How I can draw the gradients on the tangent to level curves. Shall you give me a hint? For example: $f(x,y) = x^3 - y$ and at points $(0,0)$ $∇f(0,0) = (0,-1)$ . Therefore, How to represent of the gradient whose corresponding constrains ($f(x,y) \leq 0$) are binding at the points given above $(0,0)$.
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The gradient as 0 is actually $(0,-1)$. The tangent plane at 0 is going through the origin with slope -1 along the y axis and 0 along the x axis.
dmh
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1The gradient is $3x^2, -1$ plug 0 into this to get $0,-1$ – dmh Oct 11 '20 at 15:45
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true it is my typo – Steve Josh Oct 11 '20 at 16:29