I'm having trouble finding the point of the maximum curvature of the line with parametric equations of: x = 5cos(t), and y = 3sin(t).(adsbygoogle = window.adsbygoogle || []).push({});

I know the curvature "k" is given by the eq.:

k = |vXa|/ v^3

Wherevis the derivative of the position vectorr= <5cos(t), 3sin(t) > ,ais the derivative ofv, and "v" is the norm of the velocity vectorv.

I know I first have to find x', x'', y', and y'' to get the vectorsvanda, and find the norm ofvand cube it.

After plugging in the values and taking the absolute value of the cross product above, I evaluated when the numerator of k' (the derivative of the curvature equation) is equal to zero. I came up with the values of arctan(3/5), pi, zero, and pi/2. The book says the the maximum curvature occurs at x = plus or minus 5. I'm not sure what I did wrong.

Edit: Additionally, is there any way to integrate the square root of " (25*t^2) + 9 " dt without using integration tables? Any help is appreciated.

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# Maximum curvature

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