Integration by substitution question (1 Viewer)

nrlwinner · Dec 5, 2009

Hi. I'm fairly good in doing these types of questions but this one is different because of the way it is arranged. I've tried to flip it but it still doesn't work.

$Given\ \frac{d\Theta }{dr}=\frac{3}{(1-r)^3}, and\ that\ r(0)=0, find\ r\ in\ terms\ of\ \Theta.\ Use\ the\ substitution\ u=1-r$

jet · Dec 5, 2009

I think that's right
$\frac{d\theta}{dr} = \frac{3}{\left (1 - r \right )^3} \\ \therefore \int d\theta = \int \frac{3}{\left (1 - r \right )^3} \, dr \\ \theta = \int \frac{3}{\left (1 - r \right )^3} \, dr \\ \text{We can leave out the constant on the LHS as it combines with the constant on the RHS} \\ \text{Let } u = 1 - r \\ \frac{du}{dr} = - r \Rightarrow dr = -du \\ \therefore \theta = -\int\frac{3}{u^3} \, du \\ = -3\left [\frac{u^{-2}}{-2} \right ] + C \\ = \frac{3}{2u^2} + C \\ \text{If } \theta = 0, \text{ then } r = 0 \text{ and } u = 1 \\ \therefore \frac{3}{2} + C = 0 \Rightarrow C = -\frac{3}{2} \\ \therefore \theta = \frac{3}{2u^2} - \frac{3}{2} \\ = \frac{3\left (1 - u^2 \right )}{2u^2} \\ 3 - 3u^2 = 2u^2\theta \\ \left (2\theta + 3\right)u^2= 3 \\ u^2 = \frac{3}{2\theta + 3} \\ u = \sqrt{\frac{3}{2\theta + 3}} \text{ we can ignore the negative case as } u \text{ is initially positive} \\ \text{and } r = 1 - \sqrt{\frac{3}{2\theta + 3}}$

tommykins · Dec 5, 2009

nrlwinner · Dec 5, 2009

Thanks for the help, but the answer says its

$r=1-\frac{1}{\sqrt[3]{\Theta +1}}$

Trebla · Dec 5, 2009

There may a possibility that the question was misread because from the solution you provided, we get

$1-r=\dfrac{1}{(\theta+1)^{\frac{1}{3}}}\\\\\theta=\dfrac{1}{(1-r)^3}-1\\\\\dfrac{d\theta}{dr}=\dfrac{3}{(1-r)^4}$

Integration by substitution question (1 Viewer)

nrlwinner

Member

jet

Banned

tommykins

i am number -e^i*pi

nrlwinner

Member

Trebla

Administrator

Users Who Are Viewing This Thread (Users: 0, Guests: 1)