Seperate variable and integration

[delsoo]

Homework Statement

for this Q and t are variables, 10 and surd k are constant, is my working correct?

Homework Equations

The Attempt at a Solution

 

[skrat]

No.

$\frac{\sqrt{K}}{10+(4-\sqrt{K})t}$ is NOT the same as $\frac{\sqrt{K}}{10+(4-\sqrt{K})}\cdot \frac{1}{t}$

You should think about substitution.

 

[delsoo]

i don't understand the circled part, can you expalin?

 

[skrat]

Well somebody integrated LHS and RHS of the equation, nothing more.

LHS of the equation is of course $\int \frac{dQ}{Q}=ln(Q) + C$, which you already found out. Now you have to integrate the RHS of the equation and you should get exactly the same as it is circled in your last post #3.

In your original post you already tried to compute the integral on the RHS

$-\int_{0}^{t}\frac{\sqrt{k}}{10+(4-\sqrt{k})t}dt$

but your approach is wrong because you wrote that $\frac{\sqrt{k}}{10+(4-\sqrt{k})t}=\frac{\sqrt{k}}{10+(4-\sqrt{k})}\cdot \frac{1}{t}$ which is NOT true. If anything than $\frac{\sqrt{k}}{10+(4-\sqrt{k})t}=\frac{\sqrt{k}}{\frac{10}{t}+(4-\sqrt{k})}\cdot \frac{1}{t}$ but this does not simplify the integral at all.

Again: you should consider substitution. Note that $\int \frac{dx}{x}$ is an elementary integral. Try to find such substitution that will bring you to this elementary integral.

 

[delsoo]

well, why the following steps contain (10+(4-surd k) t) /10 ... why we should divide 10?

 

[skrat]

Firstly, I never wrote that you should divide (10+(4-surd k) t) by 10.

Remember that $\frac{a}{b}\cdot \frac{c}{d}=\frac{a\cdot c}{b\cdot d}$.

Also take a look at $\frac{1}{a}\cdot \frac{1}{b}=\frac{1}{a\cdot b}$ and $\frac{1}{a+b}\cdot \frac{1}{c}=\frac{1}{(a+b)\cdot c}=\frac{1}{a\cdot c+b\cdot c}$.

Now I hope you understand a bit more.

 

[delsoo]

please refer fifth post to the post 3 . thanks.

 

[skrat]

well, why the following steps contain (10+(4-surd k) t) /10 ... why we should divide 10?

Ok, I apologize, I didn't know that we are no longer talking about the circled part. Anyhow, I hope you understand how ton integrate $-\int_{0}^{t}\frac{\sqrt{k}}{10+(4-\sqrt{k})t}dt$

Now to answer your last question:

$aln(x)=ln(x^a)$ also

$ln(x)-ln(y)=ln(\frac{x}{y})$

Note that you RHS integral goes from $0$ to $t$. Understood?

 

[delsoo]

ok! clear now!