Friday

April 18, 2014

April 18, 2014

Posted by **Jaime** on Saturday, December 8, 2007 at 11:08pm.

This is a three-part question:

a. the graph y = f(x) in the xy-plane has parametrization x=x, y=f(x), and vector formula r(x) = xi + f(x)j. Use this to show that if f is twice-differentiable, then

((absvalue f''(x))/[1+((f'(x))^2]^3/2

b. use the formula for k in part a to find the curvature of y = ln(cosx) when -pi/2 < x < pi/2.

c. show that the curvature is zero at the point of inflection.

- math -
**drwls**, Sunday, December 9, 2007 at 1:14amYou left out something in (a) There is no = sign. It looks like you are trying to compute the curvature of the y = f(x) line, which is defined as the reciprocal of the radius of curvature, or d(phi)/ds, where phi is the slope and x is the arc length.

I suggest you review derivations such as the ones at

http://books.google.com/books?id=chjoVpzLZ2UC&pg=PA203&lpg=PA203&dq=curvature+%22dy+dx%22&source=web&ots=mCjWZsZZxM&sig=5T1SzRY3HMvrctecDnY7KMfZJjc

or

http://www.everything2.com/index.pl?node_id=1290875

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