Activity 13.4.3.
Let \(\vG(x,y,z) = \langle 3e^{y^2}+z\sin(x),6xy e^{y^2} - z,3z^2-y-\cos(x)\rangle\) and \(\vH(x,y,z) = \langle 3x^2 y,x^3+2yz^3,xz+3y^2z^2\rangle\text{.}\)
(a)
If \(\vG\) and \(\vH\) are to be gradient vector fields, then there are functions \(g\) and \(h\) for which \(\vG = \nabla g\) and \(\vH=\nabla h\text{.}\) If such functions \(g\) and \(h\) exist, what would \(g_y\text{,}\) \(g_z\text{,}\) \(h_x\text{,}\) \(h_y\text{,}\) and \(h_z\) be?
(b)
Let \(g_1(x,y,z)=3xe^{y^2}+xyz-z\sin(x)\text{.}\) Calculate \(\partial g_1/\partial x\text{.}\) Could \(g_1\) be a potential function for the vector field \(\vG\text{?}\)
(c)
Find a function \(g\) so that \(\partial g/\partial x = 3e^{y^2}+z\sin(x)\text{.}\) Find a function \(h\) so that \(\partial h/\partial x = 3x^2y\text{.}\)
(d)
When finding the most general anti-derivative for a function of one variable, we add a constant of integration (usually denoted by \(C\)) to capture the fact that any constant will become \(0\) through differentiation.
(i)
When taking the partial derivative with respect to \(x\) of a function of \(x\text{,}\) \(y\text{,}\) and \(z\text{,}\) what variables can appear in terms that become \(0\) in the partial derivative because they are treated as constants?
(ii)
What does this tell you should be added to \(g\) and \(h\) in the previous part to make them the most general possible functions with the desired partial derivatives with respect to \(x\text{?}\)
(e)
Now calculate \(\partial g/\partial y\) and \(\partial h/\partial y\) based on your choices for part 13.4.3.c. Write a few sentences to explain why this tells you that we must have
\begin{equation*}
g(x,y,z) = 3xe^{y^2}-z\cos(x)-yz+m_1(z)
\end{equation*}
and
\begin{equation*}
h(x,y,z) = x^3y+y^2z^3+m_2(z)
\end{equation*}
for some functions \(m_1\) and \(m_2\) depending only on \(z\text{.}\)
(f)
Calculate \(\frac{\partial g}{\partial z}\) and \(\frac{\partial h}{\partial z}\) for the functions in the part above. Notice that \(m_1\) and \(m_2\) are functions of \(z\) alone, so taking a partial derivative with respect to \(z\) is the same as taking an ordinary derivative, and thus you may use the notation \(m'_1(z)\) and \(m'_2(z)\text{.}\)
(g)
Explain why \(\vG\) is a gradient vector field but \(\vH\) is not a gradient vector field. Find a potential function for \(\vG\text{.}\)
