Question

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

Answer 1

x^2/9+y^2/4=1
x^2/a^2+y^2/b^2=1
x(t)=acos(t)
y(t)=bsin(t)
x(t)=3cos(t)
y(t)=2sin(t)
----------------------
\[k=\frac{ \left| x'y''-y'x'' \right| }{ [(x')^2+(y')^2]^{3/2} }\]

Answer 2

x'(t)=-3sin(t)
y'(t)=2cos(t)
x"(t)=-3cos(t)
y"(t)=-2sin(t)
k=6/(5sin^2 t+4)^(3/2)
(3cos(t), 2sin(t))=(x, y)
sin^2 t=y^2/4
k=6/(5y^2/4+4)^(3/2)
y^2/4=1-x^2/9
so
\[k=\frac{ 162 }{ (81-5x^2)^{3/2} }\]

Answer 3

Now I need to find the derivative of k, so
\[k'=\frac{ 2430x(81-5x^2)^{1/2} }{ (81-5x^2)^{3} }\]

Answer 4

Simplify, I got
\[k'=\frac{ 2430x }{ (81-5x^2)^{5/2} }=0\]

Answer 5

So 2430x=0
and x=0.
So x is the critical point? So what's the correct answer?

Answer 6

@myininaya @amistre64

Answer 7

I think you might also have to consider the endpoints of the ellipse

Answer 8

So how do I find the right answer?

Answer 9

@SithsAndGiggles @thomaster @amistre64

Answer 10

Using the formula you gave, you have
\[k(t)=\frac{|x'y''-x''y'|}{\left(\left(x'\right)^2+\left(y'\right)^2\right)^{3/2}}\]
with \(x(t)=3\cos t\) and \(y(t)=2\sin t\). Your derivatives are
\[\begin{matrix}x'=-3\sin t&&x''=-3\cos t\\
y'=2\cos t&&y''=-2\sin t\end{matrix}\]
So,
\[k(t)=\frac{|6\sin^2t+6\cos^2t|}{\left(9\sin^2t+4\cos^2t\right)^{3/2}}=\frac{6}{\left(9\sin^2t+4\cos^2t\right)^{3/2}}\]
The derivative of the curvature function is
\[k'(t)=-\frac{90\cos t\sin t}{(4\cos^2t+9\sin^2t)^{5/2}}\]
so you have critical points for the values of \(t\) that give
\[\cos t\sin t=0\]
which are easy to find.

Answer 11

So t=pi/2, pi, 3pi/2, am I right?

Answer 12

@myininaya

Answer 13

And \(t=0\), or \(t=2\pi\). This means you have to check four points.

Answer 14

How did you get t=0, 2pi? And what to do after finding t?

Answer 15

When \(t=0\) you get the same point as when \(t=2\pi\) (since \(0\) and \(2\pi\) correspond to the same angle on the unit circle). You get these from the fact that \(\sin t=0\) for \(t=n\pi\).

Now you check the values of the curvature for these \(t\). For example, \(k(0)=\dfrac{6}{4^{3/2}}=\dfrac{3}{4}\) and \(k\left(\dfrac{\pi}{2}\right)=\dfrac{6}{9^{3/2}}=\dfrac{2}{9}\). Clearly the curvature at \(t=0\) is greater than at \(t=\dfrac{\pi}{2}\).

This result makes sense, since the ellipse drawn below bows out more on the major axis (the horizontal axis) than it does on the minor axis: