Question

Find the point(s) on the curve of 4x^2+9y^2=36 at which the curvature is largest.

Answer 1

@amistre64 @zepdrix

Answer 2

is curvature a representative tangent to a circle at a point?

Answer 3

\[k(x)=\frac{ \left| y'' \right| }{ [1+(y')^2]^{3/2} }\]

Answer 4

yeah, then i believe you just fill in the parts

Answer 5

by curvature, you mean rate of change of tangent vector with arc length??

Answer 6

just like a curve has a tangent line, it also has a tangent circle of a given radius

Answer 7

I need to find the curvature using the formula above, then find the derivative of curvature and set it equal to 0 to find the critical points. And plug the critical points to 4x^2+9y^2=36 to find the y values since the question is asking point(s) where the curvature is largest. But how do I find the curvature first? I'm little stucked on finding the derivatives y' and y".

Answer 8

osculating (means kissing) circles :)

Answer 9

well, what is k'?

Answer 10

You mean I supposed to find the derivative of curvature k'(x) first and then find y' and y" to plug those values into k'(x)?

Answer 11

well, its simpler to determine k' as is and then input the needed y derivatives

Answer 12

you have a quotient, and a chain rule

Answer 13

So what's k'(x)?

Answer 14

thats the question yes, but moreso k'(y)

Answer 15

\[k=y''~(1+[y']^2~)^{-3/2}\]
\[k'=y'''~(1+[y']^2~)^{-3/2}-\frac{3}{2}~2y'~y''~y''~(1+[y']^2~)^{-5/2}\]
\[k'=y'''~(1+[y']^2~)^{-3/2}-3y'~[y'']^2(1+[y']^2~)^{-5/2}\]
seems about right to me assuming some tings of course about the absolute values

Answer 16

Is it really okay to get rid of the absolute values like ^ that?

Answer 17

if you address it correctly it is

Answer 18

if y'' -s postive on the interval, then y'' = |y''|

Answer 19

4x^2+9y^2=36

8x x'+ 18y y'=0

y'= -4/9 x

y'' = -4/9

y''' = 0

Answer 20

|y''| =4/9 so im not worried about it

Answer 21

8x+18y(dy/dx)=0
18y(dy/dx)=-8x
dy/dx=-8x/(18y)=-(4x)/(9y)

Answer 22

\[k'=y'''~(1+[y']^2~)^{-3/2}-3y'~[y'']^2(1+[y']^2~)^{-5/2}\]

\[k'=0~(1+[y']^2~)^{-3/2}+3\frac49x~[\frac49]^2(1+[\frac49x]^2~)^{-5/2}\]

\[0=3\frac49x~[\frac49]^2(1+[\frac49x]^2~)^{-5/2}\]

when x=0

Answer 23

But how did you get y'=-4/9 x when I got dy/dx=-(4x)/9y?

Answer 24

-4/9 x = -4x/(9y)

Answer 25

i see, i dropped a y didnt i

Answer 26

y is still in the denominator ... so what zeros out is the top part

Answer 27

4x^2+9y^2=36

an ellipse centered at the origin

(0,+-2)
(+-3,0)

Answer 28

we would expect the largest center of the curve around an ellipse to have the biggest curvature right?

Answer 29

\[0=3\frac49\frac{x}{\color{red}y}~[\frac49]^2(1+[\frac{4x}{9y}]^2~)^{-5/2}\]
\[0=\frac{3.4.4^2~x}{9.9^2.y.(1+[\frac{4x}{9y}]^2~)^{-5/2}}\]
it still equals zero when x=0

Answer 30

@amistre64 it looks like you're saying the curvature is greatest at the points on the y-axis. (Please correct me if I'm wrong, I only read the last few posts.)

Here's a quick animation made in Mathematica that demonstrates the curvature is highest at the points along the major axis (on the x-axis).
http://imgur.com/78JtIpA

Answer 31

yeah that is what was going thru my head at the time. i might have been confusing the largest osculating circle with the 'fastest' rate of curve tho in hindsight. thnx for the insight :)