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Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

DE coursework integration?

How do i integrate mv/(-B-kv^2) dv? m B and k are constants

(edited 13 years ago)

Original post by JordanW

How do i integrate mv/(-B-kv^2) dv? m B and k are constants

\int \dfrac{mv}{-B-kv^2} dv

?

well..

If m, B and K are constant values:

\dfrac{-1}{B} \int \dfrac {mv}{-kv^2} dv =[br][br]\dfrac{-1}{B} \int \dfrac {m}{-kv} dv =[br][br]\dfrac{m}{Bk} \int \dfrac {1}{v} dv.

Think that's allowed?

_Kar.

(edited 13 years ago)

Original post by JordanW

How do i integrate mv/(-B-kv^2) dv? m B and k are constants

If you can't see it directly, use a substitution of

u=-(B+kv^2)

.

Original post by Kareir

\int \dfrac{mv}{-B-kv^2} dv

?

well..

If m, B and K are constant values:

\dfrac{-1}{B} \int \dfrac {mv}{-kv^2} dv =[br][br]\dfrac{-1}{B} \int \dfrac {m}{-kv} dv =[br][br]\dfrac{m}{Bk} \int \dfrac {1}{v} dv.

Think that's allowed?

_Kar.

The first step is wrong.

As farhan suggested, use that particular substitution and providing you are confident at integration by substitution, it should all come together quite nicely and without too much difficulty.

OP

Original post by Farhan.Hanif93

The first step is wrong.

This is true as B is not a common factor in the denominator

Original post by Farhan.Hanif93

The first step is wrong.

haha, yeah, I noticed just like a few mins after I posted. Sorry. :tongue:

:tongue:

_Kar.

OP

I dont think substitution works :/

Substitution works.

OP

Im struggling with this and its apparently pretty simple.
Im trying to find s using:

\frac{-B-kv^{2}}{m} = v\frac{dv}{ds}

hence,

\int^55_0 \frac{mv}{-B-kv^{2}}\ dv = \int 1\ ds

The integration should be between 55 and 0 but i have only just started using latex and cant seem to get it to work. I know what i have done thus far is correct.

As far as the latex goes, you need to put 55 in braces: {55} if you want it to all be superscripted.

As far as the integral goes, use the substitution Farhan suggested.

OP

Cheers for the help guys, i think it should be:

\frac{m}{-2k}ln(B+kv^2)

Original post by JordanW

Cheers for the help guys, i think it should be:

\frac{m}{-2k}ln(B+kv^2)

That's correct.

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