Question

If \(f(x,y) = xy\), find the gradient vector \(\nabla f(3,2)\) and use it to find the tangent line to the level curve \(f(x,y) = 6\) at the point \((3,2)\). Sketch the level curve, the tangent line, and the gradient vector.

Short answer

The gradient vector of the function \(f(x,y) = xy\) at the point \((3,2)\) is \(\nabla f(3,2) = \langle 2,3\rangle \). The tangent line of the surface \(f(x,y) = xy\) is \(2x + 3y = 12\).

Step-by-step solution

The tangent line of the function.

The value of\(\nabla F\left( {{x_0},{y_0},{z_0}} \right) \cdot {r^\prime }\left( {{t_0}} \right) = 0\)where\({r^\prime }\left( {{t_0}} \right)\)is tangent vector.

Find the gradient vector .

The function is, \(f(x,y) = xy\).

Compute the gradient of \(f(x,y)\) at the point \((3,2)\).

\(\begin{aligned}{l}\nabla f(x,y) = \left\langle {{f_x},{f_y}} \right\rangle \\ = \left\langle {{f_x}(xy),{f_y}(xy)} \right\rangle \\ = \langle y,x\rangle \\\nabla f(3,2) = \langle 2,3\rangle \end{aligned}\)

Thus, the gradient vector \(\nabla f(3,2) = \langle 2,3\rangle \).

Find the tangent line of the surface.

Notice that the gradient vector is perpendicular to the tangent line.

Thus, the equation of the tangent line is,

\(\begin{aligned}{l}\nabla f(3,2) \cdot \langle (x - 3),(y - 2)\rangle = \langle 2,3\rangle \cdot (x - 3),(y - 2)\\ = 2(x - 3) + 3(y - 2) = 2x - 6 + 3y - 6\\ = 2x + 3y - 12\end{aligned}\)

Thus, the tangent line of the surface \(f(x,y) = xy\) at the point \((3,2)\) is \(2x + 3y = 12\).

Plot the graph.

Use online graph calculator and draw the graph of \(f(x,y) = xy,2x + 3y = 12\) and \(\nabla f(3,2)\) as shown below in Figure 1.

From Figure 1, it can be observed that \(\nabla f(3,2){\rm{ }}\)is perpendicular to the tangent line \(\;2{\rm{ }}x + 3{\rm{ }}y = 12{\rm{ }}\)