Question

The curvature of the helix r​(t)equals(a cosine t )iplus(a sine t )jplusbt k​ (a,bgreater than or equals​0) is kappaequalsStartFraction a Over a squared plus b squared EndFraction . What is the largest value kappa can have for a given value of​ b?

Answer 1

`\kappa = (1)/(2 b)`

Step-by-step explanation:

The equation for kappa ( κ) is

`\kappa = (a)/(a^2 + b^2)`

we can find the maximum of kappa for a given value of b using derivation.

As b is fixed, we can use kappa as a function of a

`\kappa (a) = (a)/(a^2 + b^2)`

Now, the conditions to find a maximum at
`a_0` are:

`(d \kappa(a))/(da) \left | _(a=a_0) = 0`

`(d^2\kappa(a))/(da^2)  \left | _(a=a_0) < 0`

Taking the first derivative:

`(d)/(da) \kappa = (d)/(da)  ((a)/(a^2 + b^2))`

`(d)/(da) \kappa = (1)/(a^2 + b^2) (d)/(da)(a)+ a * (d)/(da)  ((1)/(a^2 + b^2) )`

`(d)/(da) \kappa = (1)/(a^2 + b^2) * 1 + a * (-1)  ((1)/((a^2 + b^2)^2) ) (d)/(da)  (a^2+b^2)`

`(d)/(da) \kappa = (1)/(a^2 + b^2) * 1 - a  ((1)/((a^2 + b^2)^2) ) (2* a)`

`(d)/(da) \kappa = (1)/(a^2 + b^2) * 1 -  2 a^2  ((1)/((a^2 + b^2)^2) )`

`(d)/(da) \kappa = (a^2+b^2)/((a^2 + b^2)^2)  -  2 a^2  ((1)/((a^2 + b^2)^2) )`

`(d)/(da) \kappa = (1)/((a^2 + b^2)^2) (a^2+b^2 -  2 a^2)`

`(d)/(da) \kappa = (b^2 -  a^2)/((a^2 + b^2)^2)`

This clearly will be zero when

`a^2 = b^2`

as both are greater (or equal) than zero, this implies

`a=b`

The second derivative is

`(d^2)/(da^2) \kappa = (d)/(da) ((b^2 -  a^2)/((a^2 + b^2)^2) )`

`(d^2)/(da^2) \kappa = (1)/((a^2 + b^2)^2) (d)/(da) ( b^2 -  a^2 ) + (b^2 -  a^2) (d)/(da) ( (1)/((a^2 + b^2)^2)  )`

`(d^2)/(da^2) \kappa = (1)/((a^2 + b^2)^2) ( -2  a ) + (b^2 -  a^2) (-2) ( (1)/((a^2 + b^2)^3)  ) (2a)`

`(d^2)/(da^2) \kappa = (-2  a)/((a^2 + b^2)^2) + (b^2 -  a^2) (-2) ( (1)/((a^2 + b^2)^3)  ) (2a)`

We dcan skip solving the equation noting that, if a=b, then

`b^2 -  a^2 = 0`

at this point, this give us only the first term

`(d^2)/(da^2) \kappa = (- 2  a)/((a^2 + a^2)^2)`

if a is greater than zero, this means that the second derivative is negative, and the point is a minimum

the value of kappa is

`\kappa = (b)/(b^2 + b^2)`

`\kappa = (b)/(2* b^2)`

`\kappa = (1)/(2 b)`