MHB Find the gradient of a function at a given point, sketch level curve

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The function \(f(x,y)=\sqrt{2x+3y}\) has a gradient at the point (-1,2) calculated as \(\frac{1}{2}\hat{i}+\frac{3}{4}\hat{j}\). To sketch the level curve passing through this point, the value of the function at (-1,2) is found to be 2, as \(f(-1,2)=\sqrt{4}=2\). This means the level curve is defined by the equation \(2=\sqrt{2x+3y}\), which simplifies to \(4=2x+3y\). The discussion clarifies the relationship between the gradient and the level curve at the specified point. Understanding this connection is crucial for accurately sketching the level curve alongside the gradient.
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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?
 
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Re: find the gradient of a function at a given point, sketch level curve

The 2 comes from:

$f(-1,2)=\sqrt{2(-1)+3(2)}=\sqrt{4}=2$
 
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