Rewriting vectors in different coordinates

[goohu]

Lets say you have a vector in spherical coordinates; how do you rewrite this vector into a cartesian one and vice versa?

Im fine with rewriting coordinates but vectors have got me confused. I've tried digging through info online but I couldn't find any good examples.

In the following task, I've found the gradient of the Point P. I'm stuck at the last step trying to find the cartesian coordinates.

View attachment 9327

We should use the following equations to transform from spherical to cartesian coordinates but plugging in our data doesn't give the right answer? And also, since out inclination angle is zero shouldn't the resulting vector be pointing on only the Z-axis?

View attachment 9328

 

[jvicens]

I can understand your confusion with transforming vectors between spherical and cartesian coordinates. It can be a tricky concept to grasp, but I am here to help you understand it better.

First, let's review the equations for transforming from spherical to cartesian coordinates:

x = r*sin(theta)*cos(phi)
y = r*sin(theta)*sin(phi)
z = r*cos(theta)

In these equations, r is the distance from the origin to the point, theta is the inclination angle (measured from the positive z-axis), and phi is the azimuthal angle (measured from the positive x-axis).

Now, let's apply these equations to the point P with coordinates (r, theta, phi) = (5, 0, 0) as given in the task. Plugging in these values, we get:

x = 5*sin(0)*cos(0) = 0
y = 5*sin(0)*sin(0) = 0
z = 5*cos(0) = 5

As you can see, the resulting vector is indeed pointing only on the z-axis, which makes sense since the inclination angle is zero.

Now, let's move on to finding the cartesian coordinates of the gradient of P. The gradient is a vector that points in the direction of the steepest increase of a scalar field. In this case, the scalar field is the function that describes P, and the gradient vector will point in the direction of the steepest increase of P.

To find the gradient, we first need to find the partial derivatives of P with respect to x, y, and z. In other words, we need to find the rate of change of P in the x, y, and z directions. The equations for the partial derivatives are:

∂P/∂x = r*cos(theta)*cos(phi)
∂P/∂y = r*cos(theta)*sin(phi)
∂P/∂z = -r*sin(theta)

Now, plugging in the values of r, theta, and phi from point P, we get:

∂P/∂x = 5*cos(0)*cos(0) = 5
∂P/∂y = 5*cos(0)*sin(0) = 0
∂P/∂z = -5*sin(0) = 0

Therefore, the gradient of P at point