How Does the Chain Rule Apply to Finding Antiderivatives in the Volume Equation?

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SUMMARY

The discussion focuses on applying the chain rule to find antiderivatives in the context of the volume equation (v^-.5)v' = k, where v represents volume. Participants clarify that since v is a function of time t, the derivative v' can be expressed as dv/dt. The correct antiderivative derived from the equation is 2(V^.5) = kt + C, confirming the application of the chain rule and elementary antiderivative formulas. A key takeaway is the importance of verifying results by differentiating the final answer.

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The text is giving me steps on how to do it but I don't get what's it is asking.

The equation is (v^-.5)v'=k where v= volume

Then says we consider the chain rule together with elementary antiderivative formulas to determine an antiderivative with respect to t of (v^-.5)v'

I haven't learned integrals yet. I'm exactly sure I'm even looking for. I don't really need an answer but explanation how chain rule can applies here. Since V is a function of t, do I expand v' to v'*dv/dt?
 
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caljuice said:
The text is giving me steps on how to do it but I don't get what's it is asking.

The equation is (v^-.5)v'=k where v= volume

Then says we consider the chain rule together with elementary antiderivative formulas to determine an antiderivative with respect to t of (v^-.5)v'

I haven't learned integrals yet. I'm exactly sure I'm even looking for. I don't really need an answer but explanation how chain rule can applies here. Since V is a function of t, do I expand v' to v'*dv/dt?

v' already is dv/dt, remember u(v(t))' = u'(v(t))*v'(t)
 
ah okay thanks. So then the antiderivative should be

2(V^.5)=kt+C ?
 
caljuice said:
ah okay thanks. So then the antiderivative should be

2(V^.5)=kt+C ?

That would be correct. If you are ever unsure, just differentiate your answer and see if you get what you started with.
 

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