Suppose T=T(r,θ)=G(x,y)(adsbygoogle = window.adsbygoogle || []).push({});

How do you prove ∇T(r,θ)=∇G(x,y)?

I can think of some arguments in favor of this equality, but I want an actual proof or a very good intuitive argument. My arguments in favor go something like this:

-Gradient vectors should be the same because if my directional derivative is taken parallel to the gradient vector then I get its maximum/minimum value and if these two gradient vectors are the same then everything will be consistent . However this proves that they should point in the same direction but it doesn't prove they're the same vector!

Also why can't ∇T(r,θ) can't just be = ∂T/∂r (unitvector-r) + ∂T/∂θ (unitvector-θ)

instead of ∇T(r,θ) = ∂T/∂r (unitvector-r) + (1/r)∂T/∂θ (unitvector-θ)... What does the (1/r) term imply? Aside from the fact that it will be useful whenever doing the dot product with some infinitesimally small displacement with a term rdθ (unitvector-θ).

Thanks.

**Physics Forums - The Fusion of Science and Community**

# How to prove gradients vectors are the same in polar and cartesian co.

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

Have something to add?

- Similar discussions for: How to prove gradients vectors are the same in polar and cartesian co.

Loading...

**Physics Forums - The Fusion of Science and Community**