09

We first compute $$\frac{d}{dt}(\vec r_1\cdot \vec r_2) =(fm,gn)' = (f'm+fm',g'n+gn').$$ We then compute $$\begin{align*}\frac{d}{dt}(\vec r_1)\cdot \vec r_2+\vec r_1\cdot \frac{d}{dt}(\vec r_2) &= (f',g')\cdot(m,n) + (f,g)\cdot (m',n')\\&=(f'm+fm',g'n+gn').\end{align*}$$

Because the length is constant, we know that $\vec r\cdot \vec r= |\vec r|^2 = c^2$. Computing the derivative of both sides of $\vec r\cdot \vec r= c^2$ gives $\frac{d\vec r}{dt}\cdot\vec r + \vec r \cdot \frac{d\vec r}{dt}=0$. The dot product is commutative, so we obtain $2\vec r \cdot \frac{d\vec r}{dt}=0$, or rather $\vec r \cdot \frac{d\vec r}{dt}=0$. The vector $\vec r$ is orthogonal to its derivative.

This is useful when we compute the derivative of a vector of constant length, such as the unit tangent vector $\vec T(t)$ whose length is 1.

The equation $ax+by+cz=0$ is equivalent to $(a,b,c)\cdot (x,y,z)=0$, namely the vector $(x,y,z)$ is orthogonal to $(a,b,c)$.

There are many ways to proceed. One option is to multiply the first equation by $d$ and the second equation by $a$, yielding $adx+bdy+cdz=0$ and $adx+aey+afz=0$. Subtracting the second equation from the first (to eliminate $x$) gives $(bd-ae)y+(cd-af)z=0$, which upon solving for $y$ yields $y=\left(\frac{cd-af}{ae-bd}\right)z$. This gives $y$ in terms of $z$.

We can eliminate $y$ in a similar manner. Multiply the first equation by $e$ and second by $b$, to obtain $aex+bey+cez=0$ and $bdx+bey+bfz=0$. Subtraction gives $(ae-bd)x+(ce-bf)z=0$, or rather $x = \left(\frac{bf-ce}{ae-bd}\right)z$. This solution, together with the previous, gives $$(x,y,z) = \left(\left(\frac{bf-ce}{ae-bd}\right)z,\left(\frac{cd-af}{ae-bd}\right)z,z\right). $$

The solution of the previous problem gave us $(x,y,z)$ in terms of $z$. We are free to pick $z$, so we let $z=ae-bd$ and the solution greatly simplifies to what we call the cross product.

The cross product of two vectors gives us a vector that is orthogonal to both of the vectors.

The cross product of the two vectors, in either order, will provide an answer. So one answer is $$(-1,2,5)\cross (3,0,4) = (8, 19, -6).$$

With Mathematica we have

u = {-1,2,5};
v = {3,0,4};
u\[Cross]v

We have $$\ds\frac{d\vec r}{dt} = (-2\sin t, 5, 2\cos t)$$ which means $\ds\left|\frac{d\vec r}{dt}\right| = \sqrt{29}$. This gives $$\ds\vec T = \frac{ (-2\sin t, 5, 2\cos t) }{\sqrt{29}}.$$ We then compute $$\ds\frac{d\vec T}{dt} = \frac{ (-2\cos t, 0,-2\sin t) }{\sqrt{29}}$$ which means $\ds\left|\frac{d\vec T}{dt}\right| = \frac{2}{\sqrt{29}}$. We then have $$\vec N = (-\cos t, 0,-\sin t).$$ The cross product gives $$\vec B = \frac{ (-5\sin t,-2,5\cos t) }{\sqrt{29}}.$$ The curvature is $$\kappa = \left|\frac{d\vec T}{ds}\right| = \frac{|d\vec T/dt|}{ds/dt} = \frac{2\sqrt{29}}{\sqrt{29}} = \frac{2}{29}.$$

Here's an animation of these topics.

• https://www.geogebra.org/m/kaGKS9BB

Here's some Mathematica code to compute all the above.

r[t_] := {2 Cos[t], 5 t, 2 Sin[t]}
T[t_] := r'[t]/Norm[r'[t]] // ComplexExpand // Simplify
Nn[t_] := T'[t]/Norm[T'[t]] // ComplexExpand // Simplify
B[t_] := T[t]\[Cross]Nn[t] // ComplexExpand // Simplify
\[Kappa][t] := Norm[T'[t]]/Norm[r'[t]] // ComplexExpand // Simplify
r'[t]
T[t]
T'[t]
Nn[t]
B[t]
\[Kappa][t]