Thread: a(t)*a'(t)=0?

how do I prove that a(t)*a'(t)=0; given that a(t) is a parametrized curve where the image of the curve lies on a sphere in R3?

You could solve the DE...

$\displaystyle \displaystyle a\,\frac{da}{dt} = 0$

$\displaystyle \displaystyle \int{a\,\frac{da}{dt}\,dt} = \int{0\,dt}$

$\displaystyle \displaystyle \int{a\,da} = C_1$

$\displaystyle \displaystyle \frac{a^2}{2} + C_2 = C_1$

$\displaystyle \displaystyle a^2 = C$ where $\displaystyle \displaystyle C = 2(C_1 - C_2)$.

I don't believe Prove It's proof is valid. He appears to be assuming that a is a real valued function when it should be vector valued. Also, he is trying to prove the converse of the statement- that if the dot product is 0 then the curve lies in a sphere.

In fact, I don't believe the theorem, as stated, is true. Consider the curve $\displaystyle \vec{a}(\theta)= Rcos(\theta)\vec{i}+ (Rsin(\theta)- R)\vec{k}$. At any point on that curve, $\displaystyle x^2+ y^2+ (z- R)^2= R^2 cos^2(\theta)+ R^2 sin^2(\theta)= R^2$ so the curve lies on the surface of the sphere with center at (0, 0, R) and radius R. But $\displaystyle \vec{a}'(\theta)= -Rsin(\theta)\vec{i}+ Rcos(\theta)\vec{k}$.

$\displaystyle \vec{a}\cdot\vec{a'}= -R^2sin(\theta)cos(\theta)+ R^2sin(\theta)cos(\theta)- R^2cos(\theta)= -R^2cos(\theta)\ne 0$.

What is true is that if $\displaystyle \vec{a}$ is a curve lieing on the surface of a sphere with center at the origin, then $\displaystyle \vec{a}\cdot\vec{a}'= 0$.

There are two ways to prove that

Analytically: Any curve lieing on the surface of a sphere with radius R and center at the origin can be written as $\displaystyle \vec{a}(t)= Rcos(\theta(t))sin(\phi(t))\vec{i}$$\displaystyle + Rsin(\theta(t))sin(\phi(t))\vec{j}+ Rcos(\phi(t))\vec{k}$
for some functions $\displaystyle \theta(t)$ and $\displaystyle \phi(t)$

Then $\displaystyle \vec{a}'(t)= (-Rsin(\theta(t))sin(\phi(t))\theta'(t)+ Rcos(\theta(t))cos(\phi(t))\phi'(t))\vec{i}$$\displaystyle + (Rcos(\theta(t))sin(\phi(t))\theta'(t)+ Rsin(\theta(t))cos(\phi(t))\phi'(t))\vec{j}+ (-Rsin(\phi(t))\phi'(t)\ve{k}$
Take the dot product of those two vectors.

Geometrically: $\displaystyle \vec{a(t)}$, the position vector, for each t, is a vector from the origin, the center of the sphere, to the sphere. It is, of course, perpendicular to the tangent plane at that point. And $\displaystyle \vec{a}'(t)$, the tangent vector to a curve lieing in the surface of the sphere, is itself in the tangent plane.