Question

Find the gradient at the point (3,-3,2) of the scaler field given by f=xy−4ze+35

Parametric equations

Find the gradient at the point (3,-3,2) of the scaler field given by $$f=xy-4ze+35$$

The gradient is given by $$\displaystyle▽{f}={\left(\begin{array}{c} {d}\frac{{f}}{{\left.{d}{x}\right.}}\\{d}\frac{{f}}{{\left.{d}{y}\right.}}\\{d}\frac{{f}}{{\left.{d}{z}\right.}}\end{array}\right)}$$
Now, $$\displaystyle{\left({d}\frac{{f}}{{\left.{d}{x}\right.}}\right)}=\frac{{d}}{{\left.{d}{x}\right.}}{\left({x}{y}-{4}{z}{e}+{35}\right)}=\frac{{d}}{{\left.{d}{x}\right.}}{\left({x}{y}\right)}+\frac{{d}}{{\left.{d}{x}\right.}}{\left(-{4}{z}{e}+{35}\right)}={y}$$ (when taking the partial derivative with respect to x, all other yariables are treating as constants). Similarly,
$$\displaystyle{d}\frac{{f}}{{\left.{d}{y}\right.}}=\frac{{d}}{{\left.{d}{y}\right.}}{\left({x}{y}\right)}+\frac{{d}}{{\left.{d}{y}\right.}}{\left(-{4}{z}{e}+{35}\right)}={x}$$ and $$\displaystyle{d}\frac{{f}}{{\left.{d}{z}\right.}}=\frac{{d}}{{\left.{d}{z}\right.}}{\left(-{4}{z}{e}\right)}+\frac{{d}}{{\left.{d}{z}\right.}}{\left({x}{y}+{35}\right)}=-{4}{e}$$ Therefore, $$\displaystyle▽{f}={\left({y},{x},-{4}{e}\right)}$$ To find the gradient at(3,-3,2), plug in $$x=3, y=-3, z=2$$ into(1): $$\displaystyle▽{f{{\left({3},-{3},{2}\right)}}}={\left(-{3},{3},-{4}{e}\right)}$$