It moves along the curve r(t)=<3cost, 2sint, 0>,0<t<2pi. Find v(pi/4) & a(pi/4). At what points is the curvature maximum or minimum?I've found the acceleration and velocity. The...

It moves along the curve r(t)=<3cost, 2sint, 0>,0<t<2pi. Find v(pi/4) & a(pi/4). At what points is the curvature maximum or minimum?

I've found the acceleration and velocity. The curvature being maximum or minimum is too confusing though?! HELP!

Asked on by binges10

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nathanshields | High School Teacher | (Level 1) Associate Educator

Posted on

After a little searching, I came up with this:

`k=(x'y''-y'x'')/((x'^2+y'^2)^(3/2))`

For your case,

`x = 3cost`

`x'=-3sint`

`x''=-3cost`

`y=2sint`

`y'=2cost`

`y''=-2sint`

So we have `((-3sint)(-2sint)-(2cost)(-3cost))/((9sin^2 t+4cos^2 t)^(3/2))`

So let's simplify this up a little

`k=(6sin^2 t+6cos^2t)/((9sin^2 t + 4cos^2 t)^(3/2))`

Wait a second - that numerator is just 6 (since sin^2 + cos^2 = 1)

But to find max or min k, we need the derivative.

`k=6(9sin^2t+4cos^2t)^(-3/2)`

` ` `k'=-(90sint cos t)/(9sin^2t+4cos^2t)^(5/2)`

Finding the roots just involves finding when sin(t)cos(t) = 0, which happens at `pi/2` and `pi` .

At `t=pi/2` , k = 2/9.

At `t = pi` , k = 3/4.

These must be the min and max curvatures, respectively.

 

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