The standard helix has parameterization

The standard helix has parameterization . Find the curvature of the helix. Does the result agree with your intuition?

The correct answer and explanation is:

Okay, let’s find the curvature of the standard helix.

The standard helix has the parameterization:

        r(t)=(acos⁡t,asin⁡t,bt)\mathbf{r}(t) = (a \cos t, a \sin t, bt)r(t)=(acost,asint,bt)
      

To find the curvature

        κ\kappaκ
      

, we can use the formula:

        κ=∣∣r′(t)×r′′(t)∣∣∣∣r′(t)∣∣3\kappa = \frac{||\mathbf{r}'(t) \times \mathbf{r}''(t)||}{||\mathbf{r}'(t)||^3}κ=∣∣r′(t)∣∣3∣∣r′(t)×r′′(t)∣∣​
      

First, we find the first and second derivatives:

        r′(t)=ddt(acos⁡t,asin⁡t,bt)=(−asin⁡t,acos⁡t,b)\mathbf{r}'(t) = \frac{d}{dt}(a \cos t, a \sin t, bt) = (-a \sin t, a \cos t, b)r′(t)=dtd​(acost,asint,bt)=(−asint,acost,b)
      

        r′′(t)=ddt(−asin⁡t,acos⁡t,b)=(−acos⁡t,−asin⁡t,0)\mathbf{r}''(t) = \frac{d}{dt}(-a \sin t, a \cos t, b) = (-a \cos t, -a \sin t, 0)r′′(t)=dtd​(−asint,acost,b)=(−acost,−asint,0)
      

Next, we calculate the magnitude of the first derivative (the speed):

        ∣∣r′(t)∣∣=(−asin⁡t)2+(acos⁡t)2+b2=a2sin⁡2t+a2cos⁡2t+b2||\mathbf{r}'(t)|| = \sqrt{(-a \sin t)^2 + (a \cos t)^2 + b^2} = \sqrt{a^2 \sin^2 t + a^2 \cos^2 t + b^2}∣∣r′(t)∣∣=(−asint)2+(acost)2+b2​=a2sin2t+a2cos2t+b2​
      

        ∣∣r′(t)∣∣=a2(sin⁡2t+cos⁡2t)+b2=a2(1)+b2=a2+b2||\mathbf{r}'(t)|| = \sqrt{a^2(\sin^2 t + \cos^2 t) + b^2} = \sqrt{a^2(1) + b^2} = \sqrt{a^2 + b^2}∣∣r′(t)∣∣=a2(sin2t+cos2t)+b2​=a2(1)+b2​=a2+b2​
      

Now, we calculate the cross product of the first and second derivatives:

        r′(t)×r′′(t)=∣ijk−asin⁡tacos⁡tb−acos⁡t−asin⁡t0∣\mathbf{r}'(t) \times \mathbf{r}''(t) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -a \sin t & a \cos t & b \\ -a \cos t & -a \sin t & 0 \end{vmatrix}r′(t)×r′′(t)=​i−asint−acost​jacost−asint​kb0​​
      

        =i((acos⁡t)(0)−b(−asin⁡t))−j((−asin⁡t)(0)−b(−acos⁡t))+k((−asin⁡t)(−asin⁡t)−(acos⁡t)(−acos⁡t))= \mathbf{i}((a \cos t)(0) - b(-a \sin t)) - \mathbf{j}((-a \sin t)(0) - b(-a \cos t)) + \mathbf{k}((-a \sin t)(-a \sin t) - (a \cos t)(-a \cos t))=i((acost)(0)−b(−asint))−j((−asint)(0)−b(−acost))+k((−asint)(−asint)−(acost)(−acost))
      

        =i(absin⁡t)−j(abcos⁡t)+k(a2sin⁡2t+a2cos⁡2t)= \mathbf{i}(ab \sin t) - \mathbf{j}(ab \cos t) + \mathbf{k}(a^2 \sin^2 t + a^2 \cos^2 t)=i(absint)−j(abcost)+k(a2sin2t+a2cos2t)
      

        =(absin⁡t,−abcos⁡t,a2(sin⁡2t+cos⁡2t))=(absin⁡t,−abcos⁡t,a2)= (ab \sin t, -ab \cos t, a^2(\sin^2 t + \cos^2 t)) = (ab \sin t, -ab \cos t, a^2)=(absint,−abcost,a2(sin2t+cos2t))=(absint,−abcost,a2)
      

Finally, we calculate the magnitude of the cross product:

        ∣∣r′(t)×r′′(t)∣∣=(absin⁡t)2+(−abcos⁡t)2+(a2)2||\mathbf{r}'(t) \times \mathbf{r}''(t)|| = \sqrt{(ab \sin t)^2 + (-ab \cos t)^2 + (a^2)^2}∣∣r′(t)×r′′(t)∣∣=(absint)2+(−abcost)2+(a2)2​
      

        =a2b2sin⁡2t+a2b2cos⁡2t+a4=a2b2(sin⁡2t+cos⁡2t)+a4= \sqrt{a^2 b^2 \sin^2 t + a^2 b^2 \cos^2 t + a^4} = \sqrt{a^2 b^2(\sin^2 t + \cos^2 t) + a^4}=a2b2sin2t+a2b2cos2t+a4​=a2b2(sin2t+cos2t)+a4​
      

        =a2b2(1)+a4=a2b2+a4=a2(b2+a2)= \sqrt{a^2 b^2(1) + a^4} = \sqrt{a^2 b^2 + a^4} = \sqrt{a^2(b^2 + a^2)}=a2b2(1)+a4​=a2b2+a4​=a2(b2+a2)​
      

Assuming

        a>0a > 0a>0
      

, this is

        aa2+b2a \sqrt{a^2 + b^2}aa2+b2​
      

.

Now, substitute these magnitudes into the curvature formula:

        κ=aa2+b2(a2+b2)3=a(a2+b2)1/2(a2+b2)3/2=a(a2+b2)1/2−3/2=a(a2+b2)−1\kappa = \frac{a \sqrt{a^2 + b^2}}{(\sqrt{a^2 + b^2})^3} = \frac{a (a^2 + b^2)^{1/2}}{(a^2 + b^2)^{3/2}} = a (a^2 + b^2)^{1/2 - 3/2} = a (a^2 + b^2)^{-1}κ=(a2+b2​)3aa2+b2​​=(a2+b2)3/2a(a2+b2)1/2​=a(a2+b2)1/2−3/2=a(a2+b2)−1
      

        κ=aa2+b2\kappa = \frac{a}{a^2 + b^2}κ=a2+b2a​
      

The curvature of the standard helix

        r(t)=(acos⁡t,asin⁡t,bt)\mathbf{r}(t) = (a \cos t, a \sin t, bt)r(t)=(acost,asint,bt)
      

is κ=aa2+b2\kappa = \frac{a}{a^2 + b^2}κ=a2+b2a​ .

Explanation and Intuition:

The result

        κ=aa2+b2\kappa = \frac{a}{a^2 + b^2}κ=a2+b2a​
      

indicates that the curvature of the standard helix is constant, meaning it bends the same amount at every point. This makes sense intuitively because the helix is symmetric and uniform; its shape doesn’t change as you move along it.

Let’s consider how the parameters

        aaa
      

and

        bbb
      

affect the curvature:

• Parameter ‘a’: This represents the radius of the cylinder the helix wraps around. If aaa is large, the helix is wide. If aaa is small, it’s narrow. Our intuition suggests a wider helix (larger aaa ) should be less curved. The formula κ=aa2+b2\kappa = \frac{a}{a^2 + b^2}κ=a2+b2a​ confirms this: as aaa increases (while bbb is fixed), the denominator a2+b2a^2+b^2a2+b2 grows faster than the numerator aaa , causing κ\kappaκ to decrease. For example, compare a=1,b=1a=1, b=1a=1,b=1 ( κ=1/2\kappa = 1/2κ=1/2 ) to a=10,b=1a=10, b=1a=10,b=1 ( κ=10/101≈0.1\kappa = 10/101 \approx 0.1κ=10/101≈0.1 ).
• Parameter ‘b’: This parameter relates to the “pitch” or vertical steepness of the helix. If bbb is large, the helix rises quickly and is more stretched out vertically. If bbb is small, it rises slowly and is more compressed vertically, becoming closer to a circle. Our intuition suggests a steeper, more stretched-out helix (larger bbb ) should be less curved. The formula confirms this: as bbb increases (while aaa is fixed), the denominator a2+b2a^2 + b^2a2+b2 increases, causing κ\kappaκ to decrease.

Consider the extreme cases:

• If b=0b=0b=0 , the parameterization becomes r(t)=(acos⁡t,asin⁡t,0)\mathbf{r}(t) = (a \cos t, a \sin t, 0)r(t)=(acost,asint,0) , which is a circle of radius aaa in the xy-plane. The curvature formula gives κ=aa2+02=aa2=1a\kappa = \frac{a}{a^2 + 0^2} = \frac{a}{a^2} = \frac{1}{a}κ=a2+02a​=a2a​=a1​ . This matches the well-known curvature of a circle of radius aaa , agreeing perfectly with intuition.
• If a=0a=0a=0 , the parameterization becomes r(t)=(0,0,bt)\mathbf{r}(t) = (0, 0, bt)r(t)=(0,0,bt) , which is a straight line along the z-axis. The curvature formula gives κ=002+b2=0\kappa = \frac{0}{0^2 + b^2} = 0κ=02+b20​=0 (as long as b≠0b \neq 0b=0 ). This also matches the intuition that a straight line has zero curvature.

The formula

        κ=aa2+b2\kappa = \frac{a}{a^2 + b^2}κ=a2+b2a​
      

therefore aligns well with the geometric intuition of how the radius and the steepness of a helix affect its bending.