Rate of change -Differentiation An inverted cone with a base area of 2500pi mm^2 and height of 300m is slowly being filled with water. Water is being poured in at a rate of 350 mm^3/s. At what rate is the height of the water changing when the water is 100mm from the top of the inverted cone?

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Rate of change -Differentiation An inverted cone with a base area of 2500pi mm^2 and height of 300m is slowly being filled with water. Water is being poured in at a rate of 350 mm^3/s. At what rate is the height of the water changing when the water is 100mm from the top of the inverted cone?

Mathematics
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I think: We are to find out: \[\frac{ dh }{ dt } \] We know \[\frac{ dV }{ dt }=350\]
|dw:1443476761216:dw|
Volume of cone: \[V=\frac{ 1 }{ 3 }\pi r^2h\] Volume of this cone:\[V=\frac{ 2500 }{ 3 }h\]

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Other answers:

  • phi
use "similar triangles" to write r in terms of h
  • phi
from the area of the base you find r=50 and you have this |dw:1443477912255:dw|
  • phi
\[ V=\frac{ 1 }{ 3 }\pi r^2h \\ V=\frac{ 1 }{ 3 }\pi \frac{h^2}{36}h \\ V= \frac{\pi}{108} h^3 \] now use implicit differentiation
\[\frac{ dv }{ dh }=36 \pi h^2\] \[\frac{ dh }{ dt }=\frac{ dv }{ dt }\times \frac{ dh }{ dv }\] am i on the right track
then \[\frac{ dh }{ dt }=350 \times \times \frac{ 1 }{ 36 \pi h^2 }\] and then sub in h=100
  • phi
I would take the derivative of each variable with respect to t on the left side you get dv/dt on the right side you get \[\frac{\pi}{108} \frac{d}{dt} h^3 \] which is \[ \frac{\pi}{108} 3 h^2\frac{d}{dt}h\] so the equation is \[ \frac{dV}{dt}= \frac{\pi}{36} h^2 \frac{dh}{dt} \]
  • phi
when the water is 100mm from the top of the inverted cone in the formula h is the height from 0 up to 300 100 mm from the top corresponds to h= 200 mm
  • phi
Your approach works except that this ***\( \frac{ dv }{ dh }=36 \pi h^2 \)*** should read \[ \frac{dV}{dh}= \frac{\pi}{36} h^2\]
then sub in h=200....
  • phi
yes
Thanks heaps!

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