# Thread: differentiation with repsect to time

1. ## differentiation with repsect to time

Hi,

How do i express this in terms of t?

a dv/dt = -g/b2(v(t)^2 + b^2)

2. ## if you mean

$\displaystyle \frac{dv}{dt}=\frac{-b}{g}(v^2+b^2)$..this is an SDE...

3. SDE? New to this thing!

4. the initial part should have said express t in terms of v,b,g, and vo(the initial speed of the ball).

5. ## Yes

SDE stands for seperable differential equation? seperate variables...integrate...solve for t...make sense? if not reply and I will try to walk you through it

6. i would appreciate the help thankyou!

7. ## Ok here is what you have to do

I will give you an example $\displaystyle \frac{dy}{dx}=9y^4$..seperate so that all the $\displaystyle y$s are with the $\displaystyle dy$s and all the $\displaystyle dx$s are with the $\displaystyle x$s...so we have then that $\displaystyle \frac{dy}{y^4}=9dx$...now we want to get rid of the dx's and dy's...so we integrate $\displaystyle \int\frac{dy}{y^4}=\int{9dx}$...so we get $\displaystyle \frac{-1}{3y^3}=9x+C$...now since y is a function of x to we are trying to find y like you are trying to find v...so we solve for y $\displaystyle y^3=\frac{-1}{27x+C_1}$....$\displaystyle y=\frac{-1}{(27x+C_1)^{\frac{1}{3}}}$..now that is an SDE...now reevaluating this...can you tell me if this an SDE?

8. think i understand so I have to take all the variables relating to t on to the left hand side of the equation and leave the other variables on the right hand side of the equation??

A(t)/v(t)^2 = (-g/b^2)*b^2

??????

9. ## No

You are looking at this too much from a physics perspective...forget that $\displaystyle a(t)=\frac{dv}{dt}$ leave it as it is seperate and integrate then solve...but make sure its an SDE

10. sorry but im confused now because my question begins with by writing a = v dv/dx ........

also why dont the constants stay together on the right hand side? Or have i misunderstood the criteria for the equation?

dv/dt = -g/b2(v(t)^2 + b^2)

separation of variables

1/v(t)^2 + b^2 = -g/b2

Integration

1/b^2arctan(v(t)^2/b^2) = ln(-g + b^2)+c

I think its totally wrong so any help would be appreciated!