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Globokim8

Answered question

2021-05-19

At what point does the curve have maximum curvature? What happens to the curvature as x tends to infinity

y = \ln x

diskusje5

Skilled2021-05-20Added 82 answers

f (x) = \ln x

f^{'} (x) = \frac{1}{x}

f^{″} (x) = - \frac{1}{x^{2}}

Plug into the curvature formula

k (x) = \frac{| f^{″} (x) |}{[1 + (f^{'} (x))^{2}]^{\frac{3}{2}}} = \frac{| - \frac{1}{x^{2}} |}{[1 + (\frac{1}{x})^{2}]^{\frac{3}{2}}} = \frac{1}{x^{2} [1 + \frac{1}{x^{2}}]^{\frac{3}{2}}}

Find k'

k (x) = \frac{1}{x^{2} [1 + \frac{1}{x^{2}}]^{3 / 2}} = x^{- 2} [1 + \frac{1}{x^{2}}]^{- 3 / 2}

k^{'} (x) = - 2 x^{- 3} [1 + \frac{1}{x^{2}}]^{3 / 2} + x^{- 2} \cdot - \frac{3}{2} [1 + \frac{1}{x^{2}}]^{- 5 / 2} (- 2 x^{- 3})

= \frac{- 2}{x^{3} [1 + \frac{1}{x^{2}}]^{3 / 2}} + \frac{3}{x^{5} [1 + \frac{1}{x^{2}}]^{5 / 2}}

= \frac{- 2 x^{2} (1 + \frac{1}{x^{2}}) + 3}{x^{5} [1 + \frac{1}{x^{2}}]^{5 / 2}}

= \frac{- 2 x^{2} - 2 + 3}{x^{5} [1 + \frac{1}{x^{2}}]^{5 / 2}}

= \frac{- 2 - 2 x^{2}}{x^{5} [1 + \frac{1}{x^{2}}]^{5 / 2}}

Find where k'=0

0 = 1 - 2 x^{2}

2 x^{2} = 1

x = \frac{1}{\sqrt{2}} \approx 0.7071

The negative root is not in the domain of ln x. We can check numbers in the surrounding intervals to make sure it is the max:

(0, 0, 7) : k^{'} (0, 1) \approx 0.96

(0, 7, \infty) : k^{'} (1) \approx - 0.18

k is increasing then decreasing, so it is the max.
As

x \to \infty

lim_{x \to \infty} \frac{1}{x^{2} [1 + \frac{1}{x^{2}}]^{3 / 2}} \to \frac{1}{\infty [1 + 0]^{3 / 2}} \to 0

the curvature approaches 0.

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