Curvature of Ellipse

1. soe236

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

2. Dick

25,853
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.

3. soe236

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)?

4. Dick

25,853
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.

5. soe236

24
Oh I see.. okay, thank you very much!