Solving this second order?

**KwonSau** · February 15th 2008, 09:53 AM

I have the following equation I am using for some air resistance calculations, I want to solve so that v becomes a function of t:

av^2 + (b/t)v = c

a, b, and c are constants.

v = f(t), v = ?...

I have no idea where to start (I took up through Calc 2 in undergrad). Thanks in advance for your help!

**topsquark** · February 15th 2008, 10:38 AM

Originally Posted by KwonSau

I have the following equation I am using for some air resistance calculations, I want to solve so that v becomes a function of t:

av^2 + (b/t)v = c

a, b, and c are constants.

v = f(t), v = ?...

I have no idea where to start (I took up through Calc 2 in undergrad). Thanks in advance for your help!

This has nothing to do with Calculus and everything to do with the quadratic formula:
$av^2 + \left ( \frac{b}{t} \right ) v = c$

$atv^2 + bv - c = 0$

$v = \frac{-b \pm \sqrt{b^2 + 4act}}{2at}$

-Dan

**TwistedOne151** · February 15th 2008, 02:19 PM

topsquark,

You made an error in your work:

$av^2+\left(\frac{b}{t}\right)v=c$
$atv^2+bv-ct=0$
(You had $atv^2+bv-c=0$ ).

This changes the answer to:
$v=\frac{-b\pm\sqrt{b^2+4act^2}}{2at}$

-Kevin C.

**topsquark** · February 15th 2008, 05:04 PM

Originally Posted by TwistedOne151

topsquark,

You made an error in your work:

$av^2+\left(\frac{b}{t}\right)v=c$
$atv^2+bv-ct=0$
(You had $atv^2+bv-c=0$ ).

This changes the answer to:
$v=\frac{-b\pm\sqrt{b^2+4act^2}}{2at}$

-Kevin C.

Whoops!

Thanks for the catch! (I knew I should have checked the units on that.)

-Dan

**KwonSau** · February 15th 2008, 10:40 PM

Wow. Thanks for the quick help guys. (I knew it wasn't calc, I just mentioned Calc 2 to show how little math experience I had). I completely forgot about the quadratic equation..dental school hasn't really been testing my algebra skills :/

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