Question

You can find everywhere that:ˆz=cos(θ)ˆr−sen(θ)ˆθz^=cos(θ)r^−sen(θ)θᵇut if i am not mistaken you cand find this kind of relations using derivates in this way:ˆz=∂r∂zˆr+∂θ∂zˆθ+∂ϕ∂zˆϕz^=∂r∂zr⁺∂θ∂zθ⁺∂ϕ∂zϕ^I got this expression from the tensorial definition of basis vectors.

My question is how is the∂θ∂z=−sen(θ)∂θ∂z=−sen(θ)ifz=rcos(θ)z=rcos(θ)and deriving in an implicit way respecto to z I got this:dzdz=drdzcos(θ)−rsen(θ)dθdzdzdz=drdzcos(θ)−rsen(θ)dθdzUsing the fact thatdrdz=cos(θ)drdz=cos(θ)anddzdz=cos2(θ)+sin2(θ)dzdz=cos2(θ)+sin2(θ)we get:cos2(θ)+sin2(θ)=cos2(θ)−rsen(θ)dθdzcos2(θ)+sin2(θ)=cos2(θ)−rsen(θ)dθdzAnd then:dθdz=−sen(θ)rdθdz=−sen(θ)r Transforming the derivative of a test function Consider the derivative of a test.

Answer 1

In spherical coordinates, the partial derivative ∂θ/∂z can be derived from the relationship z = r cos(θ), leading to the derivative dθ/dz = -sin(θ)/r. This is crucial for expressing the âz unit vector in spherical coordinates as a combination of radial and polar components.

Step-by-step explanation:

Understanding the Relationship between Spherical and Cartesian Coordinates

The relationship between spherical and Cartesian coordinates is fundamental in various areas of mathematics and physics. In spherical coordinates, a point is defined by three parameters: radial distance (r), polar angle (θ), and azimuthal angle (φ). The transformation from spherical to Cartesian coordinates is given by:

• x = r sin(θ) cos(φ)
• y = r sin(θ) sin(φ)
• z = r cos(θ)

To find the partial derivative ∂θ/∂z, we examine the relationship z = r cos(θ). Taking the derivative implicitly with respect to z, we get:

1 = -r sin(θ) dθ/dz,

which yields:

dθ/dz = -sin(θ)/r.

This result is used to express the unit vector âz in spherical coordinates, leading to the combination of âr and âθ components that comprise the âz vector in terms of the spherical unit vectors.