Question

At what point does the curve have maximum curvature? y = 9 ln(x) (x, y) =

Answer 1

y = 9ln(x) 
y' = 9x^-1 =9/x
y'' = -9x^-2 =-9/x^2

curvature k = |y''| / (1 + (y')^2)^(3/2) 

= |-9/x^2| / (1 + (9/x)^2)^(3/2) 
= (9/x^2) / (1 + 81/x^2)^(3/2) 
= (9/x^2) / [(1/x^3) (x^2 + 81)^(3/2)] 
= 9x(x^2 + 81)^(-3/2). 

To maximize the curvature,

we find where k' = 0. 
k' = 9 * (x^2 + 81)^(-3/2) + 9x * -3x(x^2 + 81)^(-5/2) 
...= 9(x^2 + 81)^(-5/2) [(x^2 + 81) - 3x^2] 
...= 9(81 - 2x^2)/(x^2 + 81)^(5/2) 

Setting k' = 0 yields x = ±9/√2. 

Since k' < 0 for x < -9/√2 and k' > 0 for x > -9/√2 (and less than 9/√2), 
we have a minimum at x = -9/√2. 

Since k' > 0 for x < 9/√2 (and greater than 9/√2) and k' < 0 for x > 9/√2, 
we have a maximum at x = 9/√2. 

x=9/√2=6.36

y=9 ln(x)=9ln(6.36)=16.66  

the answer is
(x,y)=(6.36,16.66)