# Solving this second order?

• Feb 15th 2008, 09:53 AM
KwonSau
Solving this second order?
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!
• Feb 15th 2008, 10:38 AM
topsquark
Quote:

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:
$\displaystyle av^2 + \left ( \frac{b}{t} \right ) v = c$

$\displaystyle atv^2 + bv - c = 0$

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

-Dan
• Feb 15th 2008, 02:19 PM
TwistedOne151
Error
topsquark,

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

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

-Kevin C.
• Feb 15th 2008, 05:04 PM
topsquark
Quote:

Originally Posted by TwistedOne151
topsquark,

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

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