Question

Short answer

The gradient vector field of \(f\) is \(\overline V f\left( {x,y} \right) = 2xi - j\).

Step-by-step solution

Given Information.

It is given that \(f(x,y) = {x^2} - y\).

Find the gradient vector field.

\(f(x,y) = {x^2} - y\)

\(\overline V f\left( {x,y} \right) = {f_x}\left( {x,y} \right)i + {f_y}\left( {x,y} \right)j\)

Applying the formula, we get

\(\begin{aligned}{}\overline V f\left( {x,y} \right) &= {f_x}\left( {x,y} \right)i + {f_y}\left( {x,y} \right)j\\ &= \frac{\partial }{{\partial x}}\left( {{x^2} - y} \right)i + \frac{\partial }{{\partial y}}\left( {{x^2} - y} \right)j\\ &= 2xi - j\end{aligned}\)

Therefore, the gradient vector field of \(f\) is \(\overline V f\left( {x,y} \right) = 2xi - j\).

Sketch the gradient vector field.

The sketch of the gradient vector field \(\overline V f\left( {x,y} \right) = 2xi - j\) is: