Question

(a) Find the values of \(\frac{{\partial z}}{{\partial r}}\) and \(\frac{{\partial z}}{{\partial \theta }}\) if \(z = f(x,y)\), where \(x = r\cos \theta \) and \(y = r\sin \theta \).

(b) Show the equation\({\left( {\frac{{\partial z}}{{\partial x}}} \right)^2} + {\left( {\frac{{\partial z}}{{\partial y}}} \right)^2} = {\left( {\frac{{\partial z}}{{\partial r}}} \right)^2} + \frac{1}{{{r^2}}}{\left( {\frac{{\partial z}}{{\partial \theta }}} \right)^2}\).

Short answer

(a) The value of \(\frac{{\partial z}}{{\partial r}}\) is\(\frac{{\partial z}}{{\partial x}} \cdot \cos \theta + \frac{{\partial z}}{{\partial y}} \cdot \sin \theta \).

The value of \(\frac{{\partial z}}{{\partial \theta }}\) is\(\frac{{\partial z}}{{\partial x}}( - r\sin \theta ) + \frac{{\partial z}}{{\partial y}}(r\cos \theta )\).

(b) \({\left( {\frac{{\partial z}}{{\partial x}}} \right)^2} + {\left( {\frac{{\partial z}}{{\partial y}}} \right)^2} = {\left( {\frac{{\partial z}}{{\partial r}}} \right)^2} + \frac{1}{{{r^2}}}{\left( {\frac{{\partial z}}{{\partial \theta }}} \right)^2}\).

Step-by-step solution

Chain rule

"Suppose that\(z = f(x,y)\)is a differentiable function of\(x\)and\(y\), where\(x = g(t)\)and\(y = h(t)\)are both differentiable functions of\(t\). Then,\(z\)is differentiable function of\(t\)and\(\frac{{dz}}{{dt}} = \frac{{dz}}{{dx}} \cdot \frac{{dx}}{{dt}} + \frac{{dz}}{{dy}} \cdot \frac{{dy}}{{dt}}\)”

Step 2: Find the values of \(\frac{{\partial z}}{{\partial r}}\) and \(\frac{{\partial z}}{{\partial \theta }}\) 

The given function is, \(z = f(x,y)\)

Take the partial derivative of \(x\) and \(y\) with respect to \(r\),

\(\begin{aligned}{l}\frac{{\partial x}}{{\partial r}} = \frac{\partial }{{\partial r}}(r\cos \theta )\\\frac{{\partial x}}{{\partial r}} = \cos \theta \\\frac{{\partial y}}{{\partial r}} = \frac{\partial }{{\partial r}}(r\sin \theta )\\\frac{{\partial y}}{{\partial r}} = \sin \theta \end{aligned}\)

Thus, the partial derivatives, \(\frac{{\partial x}}{{\partial x}} = \cos \theta \) and\(\frac{{\partial y}}{{\partial x}} = \sin \theta \).

Take the partial derivative of \(x\) and \(y\) with respect to\(\theta \),

\(\begin{aligned}{l}\frac{{\partial x}}{{\partial \theta }} = \frac{\partial }{{\partial \theta }}(r\cos \theta )\\\frac{{\partial x}}{{\partial \theta }} = - r\sin \theta \\\frac{{\partial y}}{{\partial \theta }} = \frac{\partial }{{\partial \theta }}(r\sin \theta )\\\frac{{\partial y}}{{\partial \theta }} = r\cos \theta \end{aligned}\)

Thus, the partial derivatives, \(\frac{{\partial x}}{{\partial \theta }} = - r\sin \theta \) and\(\frac{{\partial y}}{{\partial \theta }} = r\cos \theta \).

The partial derivative, \(\frac{{\partial z}}{{\partial r}}\) is computed as follows,

\(\begin{aligned}{l}\frac{{\partial z}}{{\partial r}} = \frac{{\partial z}}{{\partial x}} \cdot \frac{{\partial x}}{{\partial r}} + \frac{{\partial z}}{{\partial y}} \cdot \frac{{\partial y}}{{\partial r}}\\\frac{{\partial z}}{{\partial r}} = \frac{{\partial z}}{{\partial x}} \cdot \cos \theta + \frac{{\partial z}}{{\partial y}} \cdot \sin \theta \end{aligned}\)

Therefore, the value, \(\frac{{\partial z}}{{\partial r}}\) is\(\frac{{\partial z}}{{\partial x}} \cdot \cos \theta + \frac{{\partial z}}{{\partial y}} \cdot \sin \theta \).

Similarly find the value of \(\frac{{\partial z}}{{\partial \theta }}\) as follows,

\(\begin{aligned}{l}\frac{{\partial z}}{{\partial \theta }} = \frac{{\partial z}}{{\partial x}} \cdot \frac{{\partial x}}{{\partial \theta }} + \frac{{\partial z}}{{\partial y}} \cdot \frac{{\partial y}}{{\partial \theta }}\\\frac{{\partial z}}{{\partial \theta }} = \frac{{\partial z}}{{\partial x}}( - r\sin \theta ) + \frac{{\partial z}}{{\partial y}}(r\cos \theta )\end{aligned}\)

Therefore, the value of \(\frac{{\partial z}}{{\partial \theta }}\) is\(\frac{{\partial z}}{{\partial x}} \cdot ( - r\sin \theta ) + \frac{{\partial z}}{{\partial y}} \cdot (r\cos \theta )\).

Step 3: Show the equation\({\left( {\frac{{\partial z}}{{\partial x}}} \right)^2} + {\left( {\frac{{\partial z}}{{\partial y}}} \right)^2} = {\left( {\frac{{\partial z}}{{\partial r}}} \right)^2} + \frac{1}{{{r^2}}}{\left( {\frac{{\partial z}}{{\partial \theta }}} \right)^2}\)

(b)

First find the value of RHS.

From part (a) the value of \(\frac{{\partial z}}{{\partial x}} \cdot \cos \theta + \frac{{\partial z}}{{\partial y}} \cdot \sin \theta \)

Take square on both the sides

\(\begin{aligned}{l}{\left( {\frac{{\partial z}}{{\partial r}}} \right)^2} = {\left( {\frac{{\partial z}}{{\partial x}}\left( {\cos \theta } \right) + \frac{{\partial z}}{{\partial y}}\left( {\sin \theta } \right)} \right)^2}\\{\left( {\frac{{\partial z}}{{\partial r}}} \right)^2} = \left( {{{\left( {\frac{{\partial z}}{{\partial x}}} \right)}^2}{{\left( {\cos \theta } \right)}^2}} \right) + 2\left( {\left( {\frac{{\partial z}}{{\partial x}}} \right)\left( {\cos \theta } \right)\left( {\frac{{\partial z}}{{\partial y}}} \right)\left( {\sin \theta } \right)} \right) + \left( {{{\left( {\frac{{\partial z}}{{\partial y}}} \right)}^2}{{\left( {\sin \theta } \right)}^2}} \right)\\{\left( {\frac{{\partial z}}{{\partial r}}} \right)^2} = {\left( {\frac{{\partial z}}{{\partial x}}} \right)^2}{\left( {\cos \theta } \right)^2} + 2\left( {\frac{{\partial z}}{{\partial x}}} \right)\cos \theta \sin \theta + {\left( {\frac{{\partial z}}{{\partial y}}} \right)^2}{\left( {\sin \theta } \right)^2}\end{aligned}\)

Thus, the value,\({\left( {\frac{{\partial z}}{{\partial r}}} \right)^2} = {\left( {\frac{{\partial z}}{{\partial x}}} \right)^2}{\cos ^2}\theta + 2\left( {\frac{{\partial z}}{{\partial x}}} \right)\left( {\frac{{\partial z}}{{\partial y}}} \right)\cos \theta \sin \theta + {\left( {\frac{{\partial z}}{{\partial y}}} \right)^2}{\sin ^2}\theta \)

From part (a) the value of\(\frac{{\partial z}}{{\partial \theta }} = \frac{{\partial z}}{{\partial x}}( - r\sin \theta ) + \frac{{\partial z}}{{\partial y}}(r\cos \theta )\).

Take square on both sides,

\(\begin{aligned}{l}{\left( {\frac{{\partial z}}{{\partial \theta }}} \right)^2} = {\left( {\frac{{\partial z}}{{\partial x}}( - r\sin \theta ) + \frac{{\partial z}}{{\partial y}}(r\cos \theta )} \right)^2}\\{\left( {\frac{{\partial z}}{{\partial \theta }}} \right)^2} = \left( {{{\left( {\frac{{\partial z}}{{\partial x}}} \right)}^2}{{( - r\sin \theta )}^2}} \right) + 2\left( {\left( {\frac{{\partial z}}{{\partial x}}} \right)( - r\sin \theta )\left( {\frac{{\partial z}}{{\partial y}}} \right)(r\cos \theta )} \right) + \left( {{{\left( {\frac{{\partial z}}{{\partial y}}} \right)}^2}{{(r\cos \theta )}^2}} \right)\\{\left( {\frac{{\partial z}}{{\partial \theta }}} \right)^2} = {\left( {\frac{{\partial z}}{{\partial x}}} \right)^2}{r^2}{\sin ^2}\theta - 2\left( {\frac{{\partial z}}{{\partial x}}} \right)\left( {\frac{{\partial z}}{{\partial y}}} \right){r^2}\cos \theta \sin \theta + {\left( {\frac{{\partial z}}{{\partial y}}} \right)^2}{r^2}{\cos ^2}\theta \end{aligned}\)

Thus, \({\left( {\frac{{\partial z}}{{\partial \theta }}} \right)^2} = {\left( {\frac{{\partial z}}{{\partial x}}} \right)^2}{r^2}{\sin ^2}\theta - 2\left( {\frac{{\partial z}}{{\partial x}}} \right)\left( {\frac{{\partial z}}{{\partial y}}} \right){r^2}\cos \theta \sin \theta + {\left( {\frac{{\partial z}}{{\partial y}}} \right)^2}{r^2}{\cos ^2}\theta \)

Substitute the respective values in LHS and obtain,

\(\begin{aligned}{l}{\left( {\frac{{\partial z}}{{\partial r}}} \right)^2} + \frac{1}{{{r^2}}}{\left( {\frac{{\partial z}}{{\partial \theta }}} \right)^2} = \left( {{{\left( {\frac{{\partial z}}{{\partial x}}} \right)}^2}{{\cos }^2}\theta + 2\left( {\frac{{\partial z}}{{\partial x}}} \right)\left( {\frac{{\partial z}}{{\partial y}}} \right)\cos \theta \sin \theta + {{\left( {\frac{{\partial z}}{{\partial y}}} \right)}^2}{{\sin }^2}\theta } \right) + \left( {\frac{1}{{{r^2}}}} \right)\left( {{{\left( {\frac{{\partial z}}{{\partial x}}} \right)}^2}{r^2}{{\sin }^2}\theta - 2\left( {\frac{{\partial z}}{{\partial x}}} \right)\left( {\frac{{\partial z}}{{\partial y}}} \right){r^2}\cos \theta \sin \theta + {{\left( {\frac{{\partial z}}{{\partial y}}} \right)}^2}{r^2}{{\cos }^2}\theta } \right)\\{\left( {\frac{{\partial z}}{{\partial r}}} \right)^2} + \frac{1}{{{r^2}}}{\left( {\frac{{\partial z}}{{\partial \theta }}} \right)^2} = \left( {{{\left( {\frac{{\partial z}}{{\partial x}}} \right)}^2}{{\cos }^2}\theta + 2\left( {\frac{{\partial z}}{{\partial x}}} \right)\left( {\frac{{\partial z}}{{\partial y}}} \right)\cos \theta \sin \theta + {{\left( {\frac{{\partial z}}{{\partial y}}} \right)}^2}{{\sin }^2}\theta } \right) + \left( {{{\left( {\frac{{\partial z}}{{\partial x}}} \right)}^2}{{\sin }^2}\theta - 2\left( {\frac{{\partial z}}{{\partial x}}} \right)\left( {\frac{{\partial z}}{{\partial y}}} \right)\cos \theta \sin \theta + {{\left( {\frac{{\partial z}}{{\partial y}}} \right)}^2}{{\cos }^2}\theta } \right)\end{aligned}\)

Simplify further as follows,

\(\begin{aligned}{l}{\left( {\frac{{\partial z}}{{\partial r}}} \right)^2} + \frac{1}{{{r^2}}}{\left( {\frac{{\partial z}}{{\partial \theta }}} \right)^2} = {\left( {\frac{{\partial z}}{{\partial x}}} \right)^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) + {\left( {\frac{{\partial z}}{{\partial y}}} \right)^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right)\\{\left( {\frac{{\partial z}}{{\partial r}}} \right)^2} + \frac{1}{{{r^2}}}{\left( {\frac{{\partial z}}{{\partial \theta }}} \right)^2} = {\left( {\frac{{\partial z}}{{\partial x}}} \right)^2} + {\left( {\frac{{\partial z}}{{\partial y}}} \right)^2}\\{\left( {\frac{{\partial z}}{{\partial r}}} \right)^2} + \frac{1}{{{r^2}}}{\left( {\frac{{\partial z}}{{\partial \theta }}} \right)^2} = {\rm{ LHS}}\end{aligned}\)

Hence,\({\left( {\frac{{\partial z}}{{\partial x}}} \right)^2} + {\left( {\frac{{\partial z}}{{\partial y}}} \right)^2} = {\left( {\frac{{\partial z}}{{\partial r}}} \right)^2} + \frac{1}{{{r^2}}}{\left( {\frac{{\partial z}}{{\partial \theta }}} \right)^2}\)