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Chapter 12: Problem 6

The level curves of the surface \(z=x^{2}+y^{2}\) are circles in the \(x y\) -plane centered at the origin. Without computing the gradient, what is the direction of the gradient at (1,1) and (-1,-1) (determined up to a scalar multiple)?

Short Answer

Expert verified
Short answer: The direction of the gradient at (1,1) is [1,1] and at (-1,-1) is [-1,-1] (up to a scalar multiple).

Step by step solution

01

Direction of the gradient at (1,1)

At the point (1,1), imagine drawing a straight line from the origin (0,0) to the point (1,1). This line represents the direction of the gradient at this point. Since we only care about the direction of the gradient (not its magnitude), we can determine the direction of the gradient as a vector from the origin to the point (1,1). That vector is \(\begin{bmatrix}1\\1\end{bmatrix}\) (up to a scalar multiple).
02

Direction of the gradient at (-1,-1)

At the point (-1,-1), similarly, imagine drawing a straight line from the origin (0,0) to the point (-1,-1). This line represents the direction of the gradient at this point. Again, since we only care about the direction of the gradient (not its magnitude), we can determine the direction of the gradient as a vector from the origin to the point (-1,-1). That vector is \(\begin{bmatrix}-1\\-1\end{bmatrix}\) (up to a scalar multiple). So, the direction of the gradient at (1,1) is \(\begin{bmatrix}1\\1\end{bmatrix}\) and at (-1,-1) is \(\begin{bmatrix}-1\\-1\end{bmatrix}\) (both determined up to a scalar multiple).

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