# Question with the prove of r(t) ande r'(t) is perpendicular.

1. Aug 29, 2010

### yungman

In the prove of $\vec{r}(t) \;&\; \vec{r}'(t) \;$ is perpendicular:

$$\vec{r}(t) \;\cdot\; \vec{r}(t) \;=\; |\vec{r}(t)|^2 \;\Rightarrow\; \frac{d}{dt}[ \vec{r}(t) \;\cdot\; \vec{r}(t)] = \vec{r}(t) \;\cdot\; \vec{r}'(t) + \vec{r}(t) \;\cdot\; \vec{r}'(t) = \frac{d}{dt}[ |\vec{r}(t)|^2]$$

The book claimed since $\; |\vec{r}(t)|^2 \;$ is a constant, $\frac{d}{dt}[ |\vec{r}(t)|^2] = 0$.

Question is why $\; \frac{d}{dt}[ |\vec{r}(t)|^2] = 0\;$? Let $\; \vec{r}(t) = x\hat{x} + y\hat{y}$

$$\; \frac{d}{dt}[ |\vec{r}(t)|^2] = x^2 + y^2\;$$

x and y is not a constant!!! Why $\; |\vec{r}(t)|^2 \;$ is a constant?!!

2. Aug 29, 2010

### Staff: Mentor

Does r(t) describe a circular path?

3. Aug 29, 2010

this is what comes to my mind: If the path is circular then r(t) represents the radius of the circle at any point and the length of r(t) always equals a constant which is the radius. so its derivative must be zero everywhere. this is just another proof for this intuitive thing that the tangent line passing through a point on a circle is perpendicular to the radius of the circle.

4. Aug 30, 2010

### yungman

I found that the book did say if $|\vec{r}(t)|$ is constant.