Hi
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.
You 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
To answer these three-part question, we will go step by step.
a. To show that ((|f''(x)|)/[1+(f'(x))^2]^3/2 is the formula for the curvature, we need to use the given parametrization and vector formula for r(x). The curvature formula is given by k(x) = |r''(x)| / (1 + (r'(x))^2)^(3/2).
Let's start by finding the first derivative of r(x):
r'(x) = d/dx (xi + f(x)j) = 1i + f'(x)j
Now, let's find the second derivative of r(x):
r''(x) = d/dx (1i + f'(x)j) = 0i + f''(x)j = f''(x)j
With these derivatives, we can substitute them back into the curvature formula:
k(x) = |f''(x)j| / (1 + (1i + f'(x)j)^2)^(3/2)
To simplify further, we use the properties of complex numbers and note that |j| = 1 and |1i + f'(x)j| = sqrt(1^2 + (f'(x))^2) = sqrt(1 + (f'(x))^2):
k(x) = |f''(x)| / [1 + (f'(x))^2]^(3/2)
Thus, we have established that ((|f''(x)|)/[1+(f'(x))^2]^3/2 is the formula for the curvature.
b. Now let's use the formula we derived in part a to find the curvature of y = ln(cosx) when -pi/2 < x < pi/2.
To find the curvature, we need to compute f'(x) and f''(x) for the given function.
First, let's find f'(x):
f'(x) = d/dx (ln(cosx)) = -tanx
Next, let's find f''(x):
f''(x) = d/dx (-tanx) = -sec^2x
Now, substitute these derivatives back into the curvature formula:
k(x) = |f''(x)| / [1 + (f'(x))^2]^(3/2) = |-sec^2x| / [1 + (-tanx)^2]^(3/2) = sec^2x / (1 + tan^2x)^(3/2) = sec^2x / sec^3x = 1 / secx = cosx
Therefore, the curvature of y = ln(cosx) when -pi/2 < x < pi/2 is cosx.
c. To show that the curvature is zero at the point of inflection, we need to find the point(s) of inflection for the function y = ln(cosx).
Inflection points occur when the second derivative changes sign. In this case, the second derivative is f''(x) = -sec^2x.
Setting f''(x) = 0, we get -sec^2x = 0, which implies sec^2x = 0.
However, sec^2x is always positive or zero, so there are no values of x for which sec^2x equals zero.
Since there are no points of inflection for y = ln(cosx), the curvature is always zero for this function.
I hope this explanation helps you understand how to solve these three parts of the question. Let me know if you have any further questions!