Thread: Curvature of a parametric curve

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

Originally Posted by Todeezy

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

What definition of "curvature" are you using?

I'm not sure, that's the question on my problem set. My professor didn't do a very good job explaining anything this semester

I think you need absolute values in there. I'll do some of it. You want:



which is the rate of change - right out of Liethold - of the measure of the angle giving the direction of the unit tangent vector 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 :



and note:



Now try and find "The Calculus" by Leithold in the Library and finish it ok.

Thanks but I'm a little unsure where to go next. Any further hints would be appreciated.

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.

Thanks to both of you. I actually figured it out using the first method, I don't remember us doing any cross product in class but that way looks alot cleaner

ya i don't remember it either, but it's in the textbook as one of the formulas for curvature