Question

At what point does the curve have the maximum curvature?
y=7ln(x)

I know how to find the curvature but after finding the formula for the curvature I have to use that to get the critical points and the maximum... it is all very tedious. I was wondering if there was a better way.

Answer 1

@dan815

Answer 2

@Hero

Answer 3

what is you're way you started love

Answer 4

\[k=\frac{\frac{ -7 }{ x^2 }}{\sqrt{1+\frac{49}{x^2}}^3}\]

Answer 5

now to find the greatest curvature i need to take the derivative of that and set it equal to zero. then solve for x. do you see why this is a problem?

Answer 6

mhm hold on love i am thinkingi might be in corrcet about what i am ghoing t otype bhut i going to ask my friend to help wme explan it to you @jabez177

Answer 7

@jabez177 can you help me explan to this user?

Answer 8

never mind yes i do see the probleom

Answer 9

Hello! Too many people tagging me.

Answer 10

hi jabez. in this problem can solve it but not without a calculator. is there a method you can show me where it is possible to solve by hand?

Answer 11

First of all,
k(t) = |f ''(x)| / (1 + (f '(x))^2)^(3/2)
......= |-1/x^2| / (1 + (1/x)^2)^(3/2)
......= (1/x^2) / [(1/x^2) (x^2 + 1)]^(3/2)
......= (1/x^2) / [(1/x^3) (x^2 + 1)^(3/2)]
......= x(x^2 + 1)^(-3/2).

Now, we can find the maximal curvature.

k'(t) = 1(x^2 + 1)^(-3/2) + x * -3x(x^2 + 1)^(-5/2)
......= (x^2 + 1)^(-5/2) [(x^2 + 1) - 3x^2], by factoring
......= (x^2 + 1)^(-5/2) (1 - 2x^2).

Setting k' = 0 for critical points yields 1 - 2x^2 = 0 ==> x = ±1/√2.

Since k' > 0 for x in (-1/√2, 1/√2), and k' < 0 for x < -1/√2 or x > 1/√2, we conclude by the First Derivative Test that the curve has maximal curvature when x = 1/√2.

Answer 12

rosy don't forget the 7

Answer 13

7 ln(x)

Answer 14

:OOOOOO
so multiplying your answer by 7 gave me the correct answer
that must mean that we can use the curvature of JUST ln(x) as our base

Answer 15

that doesn't make sense though. doesn't 7 change the function as a whole once we find the curvature formula and its derivative?