For a position vector \overrightarrow{r}=x \hat{i}+y \hat{j}+z \hat{k} the norm of the vector can be defined as ∣r^{\to}|\sqrt{x^{2}+y^{2}+z^{2}}. Given a function \ϕ=\In|\overrightarrow{r}|, its gradient ∇ \ϕ is

Correct Answer is : r→r→⋅r→

\overrightarrow{{r}}/\overrightarrow{|{r}|}

\overrightarrow{{r}}/\overrightarrow{{r} \cdot \overrightarrow{r}}

\overrightarrow{{r}}/\overrightarrow{∣ {r}|^{3}}

GATE Mechanical Engineering (ME) 2018 Shift 2 Solved Paper

Question : 63 of 65

Marks: +1, -0

For a position vector

{r}↖{→}=x {i}↖{∧}+y {j}↖{∧}+z {k}↖{∧}

the norm of the vector can be defined as

∣r↖{→}|√{x^{2}+y^{2}+z^{2}}

. Given a function

\ϕ=\In|{r}↖{→}|

, its gradient

∇ \ϕ

Solution:

\ϕ=

r

∇ ϕ=∇(\In r)

f(r)=\In(r)

f^{'}(r)=\;{1}/{r}

∇ f(r)=\;{f^{'}(r)}/{r} ⋅ \ov{r}=(\;{1}/{r}) ×(\;{1}/{r}), {r}↖{→}

∇ f(r)=\;{{r}↖{→}}/{r^{2}}

Go to Question:

Prev Question

Next Question