Register to reply

Curvature of Ellipse

by soe236
Tags: curvature, ellipse
Share this thread:
soe236
#1
Dec10-08, 10:20 AM
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
Phys.Org News Partner Science news on Phys.org
Security CTO to detail Android Fake ID flaw at Black Hat
Huge waves measured for first time in Arctic Ocean
Mysterious molecules in space
Dick
#2
Dec10-08, 10:29 AM
Sci Advisor
HW Helper
Thanks
P: 25,246
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.
soe236
#3
Dec10-08, 11:12 AM
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)?

Dick
#4
Dec10-08, 01:27 PM
Sci Advisor
HW Helper
Thanks
P: 25,246
Curvature of Ellipse

Quote Quote by soe236 View Post
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)?
That looks ok to me. So the question is just asking you what is k(0) and k(pi/2), right? Those are the curvatures of the ellipse at the two given points.
soe236
#5
Dec10-08, 02:52 PM
P: 24
Oh I see.. okay, thank you very much!


Register to reply

Related Discussions
Ellipse in a box Calculus & Beyond Homework 15
Find the equation of a ellipse given the foci. (1,0) (3,4) General Math 2
Finding the equation of an ellipse from foci and directrices Precalculus Mathematics Homework 2
Geodesic Curvature (Curvature of a curve) Differential Geometry 8
How to find the equations of the axis of the ellipse Introductory Physics Homework 1