# Curvature of Ellipse

by soe236
Tags: curvature, ellipse
 P: 24 Find Curvature of Ellipse given x=3*cos(t) and y=4*sin(t) at the points (3,0) and (0,4) Relevant equations: curvature at r(s) is k(s)=||dT/ds|| when r(s) is arc length parametrization and T is the unit tangent vector I usually use the formula k(t)= (||r'(t) x r''(t)||)/||r'(t)||^3 So, r(t)=<3cost,4sint> and r'(t)= <-3sint,4cost> and r''(t)=<-3cost,-4sint> do I just plug them in for k(t)? And I'm clueless about how to use the given points. Someone help please. Thank you
 Sci Advisor HW Helper Thanks P: 25,235 Yes, you can just plug them into k(t). You'll need to do a cross product, so take the z component of the vectors to be zero. When you get k as a function of t, then you just need to go back and set (3*cos(t),4*sin(t))=(3,0) and figure out what t is for the first point. Ditto for the second.
 P: 24 Thanks, but I'm not sure I completely understand This is what I got so far, please correct me if I'm wrong: r'(t) x r''(t)= 12sint^2+12cost^2 = 12 ||r'(t) x r''(t)||= sqrt(144) =12 ||r'(t)||^3 = (9sint^2+16cost^2)^(3/2) k(t)= 12/(9sint^2+16cost^2)^(3/2) from what you've said, (3*cos(t),4*sin(t))=(3,0), so 3cost=3 => t= 0 or 2pi and 4sint=0 => t=0 or 2pi similarly (3*cos(t),4*sin(t))=(0,4), so 3cost=0 => t=pi/2 or 3pi/2 and 4sint=4 => t=pi/2 Sry if this is a stupid question, but how do I apply that to k(t)?