Show that the curvature of the plane parametric curve x=f(t), y = g(t) is
k(t) = (x'y'' - y'x'')/[(x')^2+(y')^2]^3/2
Again I'm still trying to figure out the notation on this website so if anything is confusing the denominator is to the power 3/2
Show that the curvature of the plane parametric curve x=f(t), y = g(t) is
k(t) = (x'y'' - y'x'')/[(x')^2+(y')^2]^3/2
Again I'm still trying to figure out the notation on this website so if anything is confusing the denominator is to the power 3/2
I think you need absolute values in there. I'll do some of it. You want:
$\displaystyle |\frac{d\phi}{ds}|$
which is the rate of change - right out of Liethold - of the measure of the angle giving the direction of the unit tangent vector $\displaystyle \bold{T}(t)$ at a point on a curve with respect to the measure of the arc length along the curve.
Then if the curve is given parametrically as $\displaystyle x=f(t), y=g(t)$:
$\displaystyle \frac{d\phi}{ds}=\frac{\frac{d\phi}{dt}}{\frac{ds} {dt}}=\frac{\frac{d\phi}{dt}}{\sqrt{f'(t)^2+g'(t)^ 2}}$
and note:
$\displaystyle \tan(\phi)=\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\fra c{dx}{dt}}$
Now try and find "The Calculus" by Leithold in the Library and finish it ok.
hey i went and asked him about this question and he said to do this:
k=|r' x r''|/|r'|^3
r= x î + y j
r'=x' i + y j
r''=x" i + y" j
r' x r'' = | i j k|
|x' y' 0|
|x'' y'' 0|
= |x'y'' - y'x''| which is the top part.
to get |r'|^3 -----> |r'|=(x'^2 + y'^2)^(1/2)
that should give you enough info.