Seperable Differentiation. Help, please. =\

[mikiritenshi]

I blanked out. Any help is greatly appreciated.

Let v(t) be the velocity, in feet per second, of a skydiver at time t seconds, t>0. After her parachute opens, her velocity satisfies the differential equation: dv/dt= -2v-32, with initial condition v(0)=-50.

(a) Use separation of variables to find an expression for v in terms of t, where t is measured in seconds.

I got up to this far:

integral [dv/(-2v-32)]= integral (dt)
u= -2v-32; du= -2dv.

Help.

 

[dextercioby]

Good.May i suggest integrating with corresponding limits...?

\int_{-50}^{v} \frac{dv}{-2v-32} = \int_{0}^{t} \ dt

The LHS integral is doable.With a substitution that u already did...So carry on.Pay attention to the way the limits are transformed.


Daniel.

 

[mikiritenshi]

Using the substitution that I did, I obtained:

(-1/2) integral(1/u)du = integral (dt)

= (-1/2) ln|u| + C = t+ C

I'm not sure on what to do next. =\

 

[RadiationX]

substitue the original variables back for the u substitutions. then use the initial conditon to find out what C is.

 

[mikiritenshi]

I know you plug in 0 for t, but what'll happen to the -50?
And do I have to exponentiate both sides?

 

[dextercioby]

From the integration i get

v(t)=-34e^{-2t}-16


Daniel.

 

[mikiritenshi]

That is the answer, but I don't know how to get it.

 

[dextercioby]

The RHS is 't',okay??

Now,the LHS can be written

-\frac{1}{2} \int _{-50}^{v} \frac{d(-v-16)}{-v-16} = -\frac{1}{2}\left[\ln (-v-16)\right]\left |_{-50}^{v}\right = -\frac{1}{2} \ln \left(\frac{-v-16}{34}\right)

Therefore

-2t=\ln\left(\frac{-v-16}{34}\right)\Rightarrow v(t)=-34e^{-2t}-16

Daniel.

 

[mikiritenshi]

Wow, thanks.