# Thread: Solving this second order?

1. ## 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!

2. 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

3. ## Error

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.

4. 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

5. 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 :/