zeesbrat3
  • zeesbrat3
Water is drained out of tank, shaped as an inverted right circular cone that has a radius of 6cm and a height of 12cm, at the rate of 3 cm3/min. At what rate is the depth of the water changing at the instant when the water in the tank is 9 cm deep? Give an exact answer showing all work and include units in your answer.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
zeesbrat3
  • zeesbrat3
@jim_thompson5910
jim_thompson5910
  • jim_thompson5910
What do you have so far?
zeesbrat3
  • zeesbrat3
r = 6cm h = 12cm r = 3cm^3/min \[v = \pi r^2h\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

jim_thompson5910
  • jim_thompson5910
`r = 3cm^3/min` is incorrect I think you meant to say dV/dt = -3 to represent the fact that the volume is decreasing by 3 cubic cm
jim_thompson5910
  • jim_thompson5910
also, you have the wrong volume formula
jim_thompson5910
  • jim_thompson5910
volume of a cone \[\Large V = \frac{1}{3}\pi*r^2*h\]
zeesbrat3
  • zeesbrat3
Oh... So then you take the derivative, yes?
jim_thompson5910
  • jim_thompson5910
yeah but first you have to somehow get everything in terms of h you have to find a connection between r and h and do a substitution
jim_thompson5910
  • jim_thompson5910
a picture might help |dw:1437352686967:dw|
zeesbrat3
  • zeesbrat3
r = 1/2h?
jim_thompson5910
  • jim_thompson5910
|dw:1437352732409:dw|
jim_thompson5910
  • jim_thompson5910
yeah that looks good, r = h/2
jim_thompson5910
  • jim_thompson5910
plug that in, derive, then isolate dh/dt
zeesbrat3
  • zeesbrat3
\[\frac{ 2 }{ 3 } \pi (\frac{ h^2 }{ 2 })\]
jim_thompson5910
  • jim_thompson5910
When you plugged r = h/2, and simplified, did you get \[\Large V = \frac{1}{12}\pi h^3\] ??
zeesbrat3
  • zeesbrat3
So you get \[\frac{ dv }{ dt } = \frac{ 1 }{ 4 } \pi h^2 \frac{ dh }{ dt }\]
jim_thompson5910
  • jim_thompson5910
good
jim_thompson5910
  • jim_thompson5910
dv/dt = -3 h = 9 solve for dh/dt
zeesbrat3
  • zeesbrat3
\[-3 = \frac{ 1 }{ 4 } \pi (9)^2 \frac{ dh }{ dt }\] \[\frac{ -12 }{ 81 \pi } = \frac{ dh }{ dt }\]
jim_thompson5910
  • jim_thompson5910
make sure you reduce the fraction as much as possible
zeesbrat3
  • zeesbrat3
\[\frac{ -4 }{ 27 \pi } = \frac{ dh }{ dt }\]
jim_thompson5910
  • jim_thompson5910
yep \[\Large \frac{dh}{dt} = -\frac{4}{27\pi} \approx -0.047157\] the units for dh/dt are cm/min the rate of the depth of the water is changing at roughly -0.047157 cm/min at exactly the depth of h = 9 cm
zeesbrat3
  • zeesbrat3
Thank you!!
jim_thompson5910
  • jim_thompson5910
glad to be of help

Looking for something else?

Not the answer you are looking for? Search for more explanations.