Log & Rates Q help (1 Viewer)

Drewk · Jun 6, 2013

http://snag.gy/8jrug.jpg
Not sure if you can do this^ as i get 1/6 not 6??
Please check

HeroicPandas · Jun 6, 2013

You can

$\frac{12t}{t^2 +4}L/min $means: \\ \\ $\frac{dV}{dt} = \frac{12t}{t^2 + 4} \\ dV = \frac{12t}{t^2 +4}dt \\ \therefore \int dV = 12\int \frac{t}{t^2 +4}dt \\ \int dV = \frac{12}{2}\int \frac{2t}{t^2 +4}dt \\ \therefore V = 6ln(t^2 +4)+ C \\ \\$ $At $ t = 0, v = 0, \therefore C = -6ln(4) \\ \ \therefore V = 6ln(t^2 + 4) - 6ln(4) \\\\ \therefore V = 6ln(\frac{t^2 +4}{4})\sqsubset| \sqsupset$

Drewk · Jun 6, 2013

HeroicPandas said:
You can

$\frac{12t}{t^2 +4}L/min $means: \\ \\ $\frac{dV}{dt} = \frac{12t}{t^2 + 4} \\ dV = \frac{12t}{t^2 +4}dt \\ \therefore \int dV = 12\int \frac{t}{t^2 +4}dt \\ \int dV = \frac{12}{2}\int \frac{2t}{t^2 +4}dt \\ \therefore V = 6ln(t^2 +4)+ C \\ \\$ $At $ t = 0, v = 0, \therefore C = -6ln(4) \\ \ \therefore V = 6ln(t^2 + 4) - 6ln(4) \\\\ \therefore V = 6ln(\frac{t^2 +4}{4})\sqsubset| \sqsupset$

Thanks a lot, i forgot to take out the constant at the start

Log & Rates Q help (1 Viewer)

Drewk

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HeroicPandas

Heroic!

Drewk

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