Question

Find the gradient of =√x²+y²+z² where is magnitude of position vector ⃗ = + j+k

Answer 1

Answer:

The position vector of a point P(x, y) in two dimensions is xi + ... j +. ∂f. ∂z k . Exercise 1. Calculate the gradient of the ... The vector 4i−3k has magnitude √42 + (−3)2 =.

Answer 2

Answer:

The gradient of the given vector is

$\dfrac{x}{ \sqrt{x^2+y^2+z^2} }\hat i+ \dfrac{y}{ \sqrt{x^2+y^2+z^2}} \hat j+ \dfrac{z}{ \sqrt{x^2+y^2+z^2}} \hat k$

Explanation:

Assuming the given vector is

$R(x,y,z)=\sqrt{x^2+y^2+z^2}$

which represents the magnitude of position vector.

Need to find the gradient of the given vector.

If $f(\vec r) = f(x, y, z)$ is defined as a scalar field, then a scalar function of position $\vec r = (x, y, z)$, then its gradient at any point in three dimensions, $f(x, y, z)$ is given by

$\nabla f(x,y,z)=\dfrac{\partial f}{\partial x} \hat i+\dfrac{\partial f}{\partial y} \hat j+\dfrac{\partial f}{\partial z} \hat k$

Rewriting the vector equation,

$R(x,y,z)=\big(x^2+y^2+z^2\big)^\frac{1}{2}$

Substituting the given vector

$\nabla R(x,y,z)=\dfrac{\partial \big(x^2+y^2+z^2\big)^\frac{1}{2}}{\partial x} \hat i+\dfrac{\partial \big(x^2+y^2+z^2\big)^\frac{1}{2}}{\partial y} \hat j+\dfrac{\partial \big(x^2+y^2+z^2\big)^\frac{1}{2}}{\partial z} \hat k$

$=\dfrac{1}{2} \big(x^2+y^2+z^2\big)^{\frac{1}{2}-1}\times \dfrac{\partial \big(x^2\big)}{\partial x} \hat i+\dfrac{1}{2} \big(x^2+y^2+z^2\big)^{\frac{1}{2}-1}\times \dfrac{\partial \big(y^2\big)}{\partial y} \hat j+\dfrac{1}{2} \big(x^2+y^2+z^2\big)^{\frac{1}{2}-1}\times \dfrac{\partial \big(z^2\big)}{\partial z} \hat k$

$=\dfrac{1}{2} \big(x^2+y^2+z^2\big)^{-\frac{1}{2}}\times 2x^{2-1}\hat i+\dfrac{1}{2} \big(x^2+y^2+z^2\big)^{-\frac{1}{2}}\times 2y^{2-1} \hat j+\dfrac{1}{2} \big(x^2+y^2+z^2\big)^{-\frac{1}{2}}\times 2z^{2-1} \hat k$

$= \big(x^2+y^2+z^2\big)^{-\frac{1}{2}} x\hat i+ \big(x^2+y^2+z^2\big)^{-\frac{1}{2}}\times y \hat j+ \big(x^2+y^2+z^2\big)^{-\frac{1}{2}} z \hat k$

$= \dfrac{x}{ \big(x^2+y^2+z^2\big)^{\frac{1}{2}}}\hat i+ \dfrac{y}{ \big(x^2+y^2+z^2\big)^{\frac{1}{2}}} \hat j+ \dfrac{z}{ \big(x^2+y^2+z^2\big)^{\frac{1}{2}}} \hat k$

$= \dfrac{x}{ \sqrt{x^2+y^2+z^2} }\hat i+ \dfrac{y}{ \sqrt{x^2+y^2+z^2}} \hat j+ \dfrac{z}{ \sqrt{x^2+y^2+z^2}} \hat k$

Therefore, the gradient of the given vector is

$\dfrac{x}{ \sqrt{x^2+y^2+z^2} }\hat i+ \dfrac{y}{ \sqrt{x^2+y^2+z^2}} \hat j+ \dfrac{z}{ \sqrt{x^2+y^2+z^2}} \hat k$

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