Question

Problem Set IJ-2. (a) Why does

ds/dt = 1 / (1+t^2) ?

I thought it should equal

ds/dt = 1 / sqrt( (1+t^2) )

I set up the arithmetic like this...

ds/dt = sqrt( (1/(1+t^2))^2 + (t/(1+t^2))^2 )
= sqrt( 1/(1+t^2)^2 + (t^2)/(1+t^2)^2 )
= sqrt ( (1 + t^2) / (1+t^2)^2 )
=sqrt( 1 / (1+t^2) )
= 1 / sqrt(1+t^2) <--- my final answer

Am I missing something?

Answer 1

I'm not sure how to directly calculate ds/dt
Instead we use the idea that
\[ \frac{ds}{dt} = | \vec{v} | = \bigg| \frac{dr}{dt} \bigg| \]
where r is the position vector
\[ \vec{r} = \bigg<\frac{1}{1+t^2}, \frac{t}{1+t^2} \bigg> \]

Answer 2

sqrt( (1/(1+t^2))^2 + (t/(1+t^2))^2 )
= sqrt( 1/(1+t^2)^2 + (t^2)/(1+t^2)^2 )
= sqrt ( (1 + t^2) / (1+t^2)^2 )
=sqrt( 1 / (1+t^2) )
= 1 / sqrt(1+t^2)

the above steps you posted is taking magnitude of the position vector, this tells us the straight line displacement from the origin. (not ds/dt)

taking its derivative with respect to time, tells us rate at which displacement from the origin is changing. ( not speed which is rate at which distance is changing along the path followed by the point)

Answer 3

This is a hard problem. I am also stuck, but I can tell you why you're wrong (lol)
Speed = the norm of the velocity vector.

\[\frac{ |ds| }{ |dt| } = |v| \]

The velocity vector = the derivative of the position vector.

\[v = \frac{ dr }{ dt }\]


In this case, the position vector is
\[r = OP = \ < \frac{ 1 }{ 1 + t^{2} } , \ \frac{ t }{ 1 + t^ {2} } >\]

The velocity vector is the derivative of r (or OP, same thing): Note: I use parentheses instead of vector brackets, because i don't know how to add vector brackets.

\[v = \frac{ dr }{ dt } = \left( \frac{(1+t^{2})(0)-(1)(2t)}{(1 + t^{2})^{2}}, \frac{(1+t^{2})(1)-(t)(2t)}{(1 + t^{2})^{2}}\right)\]
\[v = \frac{ dr }{ dt } = \left( \frac{-2t}{(1 + t^{2})^{2}}, \frac{1-t^{2}}{(1 + t^{2})^{2}}\right)\]


Once you have v (or dr/dt, same thing), just take its norm:

\[\frac{ds}{dt} = |v| = \sqrt{\left( \frac{-2t}{(1 + t^{2})^{2}} \right)^{2} + \left( \frac{1-t^{2}}{(1 + t^{2})^{2}} \right)^{2}}\]
\[\frac{ds}{dt} = |v| = \sqrt{\left( \frac{-4t^{2}}{(1 + t^{2})^{4}} \right) + \left( \frac{(1-t^{2})^{2}}{(1 + t^{2})^{4}} \right)}\]

Beyond this point there be dragons ;)