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Mathematics 22 Online
OpenStudy (ksaimouli):

Curvature

OpenStudy (ksaimouli):

When. To use |T'(t)|/|r'(t)| and this |r'(t)| x r''(t)|/|r'(t)|^3

OpenStudy (ksaimouli):

@Loser66

OpenStudy (ksaimouli):

Both are the formulas of the curvature but got confused when to implement them

OpenStudy (loser66):

Not sure what you mean :)

OpenStudy (ksaimouli):

For instance find the curvature r(t) = <t,t^2,t^3> at the point (1,1,1)

OpenStudy (ksaimouli):

I got the answer

OpenStudy (ksaimouli):

But why can't we use the 1eqyation

OpenStudy (loser66):

r'(t) = <1,2t, 3t^2> r"(t) =<0,2,6t> cross them to get <6t^2,-6t,2>

OpenStudy (loser66):

My prof said that sometime this way is more convenient: K(x) = \(\dfrac{|f"(x)|}{(1+(f'(x))^2)^3/2}\)

OpenStudy (loser66):

if we are given y = f(x) directly, not a parametric equation

OpenStudy (loser66):

We flexibly use one of them. No need to worry. Like derivative, we can find by limit but why , right? when using theorem is much more convenient

OpenStudy (ksaimouli):

but my question is in paramtric form

OpenStudy (ksaimouli):

u said to use \[|r'(t)| X |r''(t)|/|r'(t)|^3\]

OpenStudy (ksaimouli):

because it is convenient?

OpenStudy (loser66):

yes.

OpenStudy (loser66):

you see, just derivatives

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