Question

Consider the following. y=(x−5) 2 +1 (a) Find the point on the curve at which the curvature K is a maximum. (x,y)=((x)) (b) Find the limit of the curvature K as x→[infinity]. lim x→[infinity] ​ K=

Answer 1

(a) The point on the curve where the curvature K is a maximum is (x, y) = (5, 1).

(b) The limit of the curvature K as x approaches infinity is 0.

Step-by-step explanation:

(a) The curvature (K) of a curve given by
`y = (x - 5)^2 + 1` can be expressed as

`K = |y''| / (1 + (y')^2)^(3/2)`,

where y' and y'' denote the first and second derivatives of y with respect to x. To find the maximum curvature, we set y'' equal to zero and solve for x. Taking the second derivative of y, y'' = 2.

The curvature equation becomes
`K = 2 / (1 + (2(x - 5))^2)^(3/2)`.

To maximize K, we set the denominator to its minimum value, which occurs when
`(2(x - 5))^2 = 0`. Solving this yields x = 5. Substituting x = 5 back into the original equation gives the point (5, 1) as the location where the curvature is a maximum.

(b) To find the limit of K as x approaches infinity, we examine the curvature equation. As x becomes very large, the term
`(2(x - 5))^2` dominates. Therefore, K approaches 0 as x goes to infinity. This is because the denominator in the curvature formula becomes extremely large compared to the numerator, resulting in a limit of 0. Thus, lim (x → ∞) K = 0.

In summary, the maximum curvature occurs at the point (5, 1), and as x approaches infinity, the curvature approaches zero due to the dominance of the squared term in the denominator of the curvature formula.