Hi all, i am just having a little bit of a problem with part © of this question....

 

 

i get [math] \frac{dr}{dt} = r^3 [/math]

 

and so [math] r(t) = -\frac{1}{\sqrt{2t}} [/math]

 

this doesnt seem right to me....but what do i know!

 

also i can't see how to get [math] \frac{d \theta}{dt} [/math]

 

any ideas guys?

 

Thanks

 

Sarah

i get [math] \frac{dr}{dt} = r^3 [/math]

 

This looks right.

 

and so [math] r(t) = -\frac{1}{\sqrt{2t}} [/math]

 

but this isn't (you screwed up the algebra after the integration I think). Also' date=' be careful of your intergration endpoints.

 

also i can't see how to get [math] \frac{d \theta}{dt} [/math]

 

You started with 2 equations and converted them into r and theta. This must have given you two equations which you solved for dr/dt to get the dr/dt equation you have above. Now you can put this dr/dt back in and solve for dtheta/dt. (The point of this exercise is to show you how problems can be made much much simpler by choosing the correct reference frame.)

ok i get [math] \frac{d\theta}{dt} = -1 [/math]

 

and so [math] \theta = -t + t_0 [/math]

 

but for the dr/dt part i don't quite see how i can get a different answer...

 

if [math] \frac{dr}{dt} = r^3 [/math]

 

then [math] \frac{1}{r^3}dr = dt [/math]

 

right?

 

but integrating [math] \frac{1}{r^3}dr [/math] gives [math] -\frac{1}{2r^2} [/math] .... which seems to be my problem...

 

also when you are say the limits of integration... that should be r = 0 to r = infinity?

how about

for (i)

[math]

 

r(t) = \frac{1}{\sqrt{2(t_0-t)}}

 

[/math]

off topic, but...

 

I love polar coordinates.

off topic' date=' but...

 

I love polar coordinates.[/quote']

 

 

lol, fair enough whatever turns you on

ok i get [math'] \frac{d\theta}{dt} = -1 [/math]

 

Edit: I just checked and do agree with [math] \frac{d\theta}{dt} = -1 [/math]

 

how about

for (i)

[math]

 

r(t) = \frac{1}{\sqrt{2(t_0-t)}}

 

[/math]

 

Yes, that's better.

yep,, got it all now. thanks Severian.