. Homework Statement
The tension T in the string of the yo-yo is given by:
T=(mgR)/(2r^2+R^2)
where m is the mass of the yo-yo and g is the acceleration due to gravity. Use differentials to estimate the change in tension if R is increased from 3cm to 3.1cm and r is increased from 0.7cm...
I got:
lim r->0^+ [sin(r^2)(2r)-r^2cos(r^2)(2r)]/r^4
which simplifies to:
lim r->0^+ [2sin(r^2)-2r^2cos(r^2)]/r^3
but that's still 0/0 form, so it got me nowhere useful as far as I can see.
Homework Statement
Using polar coordinates, show that lim (x,y)->(0,0) [sin(x^2+y^2)]/[x^2+y^2] = 1
Homework Equations
r^2=x^2+y^2
The Attempt at a Solution
I was able to get the limit into polar coordinates:
lim r->0^+ [sin(r^2)]/r^2
but I'm not sure how to take this limit. I tried...
Oh man...I totally get it now...i feel kinda dumb :P
the derivative would just be:
2|r(t)|d/dt|r(t)|= r'(t) dot r(t) + r(t) dot r'(t)
2|r(t)|d/dt|r(t)|= 2(r'(t) dot r(t))
|r(t)|d/dt|r(t)|= r'(t) dot r(t)
d/dt|r(t)|=(1/|r(t)|) ( r'(t) dot r(t))
Thanks for letting me waste your time :)
Okay, now I'm confused... in this situation, wouldn't f'(g(x)) be equivalent to 2|r(t)| and g'(x) be equivalent to 0 because |r(t)| is a length and the derivative of a number is 0? So then it would be 2|r(t)|(0)=r'(t) dot r(t) + r(t) dot r'(t) --> 0=r'(t) dot r(t) + r(t) dot r'(t)?
If this is...
Homework Statement
If vector r(t) is not 0, show that d/dt |r(t)| = (1/|r(t)|) r(t)dot r'(t).
Homework Equations
The hint given was that |r(t)|^2 = r(t) dot r(t)
The Attempt at a Solution
the post I saw about this question said that you should take the derivatives of both sides of the hint...