Question

1) If a scalar field φ(x,y,z)=3x2y-y3z2, calculate the gradient Ñφ at point M(1,-2,1).

Answer 1

If we know the scalar field is $φ(x,y,z)=3x^2y-y^3z^2$, we only have to find the first partial derivatives to have the gradient of the function that describes the scalar field:

$\bigtriangledown φ(x,y,z)= \begin{pmatrix} \frac{\partial φ(x,y,z)}{\partial x}, & \frac{\partial φ(x,y,z)}{\partial y}, & \frac{\partial φ(x,y,z)}{\partial z} \end{pmatrix} = \begin{pmatrix} 6xy, & 3(x^2-y^2z^2) , & -2y^3z \end{pmatrix}$

The gradient can be calculated by substituting (x, y, z) = (1, -2, 1) and we find

$\bigtriangledown φ(1,-2,1)= \begin{pmatrix} 6(1)(-2), & 3((1)^2-(-2)^2(1)^2) , & -2(-2)^3(1) \end{pmatrix}$

$\implies \bigtriangledown φ(1,-2,1)= \begin{pmatrix} -12, & -9 , & 16 \end{pmatrix}$ .

In conclusion, at the point (1, -2, 1) we find the gradient▽φ_{(1, -2, 1)} to be (-12, -9, 16).

Reference:

• Serway, R. A., & Jewett, J. W. (2018). Physics for scientists and engineers. Cengage learning.