Differential calculus

**Rydbirk** · December 19th 2007, 12:14 PM

How do I go from

$ \int \frac{dv}{cv^2 - g} = \int~dt $

to

$ \frac{1}{2\sqrt{cg}}\ln\left(\frac{\sqrt g + v\sqrt c}{\sqrt g - v\sqrt c}\right)=a-t $

It's not just urgent - it's VERY urgent! :/ Big assignment..

**colby2152** · December 19th 2007, 12:23 PM

Shouldn't it be a + t?

**Rydbirk** · December 19th 2007, 12:25 PM

My calculator gives me a-t..

**Soroban** · December 19th 2007, 01:19 PM

Hello, Rydbirk!

How do I go from: . $ \int \frac{dv}{cv^2 - g} \:= \:\int \,dt$

. . . to: . $\frac{1}{2\sqrt{cg}}\ln\left|\frac{\sqrt g + v\sqrt c}{\sqrt g - v\sqrt c}\right|\;=\;C-t$

The right side is: . $\int dt \;=\;t + C$

Let: $cv^2 \:=\:u^2\quad\Rightarrow\quad v \:=\:\frac{1}{\sqrt{c}}u\quad\Rightarrow\quad dv \:=\:\frac{du}{\sqrt{c}}$
Also let: $g \:=\:a^2$

The left side becomes: . $\int\frac{1}{u^2-a^2}\left(\frac{du}{\sqrt{c}} \right) \;=\;\frac{1}{2a\sqrt{c}}\ln\left|\frac{u-a}{u+a}\right| + C$

Back-substitute: . $u \:=\:v\sqrt{c}\;\text{ and }\; a = \sqrt{g}$

. . $\frac{1}{2\sqrt{g}\sqrt{c}} \ln\left|\frac{v\sqrt{c} - \sqrt{g}}{v\sqrt{c} + \sqrt{g}}\right|$

And we have: . $\frac{1}{2\sqrt{cg}}\ln\left|\frac{v\sqrt{c} - \sqrt{g}}{v\sqrt{c} + \sqrt{g}}\right| \;=\;t + C$

This answer is equivalent to the one your calculator gave.

**Rydbirk** · December 19th 2007, 01:28 PM

Thank you!! You saved my day

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