Question

If \({x^2} + {y^2} + {z^2} = 9\), \(\frac{{dx}}{{dt}} = 5\), and \(\frac{{dy}}{{dt}} = 4\), find \(\frac{{dz}}{{dt}}\) when \(\left( {x,y,z} \right) = \left( {2,2,1} \right)\).

Short answer

Thus, the required answer is \(\frac{{dz}}{{dt}} = - 18\).

Step-by-step solution

The chain rule

If \(g\) is differentiable at \(x\) and \(f\) is differentiable at \(g\left( x \right)\), then the composite function \(F = f \circ g\) defined by \(F\left( x \right) = f\left( {g\left( x \right)} \right)\) is differentiable at \(x\) and \(F'\) is given by the product \(F'\left( x \right) = f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)\). In Leibniz notation, if \(y = f\left( u \right)\) and \(u = g\left( x \right)\) are both differentiable functions, then \(\frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \cdot \frac{{du}}{{dx}}\).

The derivative of the given function

Differentiate \({x^2} + {y^2} + {z^2} = 9\) both the sides with respect to \(t\) as follows:

\(\begin{aligned}\frac{d}{{dt}}\left( {{x^2} + {y^2} + {z^2}} \right) &= \frac{d}{{dt}}\left( 9 \right)\\\left( {2x} \right) \cdot \frac{{dx}}{{dt}} + \left( {2y} \right) \cdot \frac{{dy}}{{dt}} + \left( {2z} \right) \cdot \frac{{dz}}{{dt}} &= 0\\2\left( {x\frac{{dx}}{{dt}} + y\frac{{dy}}{{dt}} + z\frac{{dz}}{{dt}}} \right) &= 0\\x\frac{{dx}}{{dt}} + y\frac{{dy}}{{dt}} + z\frac{{dz}}{{dt}} &= 0\end{aligned}\)

Substitute the given values

Given that \(\frac{{dx}}{{dt}} = 5\), \(\frac{{dy}}{{dt}} = 4\), and \(\left( {x,y,z} \right) = \left( {2,2,1} \right)\). Substitute these values in the above equation as follows:

\(\begin{aligned}2\left( 5 \right) + 2\left( 4 \right) + 1\frac{{dz}}{{dt}} &= 0\\10 + 8 + \frac{{dz}}{{dt}} &= 0\\\frac{{dz}}{{dt}} &= - 18\end{aligned}\)

Hence, \(\frac{{dz}}{{dt}} = - 18\).