In the prove of [itex]\vec{r}(t) \;&\; \vec{r}'(t) \;[/itex] is perpendicular:(adsbygoogle = window.adsbygoogle || []).push({});

[tex] \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] [/tex]

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

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

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

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

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

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