Differential Equation, not getting correct answer can't see why ...

[pazzy78]

Hello people..

Here is the question :

Here is the correct solution :

Here is mine, I didn't do the nice re write of -mg - 0.098v^2 ... but it still should work :

This will not be 1.5, cannot see where I went wrong ...

 

[logistic_guy]

\(\displaystyle t \neq \frac{1}{\sqrt{g}} \tan^{-1}12\)

You forgot to divide by \(\displaystyle \sqrt{0.098}\).

\(\displaystyle t = \frac{1}{\sqrt{g}\sqrt{0.098}} \tan^{-1}12\)

 

[pazzy78]

I still don't understand ...

where do i need to divide by sqrt(.098)..?


1/(root g) = 0.3194....

(0.313(v)/root g) when v = 120 = [0.313*120]/3.1304....

= 37.56/3.13 = 12

so

1/root g * tan(inv) 12 [radians]

0.314 * 1.4876

0.46712....



your method gets the right answer , i just don't know where the extra dividing by root(0.098) comes from ...

 

[logistic_guy]

I still don't understand ...

where do i need to divide by sqrt(.098)..?

\(\displaystyle \int\frac{1}{g + 0.098v^2} \ dv\)

How to solve this integral?

First, you get rid of the \(\displaystyle g\).

\(\displaystyle \int\frac{1}{g + 0.098v^2} \ dv = \int\frac{1}{g + \frac{0.098gv^2}{g}} \ dv = \frac{1}{g}\int\frac{1}{1 + \frac{0.098v^2}{g}} \ dv\)

Now you square everything in the second term of the denominator.

\(\displaystyle \frac{1}{g}\int\frac{1}{1 + \frac{0.098v^2}{g}} \ dv = \frac{1}{g}\int\frac{1}{1 + \left(\frac{\sqrt{0.098}v}{\sqrt{g}}\right)^2} \ dv\)

Now you need to multiply and divide the integral by \(\displaystyle \frac{\sqrt{0.098}}{\sqrt{g}}\) before you use a \(\displaystyle u\) substitution.

So, the denominator of the integral will be \(\displaystyle \frac{1}{g\frac{\sqrt{0.098}}{\sqrt{g}}}\).

Simplify.

\(\displaystyle \frac{1}{g\frac{\sqrt{0.098}}{\sqrt{g}}} = \frac{1}{\sqrt{g}\sqrt{0.098}}\)

 

[pazzy78]

^^ Why do you need to get rid of the g ? surely it is just a constant ?

 

[logistic_guy]

^^ Why do you need to get rid of the g ? surely it is just a constant ?

We need to get rid of \(\displaystyle g\) because we want the integral to look like this after the substitution:

\(\displaystyle \int \frac{1}{1 + u^2} \ du = \tan^{-1} u + C\)

 

[pazzy78]

OK seems I need to do a u sub as the coefficient of x^2 [in this case v^2] needs to be 1...

 

[logistic_guy]

OK seems I need to do a u sub as the coefficient of x^2 [in this case v^2] needs to be 1...

Yes.
Basically, you need to do this:

\(\displaystyle \int\frac{1}{g + 0.098v^2} \ dv = \frac{1}{g\frac{\sqrt{0.098}}{\sqrt{g}}}\int\frac{\frac{\sqrt{0.098}}{\sqrt{g}}}{1 + \left(\frac{\sqrt{0.098}v}{\sqrt{g}}\right)^2} \ dv\)

\(\displaystyle u = \frac{\sqrt{0.098}}{\sqrt{g}}v\)

\(\displaystyle du = \frac{\sqrt{0.098}}{\sqrt{g}} \ dv\)

Therefore,

\(\displaystyle \int\frac{1}{g + 0.098v^2} \ dv = \frac{1}{g\frac{\sqrt{0.098}}{\sqrt{g}}} \int\frac{1}{1 + u^2} \ du\)

 

[pazzy78]

Thanks, sorry got it out when I did the u sub ... sorry silly question should have known the coefficient of variable should be 1...