ExamVeda
Login
Home

Let phi be an arbitrary smooth real valued scalar function and...

This question belongs to Engineering Maths Calculus
Calculus
?

Let \[\phi \] be an arbitrary smooth real valued scalar function and V be an arbitrary smooth vector valued function in a three-dimensional space. Which one of the following is an identity?

Answer & Solution
Correct Answer: Option C
Let's discuss the solution. Join the discussion
Examveda
Question posted by Examveda
Community

Join the Discussion

No comments yet Be the first to discuss this question.
Keep Learning

Related Questions on Calculus

What is the value of \[\mathop {\mathop {\lim }\limits_{{\text{x}} \to 0} }\limits_{{\text{y}} \to 0} \frac{{{\text{xy}}}}{{{{\text{x}}^2} + {{\text{y}}^2}}}?\]
A. 1 B. -1 C. 0 D. Limit does not exit
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 + \[\frac{{{{\text{x}}^3}}}{2}\] + ...
\[\mathop {\lim }\limits_{{\text{x}} \to \infty } \sqrt {{{\text{x}}^2} + {\text{x}} - 1} - {\text{x is}}\]
A. 0 B. \[\infty \] C. \[\frac{1}{2}\] D. \[ - \infty \]
The Taylor series expansion of \[\frac{{\sin {\text{x}}}}{{{\text{x}} - \pi }}\]  at \[{\text{x}} = \pi \]  is given by
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. \[ - 1 + \frac{{{{\left( {{\text{x}} - \pi } \right)}^2}}}{{3!}} + ...\]
More Related Questions on Calculus