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January 31, 2015

January 31, 2015

Posted by **math whizz** on Monday, December 5, 2011 at 9:12pm.

- Calculus -
**MathMate**, Monday, December 5, 2011 at 9:21pmdy/dx=ksqrt(t)

dy=ksqrt(t)dt

Integrate both sides:

y=k(2/3)t^(3/2) + C

where C is an integration constant determined from initial conditions.

Thus

y = (2k/3)t^(3/2) + C

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