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

[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?!

 

[Mark44]

Does r(t) describe a circular path?

 

[AdrianZ]

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.

 

[yungman]

Does r(t) describe a circular path?

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

My bad.