Question

Define the derivative of the function \(z=f(x, y)\) in the direction \(\mathbf{u}=\cos \theta \mathbf{i}+\sin \theta \mathbf{j}\).

Short answer

The derivative of the function \(f(x, y)\) in the direction \(\mathbf{u} = \cos \theta \mathbf{i} + \sin \theta \mathbf{j}\) is \(D_{\mathbf{u}}f = f_x \cdot \cos\theta + f_y \cdot \sin\theta\), where \(f_x\) and \(f_y\) are the partial derivatives of \(f\).

Step-by-step solution

Understand the Directional Derivative

The directional derivative of a function in the direction of a vector \(\mathbf{u}\) gives the rate at which the function changes at a point in that direction. It's a generalization of the concept of derivative, but in multiple directions rather than just one.

Directional Derivative Formula

The general formula for the directional derivative of a function \(f(x, y)\) in the direction of a unit vector \(\mathbf{u} = \cos \theta \mathbf{i} + \sin \theta \mathbf{j}\) is \(D_{\mathbf{u}}f = f_x \cdot u_x + f_y \cdot u_y\), where \(f_x\), \(f_y\) are the partial derivatives of \(f\) with respect to \(x\) and \(y\) respectively, and \(u_x = \cos\theta\) and \(u_y = \sin\theta\) are the components of the unit vector \(\mathbf{u}\).

Define the Derivatives of \(f\)

To apply the formula, we need the partial derivatives of \(f\). We can write them as \(f_x = \frac{df}{dx}\) and \(f_y = \frac{df}{dy}\). These values are specific to the function \(f\), and as they are not given in the exercise, we'll keep them in this general form.

Substitute into the Formula

Now we substitute into the formula \(D_{\mathbf{u}}f = f_x \cdot u_x + f_y \cdot u_y\). Using \(u_x = \cos\theta\) and \(u_y = \sin\theta\), we get \(D_{\mathbf{u}}f = f_x \cdot \cos\theta + f_y \cdot \sin\theta\).