Question

Short answer

The gradient vector field of \(f\) is \(\overline V f\left( {x,y} \right) = 3{\sec ^2}\left( {3x - 4y} \right)i - 4{\sec ^2}\left( {3x - 4y} \right)j\).

Step-by-step solution

Given Information.

It is given that \(f(x,y){\rm{ }} = {\rm{ }}\tan \left( {3x - 4y} \right)\).

Find the gradient vector field.

\(f(x,y){\rm{ }} = {\rm{ }}\tan \left( {3x - 4y} \right)\)

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

To construct gradient vector field, we will first find the partial derivatives.

At first, we will find \({f_x}\).

\(\begin{aligned}{l}{f_x} & = \frac{\partial }{{\partial x}}\left( {\tan \left( {3x - 4y} \right)} \right)\\ & = {\sec ^2}\left( {3x - 4y} \right) \cdot \frac{\partial }{{\partial x}}\left( {3x - 4y} \right)\\ & = {\sec ^2}\left( {3x - 4y} \right) \cdot \left( {3 - 0} \right)\\ & = 3{\sec ^2}\left( {3x - 4y} \right)\end{aligned}\)

Now, we will find \({f_y}\).

\(\begin{aligned}{l}{f_y} & = \frac{\partial }{{\partial x}}\left( {\tan \left( {3x - 4y} \right)} \right)\\ & = {\sec ^2}\left( {3x - 4y} \right) \cdot \frac{\partial }{{\partial y}}\left( {3x - 4y} \right)\\ & = {\sec ^2}\left( {3x - 4y} \right) \cdot \left( {0 - 4} \right)\\ & = - 4{\sec ^2}\left( {3x - 4y} \right)\end{aligned}\)

Therefore, the gradient vector field of \(f\) is \(\overline V f\left( {x,y} \right) = 3{\sec ^2}\left( {3x - 4y} \right)i - 4{\sec ^2}\left( {3x - 4y} \right)j\).