# If a curve has the property that the position vector

If a curve has the property that the position vector $r\left(t\right)$ is always perpendicular to the tangent vector ${r}^{\prime }\left(t\right)$, show that the curve lies on a sphere with center the origin.
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kaluitagf

Step 1
It is given that $r\left(t\right)×{r}^{\prime }\left(t\right)=0$ for all $t\in \mathbb{R}$
Now, let us consider,
$\frac{d\left(r\left(t\right)×{r}^{\prime }\left(t\right)\right)}{dt}=r\left(t\right)×r\left(t\right)+r\left(t\right)×{r}^{\prime }\left(t\right)$
$=2×r\left(t\right)×{r}^{\prime }\left(t\right)$
(given $r\left(t\right)×{r}^{\prime }\left(t\right)=0$)
$=0$
This implies that $\frac{d\left(\parallel r\left(t\right){\parallel }^{2}\right)}{dt}=0$
$⇒\parallel r\left(t\right){\parallel }^{2}$ is constant and therefore, $\parallel r\left(t\right)\parallel$ should also be a constant. Let K be the constant such that $\parallel r\left(t\right)\parallel =K$
This implies that, $\parallel r\left(t\right){\parallel }^{2}={K}^{2}$
Let us say x, y, z are the components of vector function $r\left(t\right)$, this implies that, we have,
${x}^{2}+{y}^{2}+{z}^{2}={K}^{2}$
which is a sphere centered at and with radius K
Thus, we have such that,
$\parallel r\left(t\right)\parallel =\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}$
$⇒\parallel r\left(t\right){\parallel }^{2}={x}^{2}+{y}^{2}+{z}^{2}$
Hence, the curve $r\left(t\right)$ must be a sphere centered at origin.

zurilomk4
Step 1
Given that if a curve has the property that the position vector $r\left(t\right)$ is always perpendicular to the tangent vector ${r}^{\prime }\left(t\right)$. It is required to show that the curve lies on a sphere with centre the origin.
Let $r\left(t\right)$ be then,
$|r\left(t\right)|=\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}$
${|r\left(t\right)|}^{2}={x}^{2}+{y}^{2}+{z}^{2}$
Step 2
Since the position vector $r\left(t\right)$ is always perpendicular to the tangent vector ${r}^{\prime }\left(t\right)$
$r\left(t\right)×{r}^{\prime }\left(t\right)=0$
$2×r\left(t\right)×{r}^{\prime }\left(t\right)=2×0$
$2×r\left(t\right)×{r}^{\prime }\left(t\right)=0$
$\frac{d}{dt}\left({|r\left(t\right)|}^{2}\right)=0$
Now integrate on both sides,
$\int \frac{d}{dt}\left({|r\left(t\right)|}^{2}\right)=\int 0$
${|r\left(t\right)|}^{2}=c$
It is observed that as $|r\left(t\right)|=c$ then ${x}^{2}+{y}^{2}+{z}^{2}=c$, the position vector of the curve is a constant that is the distance is same in all directions.
Thus, the curve lies on a sphere whose centre is origin.

nick1337

Step 1
So we have
$r\left(t\right)×{r}^{\prime }\left(t\right)=0$
which gives us;
$\left(r\left(t\right)×r\left(t\right){\right)}^{\prime }=2r\left(t\right)×{r}^{\prime }\left(t\right)=0$
Thus:
$\parallel r\left(t\right){\parallel }^{2}=r\left(t\right)×r\left(t\right)=c$
for some constant c
So the curve lies on:
${x}^{2}+{y}^{2}+{z}^{2}=c$