Equation of Motion for a Projectile Under Quadratic Air Resistance

[learningphysics]

I'm not sure if that's how you meant me to use a substitution.

It's a method of solving integrals, called integration by substitution.

u = \sqrt{\frac{c}{mg}}*v

du = \sqrt{\frac{c}{mg}}*dv

solve for v and dv in the above equations, and substitute into the integral:

tf-ti = \frac{-m}{mg}*\frac{1}{\sqrt{\frac{c}{mg}}}\int\frac{du}{1+u^2}

now you can use the arctan for the right side...

once you get the integral in terms of u... you can substitute \sqrt{\frac{c}{mg}}*v for u

 

[Oblio]

It's a method of solving integrals, called integration by substitution.

u = \sqrt{\frac{c}{mg}}*v

du = \sqrt{\frac{c}{mg}}*dv

Not dv/dt?

 

[Oblio]

It's a method of solving integrals, called integration by substitution.

u = \sqrt{\frac{c}{mg}}*v

du = \sqrt{\frac{c}{mg}}*dv

solve for v and dv in the above equations, and substitute into the integral:

tf-ti = \frac{-m}{mg}*\frac{1}{\sqrt{\frac{c}{mg}}}\int\frac{du}{1+u^2}

now you can use the arctan for the right side...

once you get the integral in terms of u... you can substitute \sqrt{\frac{c}{mg}}*v for u

Ok, we went from tf-ti = \int \frac{-mdv}{mg(1+(c/mg)v^2}

Bring out the constants \frac{-m}{mg}


=\frac{-m}{mg} \int\frac{dv}{1+(c/mg)v^2}

the numerator and denominator are factored out above...


Am I correct in say that \sqrt{\frac{c}{mg}} in the numerator and denominator in the integral cancel out leaving just v and dv?

I'm not following the step in getting \frac{1}{\sqrt{\frac{c}{mg}}} outside...

 

[learningphysics]

Do you have a calculus textbook handy by any chance?

 

[Oblio]

at home... not here at school. Sorry. Why?

 

[learningphysics]

at home... not here at school. Sorry. Why?

Just wanted you to read the substitution method for solving integrals... Have a look at this example online:

http://www.sosmath.com/calculus/integration/substitution/Example1/Example1.html

Let me know if it makes sense.

 

[Oblio]

Why is du divided by 2?

 

[learningphysics]

Ok, we went from tf-ti = \int \frac{-mdv}{mg(1+(c/mg)v^2}

Bring out the constants \frac{-m}{mg}


=\frac{-m}{mg} \int\frac{dv}{1+(c/mg)v^2}

the numerator and denominator are factored out above...

Yeah, let's cancel out the m's... leaving:

\frac{-1}{g} \int\frac{dv}{1+(c/mg)v^2}

The idea is now I want this integral to look like:

\int\frac{dx}{1+x^2}

for that we need to substitute a new variable instead of v...

If I let u = \sqrt{c/mg}*v, then the denominator is 1+(c/mg)v^2 = 1+u^2... does this part make sense?

 

[learningphysics]

Why is du divided by 2?

du = 2xdx

du/2 = xdx

so they substitute in du/2 instead of xdx.

 

[Oblio]

Yeah, let's cancel out the m's... leaving:

\frac{-1}{g} \int\frac{dv}{1+(c/mg)v^2}

The idea is now I want this integral to look like:

\int\frac{dx}{1+x^2}

for that we need to substitute a new variable instead of v...

If I let u = \sqrt{c/mg}*v, then the denominator is 1+(c/mg)v^2 = 1+u^2... does this part make sense?

u = \sqrt{c/mg}*v
= u^2 = (c/mg)v^2

Yep. I get it!

 

[Oblio]

I just changed it, you may have already started replying to my dumb mistake before the edit. :P

 

[Oblio]

du = 2xdx

du/2 = xdx

so they substitute in du/2 instead of xdx.

right. duh.
I'm going to stab my brain with a q-tip.

 

[Oblio]

Shouldn't the du integrate into a u?

 

[learningphysics]

Shouldn't the du integrate into a u?

write out the integral after substituting in u...

 

[learningphysics]

right. duh.
I'm going to stab my brain with a q-tip.

lol. don't be so hard on yourself.

 

[Oblio]

so you'd have:

u^76*u*1/152

=(u^76*x^2+5) / 152

 

[learningphysics]

so you'd have:

u^76*u*1/152

=(u^76*x^2+5) / 152

no... du doesn't integrate to u...

When you have an integral:

let's say:

\int5xdx

What you are taking the anti-derivative of is 5x... so the answer would be (5/2)x^2.

same way:

\int u^{75}(du/2) = \int \frac{u^{75}}{2} du

You take the anti-derivative of u^75/2 to get your answer.

 

[Oblio]

Ok, got it. It's more of a notation.

Yeah I follow your online example now.

 

[Oblio]

Yeah, let's cancel out the m's... leaving:

\frac{-1}{g} \int\frac{dv}{1+(c/mg)v^2}

The idea is now I want this integral to look like:

\int\frac{dx}{1+x^2}

for that we need to substitute a new variable instead of v...

If I let u = \sqrt{c/mg}*v, then the denominator is 1+(c/mg)v^2 = 1+u^2... does this part make sense?

So, if u = \sqrt{c/mg}*v

How do we still end up with \sqrt{c/mg} outside the integral as well?

 

[learningphysics]

So, if u = \sqrt{c/mg}*v

How do we still end up with \sqrt{c/mg} outside the integral as well?

we not only need to substitute u... we also need to take care of the dv, and get du instead...

so:

du = \sqrt{c/mg}*dv

dv = \frac{du}{\sqrt{c/mg}}

So substitute that into the integral for dv...

what do you get for the integral now?

 

[learningphysics]

Ok, got it. It's more of a notation.

Yeah, for the most part...

Yeah I follow your online example now.

cool.

 

[Oblio]

we not only need to substitute u... we also need to take care of the dv, and get du instead...

so:

du = \sqrt{c/mg}*dv

dv = \frac{du}{\sqrt{c/mg}}

So substitute that into the integral for dv...

what do you get for the integral now?

\frac{-m}{mg} \int \frac{\frac{du}{sqrt[c/mg]}}{1 + c/mg* v^2}

 

[learningphysics]

\frac{-m}{mg} \int \frac{\frac{du}{sqrt[c/mg]}}{1 + c/mg* v^2}

yes, but you also need the u substitution...

 

[Oblio]

You mean du?

 

[learningphysics]

You mean du?

you've taken care of the du... but you still have v in the integral... you want u... we had it before

1+u^2

 

[Oblio]

\frac{-m}{mg} \int \frac{\frac{du}{sqrt[c/mg]}}{1 + c/mg* v^2}

right...

\frac{-m}{mg} \int \frac{\frac{du}{sqrt[c/mg]}}{1+ u^2}

 

[learningphysics]

right...

\frac{-m}{mg} \int \frac{\frac{du}{sqrt[c/mg]}}{1+ u^2}

exactly... now take all the constants outside the integral... and we get:

-\sqrt{\frac{m}{gc}}\int\frac{du}{1+u^2}

now we can apply arctan...

 

[Oblio]

Do you treat the integral sign almost as an equality when moving things in and out?

 

[learningphysics]

Do you treat the integral sign almost as an equality when moving things in and out?

not like an equal sign... more like a parentheses...

ie: 4x^2y+8xy = 4(x^2y + 2xy) = 4xy(x+2)

it's the same way I'm moving out the constants...

 

[Oblio]

Ok because I am having trouble matching what you got taking out the last constants