanks for the try, i got the answer .125

Answer 5

The similar triangle may help with the relation between r and h

Answer 6

r / 2 = h / 3
r = 2/3h

V = 1/3(pi)r^2h . Put r = 2/3h
V = 1/3(pi)(4/9)h^3 = 4(pi)/27h^3

dV/dt = 4(pi)/27(3h^2)dh/dt = 4/9(pi)(h^2)dh/dt

Plug in the numbers and solve for dh/dt at h = 0.8

Answer 7

Hmm i got this as my answer now (4/9)pi3^2(.7) and got 8.79 but still wrong

Answer 8

i got 10.05 now let me check

Answer 9

dV/dt = 4/9(pi)(h^2)dh/dt = 1.396h^2dh/dt

dV/dt = 0.7, h = 0.8
0.7 = 1.39(.8^2)dh/dt
dh/dt = 0.78 meters / min

I haven't double checked my numbers.

Answer 10

I missed I was suppose to plug in .8 for h I kept using the given H in the problem, that is the right answer thanks a ton!

Answer 11

You are welcome.