Here's the graph of the scalar field $f(x, y)=-\left(x-\frac{1}{2}\right)^{2}-y^{2}+1$, where $-1

Q: 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

∻Solution by Steps∻ ‖ step 1 ⋮ To find the points where the gradient of $f(x, y)$ is zero, we calculate the partial derivatives of $f(x, y) = -\left(x - \frac{1}{2}\right)^2 - y^2 + 1$. ‖ ‖ 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.⚹◊What is the formula to calculate the gradient of a scalar field $f(x, y)$?⍭ Generate me a similar question◊

Question

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

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the points where the gradient of $f(x, y)$ is zero, we calculate the partial derivatives of $f(x, y) = -\left(x - \frac{1}{2} ight)^2 - y^2 + 1$. ‖ ‖ step 2 ⋮ The partial derivative with respect to $x$ is $\frac{\partial f}{\partial x} = -2\left(x - \frac{1}{2} ight)$. ‖ ‖ 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: $\begin{cases} -2\left(x - \frac{1}{2} ight) = 0 \ -2y = 0 \end{cases}$. ‖ ‖ step 5 ⋮ Solving the system of equations, we find $x = \frac{1}{2}$ and $y = 0$. ‖ ‖ step 6 ⋮ There is only one point $(x, y) = \left(\frac{1}{2}, 0 ight)$ 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.⚹◊What is the formula to calculate the gradient of a scalar field $f(x, y)$?⍭ Generate me a similar question◊

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