Unraveling the Mystery of Directional Derivatives

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SUMMARY

The discussion focuses on calculating the gradient of a function of two variables, f(x,y), at a specific point P using directional derivatives. The tangent vector to the level curve at P is given as (2,1), leading to the conclusion that the gradient of f at P, when dotted with this vector, equals zero. Additionally, the directional derivative of f at P in the direction (3/5, 2/5) is stated to be 1/5. The correct gradient of f at P is determined to be (1, -2) through the application of simultaneous equations and careful algebraic manipulation.

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cooev769
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I've come across a question which as really stumped me because I thought I knew how to do this but apparently not.

The question is that we have a tangent vector to a level curve of a function of two variables f(x,y) at a point P is (2,1). Hence by my logic this means grad of f x unit vector of (2,1) is 0. The next part is that the directional derivative of f at P in the direction (3/5, 2/5) is 1/5. Hence by applying the same operation above. Find the gradient of f at P. I would just evaluate both and use simultaneous equations. But the answer they arrive at is really nice at (1, -2) I can't get it myself. Any help would be appreciated thanks.
 
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Looks like the method I'd use. Please show your working - step by step.
Double check the algebra and arithmetic.
 

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