Question

The curvature of the helix r(t)=(acost)i+(asint)j+(bt)k (a, b is greater than or equal to 0) is k=a/(a^2+b^2). What is the largest value k can have for a given value of b?

Answer 1

Answer:

  1/(2b)

Step-by-step explanation:

The value of k will be a maximum where its derivative with respect to 'a' is zero. The derivative of k with respect to 'a' is ...

  $k'=\dfrac{(a^2+b^2)(1)-a(2a)}{(a^2+b^2)^2}=\dfrac{b^2-a^2}{(a^2+b^2)^2}$

We want to find 'a' when k' = 0. k' will be zero when the numerator of the rational function is zero:

  b^2 -a^2 = 0

  (b -a)(b +a) = 0

  a = ±b

Then the value of k is ...

  k = ±b/(b^2 +b^2) = ±1/(2b)

The largest value k can have is 1/(2b).

Answer 2

The largest value k is 1/2b can have for a given value of b.

Given

The curvature of the helix $\rm r(t)=(acost)i+(asint)j+(bt)k$

Where a, b is greater than or equal to 0 is k=a/(a^2+b^2).

How to determine the largest value of the function?

The largest value of k is determined by using differentiating the value of the function.

The function is;

$\rm k=\dfrac{a}{(a^2+b^2)}$

Differentiate both sides with respect to a.

$\rm \dfrac{dk}{da}=\dfrac{d\left (\dfrac{a}{a^2+b^2\right )}}{da}\\\\\rm \dfrac{dk}{da}=\dfrac{1}{a^2+b^2} \times \dfrac{d(a)}{da}+ a \times \dfrac{d \left (\dfrac{1}{a^2+b^2} \right )}{da}\\\\\rm \dfrac{dk}{da}=\dfrac{1}{a^2+b^2} \times 1 + a \times (-2a) \times \dfrac{1}{(a^2+b^2)^2}\\\\\rm \dfrac{dk}{da}=\dfrac{1}{a^2+b^2} -\dfrac{2a^2}{(a^2+b^2)^2}\\\\\dfrac{d^2k}{da^2} =\dfrac{-2a}{(a^2+b^2)^2}+(b^2-a^2) (-2) \dfrac{1}{(a^2+b^2)^3}(2a)$

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

Then,

The value of k is;

$\rm k =\dfrac{b}{b^2+b^2}\\\\k = \dfrac{b}{2b^2}\\\\k = \dfrac{1}{2b}$

Hence, the largest value k is 1/2b can have for a given value of b.

To know more about Differentiation click the link given below.

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