Question

Math

Posted 6 months ago

```
Here's the graph of the scalar field $f(x, y)=-\left(x-\frac{1}{2}\right)^{2}-y^{2}+1$, where $-1<x<1$ and $-1<y<1$.
How many points $(x, y)$ within the graphed range have a zero gradient with $f$ ?
Choose 1 answer:
(A) 0
(B) 1
(C) 2
(D) infinite
```

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Answer from Sia

Posted 6 months ago

Solution by Steps

step 2

The partial derivative with respect to $x$ is $\frac{\partial f}{\partial x} = -2\left(x - \frac{1}{2}\right)$

step 3

The partial derivative with respect to $y$ is $\frac{\partial f}{\partial y} = -2y$

step 4

Setting the partial derivatives equal to zero gives us the system of equations: $\begin{cases} -2\left(x - \frac{1}{2}\right) = 0 \\ -2y = 0 \end{cases}$

step 5

Solving the system of equations, we find $x = \frac{1}{2}$ and $y = 0$

step 6

There is only one point $(x, y) = \left(\frac{1}{2}, 0\right)$ within the graphed range that has a zero gradient

B

Key Concept

Gradient and Critical Points

Explanation

The gradient of a function gives the direction of the steepest ascent. Setting the gradient to zero helps us find critical points, which are candidates for local maxima, minima, or saddle points. In this case, there is only one critical point within the specified domain.

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