- #1

- 434

- 0

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).

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

k = |

Where

I know I first have to find x', x'', y', and y'' to get the vectors

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.

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

k = |

**v**X**a**|/ v^3Where

**v**is the derivative of the position vector**r**= <5cos(t), 3sin(t) > ,**a**is the derivative of**v**, and "v" is the norm of the velocity vector**v**.I know I first have to find x', x'', y', and y'' to get the vectors

**v**and**a**, and find the norm of**v**and 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.

Last edited: