Take R2. Take a function f(x,y) defined on R2 which maps every point to a real number. The gradient of this at any point mean a vector which points in the direction of steepest incline. The magnitude of the vector is the value of the derivative of the function in that direction. Both of these things are very real. This vector is solid and is surely there, so why doesn't it transform contravariantly? I had a go to explore this take x coordinate as Cartesian and x bar as the polars. Then define the function f(x,y) as x^2 +y^2. And if you work out the gradient vector and transform this contarvariant it does give you (2r, 0) which is thee gradient vector relative to polars.(adsbygoogle = window.adsbygoogle || []).push({});

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# Why is grad(f) a covariant vector

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