Question

Exercises: A simple field which we often use is one whose value at any point is just equal to the distance r of that point from the origin: T(x,y,z) = \x2 + y2 +:2 =r Sketch a contour map of this field on a plane through the origin. Calculate the gradient of this field and represent it as a vector quantity at a few points on your contour map

Full working out and answers please.

Answer 1

Contour map of this field   $T(x,y,z) = \sqrt{x^{2} + y^{2} + z^{2}} = r$ , on a plane is simply circle of radius r.

у X-y plane , z=0 o

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

$\triangledown T = \frac{\partial T}{\partial x} \hat{i} + \frac{\partial T}{\partial y} \hat{j} + \frac{\partial T}{\partial z} \hat{k}$

$\triangledown T = \frac{1}{2} \frac{2x}{\sqrt{x^{2}+y^{2}+z^{2}}} .\hat{i} + \frac{1}{2} \frac{2y}{\sqrt{x^{2}+y^{2}+z^{2}}} .\hat{j} + \frac{1}{2} \frac{2z}{\sqrt{x^{2}+y^{2}+z^{2}}} .\hat{k}$

$\triangledown T = \frac{ x \hat{i} + y \hat{j} + z \hat{k} }{\sqrt{x^{2}+y^{2}+z^{2}}} = \frac{\vec{r}}{r} = \hat{r}$

Gradient of given function T is unit vector in radial direction. To represent the gradient on any point on contour , we just draw a unit vector in radial direction at any point on the contour .