- #1
skate_nerd
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So I have a function \(f(x,y)=\sqrt{2x+3y}\) and need to find the gradient at the point (-1,2). I got this part already, its \(\frac{1}{2}\hat{i}+\frac{3}{4}\hat{j}\). The part I'm having trouble with is when it asks me to sketch the gradient with the level curve that passes through (-1,2).
The back of the book has the answer and it calls the line passing through the point given point \(4=2x+3y\). I know that in relation to the given function this would also equal \(2=\sqrt{2x+3y}\). So where does that 2 come from?
The back of the book has the answer and it calls the line passing through the point given point \(4=2x+3y\). I know that in relation to the given function this would also equal \(2=\sqrt{2x+3y}\). So where does that 2 come from?