Question

Find a function \(f\) such that \(\nabla f=e^{x} \cos y \mathbf{i}-e^{x} \sin y \mathbf{j}+z \mathbf{k}\).

Short answer

The function \(f(x, y, z)\) that has the given gradient is \(f(x, y, z) = e^{x} \cos y + \frac{z^2}{2}\).

Step-by-step solution

Integrate the first component

Integrate the first component \(e^{x} \cos y\) with respect to \(x\), while treating \(y\) as a constant. We get \(\int{e^x \cos y} \, dx = e^x \cos y + C\). As we have to find a scalar function, the constant of integration \(C\) is a function of other variables, i.e., \(y\), \(z\). So, \(C = C(y, z)\).

Integrate the second component

Next, integrate the second component \(-e^{x} \sin y\) with respect to \(y\), while treating \(x\) as a constant. This gives \(\int{-e^x \sin y} \, dy = e^x \cos y + D\). As before, the constant of integration \(D\) is a function of the remaining variable, i.e., \(x\), \(z\). Therefore, \(D = D(x, z)\).

Integrate the third component

Finally, integrate the third component \(z\) with respect to \(z\). This gives \(\int z \, dz = \frac{z^2}{2} + E\). The constant of integration \(E\) is a function of \(x\) and \(y\), thus \(E = E(x, y)\). In comparing all three results, we can observe that the function \(f(x, y, z)\) common in all is \(e^{x} \cos y\). So, one possible function \(f(x, y, z)\) that fulfills the given conditions is \(f(x, y, z) = e^{x} \cos y\). However, as the gradient was given with a \(z\)-component, it's necessary to add the integral of the \(z\)-component to the final function, which results in \(f(x, y, z) = e^{x} \cos y + \frac{z^2}{2}\). Note that the constants of integration are not included because we are looking for solutions that satisfy the given gradient and are consistent for all three components.