Need HELP! Calculus / Differential Equations

Edited: 12/1/2008 5:04:07 PM by MJAVY7

Hey Trent I agree with what you saying, but in the industry that I am in derivatives and integrals are not used as much or at all like in other diciplines of engineering. And to be realistic now and days a calculator and a computer will do it for you.

With that said I still need a littler more help

this goes back to algebra something I took in high school 10 - 11 years ago...

1/y = tan(t^2/2)

How do I solve for y? do I square root the right hand side with a -1 in front of the square root?

multiply both sides by y

1 = y tan(t^2/2)

Then divide by the tangent function

1/tan(t^2/2)= y

Edited: 12/1/2008 5:49:12 PM by Layumon22s

Edited: 12/1/2008 5:41:38 PM by Layumon22s

cross multiply and divide for yy = 1/tan(t^2/2) OR y = cot(t^2/2) (thanks for the correction, I don't know what made me think that 1/tan(x) = tan^-1(x) can't continue to use up all my posts for the month so I just edited this one)

and yes Gump GT puts life in a whole new perspective. I no longer feel like the biggest nerd on campus when I'm there because there are some serious uber nerds all over the place. I have to be cooler than at least 1 or 2 of them

Isn't 1/tan(x) = cot(x)

all right here is my final answer...

y'(t) = ty^2+t
dy/dt = t(y^2+1)
dy/(y^2+1) = t dt Integrate both sides
arctan y = t^2/2 + C Solve for "y"
y = tan t^2/2 + C

Now I am moving on to the next problem

dv/dt = 9.8-kv^2 Multiply both sides by "dt"
dv = (9.8-kv^2)dt Divide both sides by "(9.8-kv^2)"
1/(9.8-kv^2)*dv = dt Integrate both sides
Now I am looking thru my book for a formula any ideas?