Question

r×a/r.a differentiate

Answer 1

Step-by-step explanation:

(r×a) /( r.a) = ra sinx. n^ / ra cosx = tanx

=> d/dx tanx(n^)

= sec²x( n^)

where n^ is direction vector

hope this helps you

Answer 2

Differentiation of vector products

To find: differentiate $\frac{\vec{r}\times\vec{a}}{\vec{r}.\vec{a}}$

Solution:

• We know that, $\vec{r}\times\vec{a}=|\vec{r}||\vec{a}|sin\theta\hat{n}$ where the two vectors $\vec{r}$ and $\vec{a}$ are inclined at an angle $\theta$, and $\hat{n}$ is the unit vector orthogonal to both $\vec{r}$ and $\vec{a}$.

• Similarly, $\vec{r}.\vec{a}=|\vec{r}||\vec{a}|cos\theta$ where the two vectors $\vec{r}$ and $\vec{a}$ are inclined at an angle $\theta$.

Now, $\frac{\vec{r}\times\vec{a}}{\vec{r}.\vec{a}}$

$\quad=\frac{|\vec{r}||\vec{a}|sin\theta\hat{n}}{|\vec{r}||\vec{a}|cos\theta}$

$\quad=tan\theta\hat{n}$

We see that, $\theta$ is the variable here and thus we differentiate with respect to $\theta$

$\quad\frac{d}{d\theta}(\frac{\vec{r}\times\vec{a}}{\vec{r}.\vec{a}})=\frac{d}{d\theta}(tan\theta\hat{n})$

$\Rightarrow \frac{d}{d\theta}(\frac{\vec{r}\times\vec{a}}{\vec{r}.\vec{a}})=\frac{d}{d\theta}(tan\theta)\hat{n}$

$\Rightarrow \frac{d}{d\theta}(\frac{\vec{r}\times\vec{a}}{\vec{r}.\vec{a}})=sec^{2}\theta\hat{n}$

This is the required differentiation.