- #1
james brug
- 34
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I'm having trouble understanding this post: https://www.physicsforums.com/showthread.php?t=81157
Specifically, part (a)
(1) [tex]\frac{dV}{dt} = k - \alpha \pi \left( \frac{3a}{\pi h} \right) ^{2/3} V ^{2/3} [/tex]
Leaving out k, and considering a cone with no water coming in:
(2) [tex]\frac{dV}{dt} = - \alpha \pi \left( \frac{3a}{\pi h} \right) ^{2/3} V ^{2/3} [/tex]
How would I solve this for an equation to determine the volume in terms of time t? How do I interpret V in terms of t?
(3) [tex]\frac{dV}{dt} = - \alpha \pi \left( \frac{3a}{\pi h} \right) ^{2/3} {\left(\frac{1}{3}\pi r^2 h\right)} ^{2/3} [/tex]
I can't just solve eq. 3, that doesn't seem to make sense.
Specifically, part (a)
(1) [tex]\frac{dV}{dt} = k - \alpha \pi \left( \frac{3a}{\pi h} \right) ^{2/3} V ^{2/3} [/tex]
Leaving out k, and considering a cone with no water coming in:
(2) [tex]\frac{dV}{dt} = - \alpha \pi \left( \frac{3a}{\pi h} \right) ^{2/3} V ^{2/3} [/tex]
How would I solve this for an equation to determine the volume in terms of time t? How do I interpret V in terms of t?
(3) [tex]\frac{dV}{dt} = - \alpha \pi \left( \frac{3a}{\pi h} \right) ^{2/3} {\left(\frac{1}{3}\pi r^2 h\right)} ^{2/3} [/tex]
I can't just solve eq. 3, that doesn't seem to make sense.