when the depth of the liquid in a container is x, volume is x(x^2 + 36). Liquid is added at 3 cm^3/s.
What is the rate of change of x, when x = 11?
thanks
2007-05-11 17:51:07 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
doesn't say what container
2007-05-11 18:03:03 · update #1
thank you, it helped
2007-05-11 18:42:25 · update #2
Hello.
Let v = x(x^2 + 36) then v = x^3 + 36x
Take the derivative of both sides with respect to t and we have
dv/dt = 3x^2 dx/dt + 36 dx/dt
Now dv/dt = 3 and x = 11 giving us 3 = (3*121 + 36) dx/dt solving for dx/dt we have dx/dt = 3/399 or 1/133 cm/s.
Hope This Helps!!
2007-05-11 18:18:59 · answer #1 · answered by CipherMan 5 · 0⤊ 0⤋
I think you are still missing a vital piece of information. The only way it could work is if you used a box with a square base, then it could work, and if you do it that way, you get your answer to be .120 cm/seconds.
But I am not sure, so dont use my answer completely.
Good Luck!
2007-05-12 00:58:18 · answer #2 · answered by Anonymous · 0⤊ 0⤋