For a position vector \[{\rm{r}} = {\rm{x\hat i}} + {\rm{y\hat j}} + {\rm{z\hat k}}\] the norm of the vector can be defined as $$\left| {\overrightarrow {\text{r}} } \right| = \sqrt {{{\text{x}}^2} + {{\text{y}}^2} + {{\text{z}}^2}} .$$ Given a function $$\phi = \ln \left| {\overrightarrow {\text{r}} } \right|,$$ its gradient $$\nabla \phi $$ is
A. $$\overrightarrow {\text{r}} $$
B. $$\frac{{\overrightarrow {\text{r}} }}{{\left| {\overrightarrow {\text{r}} } \right|}}$$
C. $$\frac{{\overrightarrow {\text{r}} }}{{\overrightarrow {\text{r}} \cdot \overrightarrow {\text{r}} }}$$
D. $$\frac{{\overrightarrow {\text{r}} }}{{{{\left| {\overrightarrow {\text{r}} } \right|}^3}}}$$
Answer: Option C
This Question Belongs to Engineering Maths >> Calculus
The Taylor series expansion of 3 sinx + 2 cosx is . . . . . . . .
A. 2 + 3x - x2 - \[\frac{{{{\text{x}}^3}}}{2}\] + ...
B. 2 - 3x + x2 - \[\frac{{{{\text{x}}^3}}}{2}\] + ...
C. 2 + 3x + x2 + \[\frac{{{{\text{x}}^3}}}{2}\] + ...
D. 2 - 3x - x2 + \[\frac{{{{\text{x}}^3}}}{2}\] + ...
B. \[\infty \]
C. \[\frac{1}{2}\]
D. \[ - \infty \]
A. \[1 + \frac{{{{\left( {{\text{x}} - \pi } \right)}^2}}}{{3!}} + ...\]
B. \[ - 1 - \frac{{{{\left( {{\text{x}} - \pi } \right)}^2}}}{{3!}} + ...\]
C. \[1 - \frac{{{{\left( {{\text{x}} - \pi } \right)}^2}}}{{3!}} + ...\]
D. \[ - 1 + \frac{{{{\left( {{\text{x}} - \pi } \right)}^2}}}{{3!}} + ...\]
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