5.11 Predicting Change (Differentials)
For a function of two variables \(z = f(x,y)\), the differentials \(dx\) and \(dy\) and \(dz\), also called the total differential is defined by \begin {align*} dz & = f_x(x,y) dx + f_y (x,y) dy\\ & = \frac {\partial z}{\partial x}\hspace {0.1cm}dx + \frac {\partial z}{\partial y}\hspace {0.1cm}dy \end {align*}
Sometimes the notation \(df\) is used in place of \(dz\).
For a function of two variables \(z = f(x,y)\) suppose that \(x\) changes from \(a\) to \(a + \Delta x\) and \(y\) changes from \(b\) to \(b + \Delta y\), then the corresponding increment in \(z\) is \[\Delta z = f(a + \Delta x, b + \Delta y ) - f(x,y)\] Thus the increment \(\Delta z\) represents the change in the value of \(f\) when \((x,y)\) changes from \((a,b)\) to \((a+ \Delta x, b + \Delta y)\). We shall approximate \(\Delta z\) using \(dz\).
- 1.
- If \(z = f(x,y) = x^2 + 3xy - y^2\), find the differential \(dz\).
- 2.
- If \(x\) changes from 2 to 2.05 and \(y\) changes from 3 to 2.96, compare the values of \(\Delta z\) and \(dz\)
Solution.
- 1.
- \( dz = f_x(x,y) dx + f_y (x,y) dy\)
\(dz = (2x + 3y)dx + (3x -2y)dy\)
- 2.
- \( dz = f_x(a,b)\Delta x + f_y(a,b) \Delta (a,b) \Delta \)
\[\Delta x = 2.05 - 2= 0.05\hspace {0.4cm} , \hspace {0.4cm} \Delta y = 2.96 - 3 = -0.04\] \begin {align*} dz & = f_x(2,3)(0.05) + f_y(2,3) (-0.04)\\ & = (2(2) + 3(3))(0.05) + (3(2) - 2(3))(-0.04)\\ & = 0.65 \end {align*}
\begin {align*} \Delta z & = f(a+ \Delta x, b + \Delta y) - f(a,b)\\ & = f(2.05,2.96) - f(2,3)\\ & = (2.05)^2 + 3(2.96) - (2.96)^2 - [2^2 + 3(2)(3)-(3)^2]\\ & = 0.6449\\ \end {align*}
\(\Delta z\) is called an absolute change in \(z\). When \((x,y)\) changes from \((a,b)\) to \(( a + \Delta x, b + \Delta y)\).
| Change | True | Estimate |
| Absolute | \(\displaystyle {\Delta z = \Delta f}\) | \(dz = df\) |
| Relative | \(\displaystyle {\dfrac {\Delta f}{f(a,b)}}\) | \(\dfrac {df}{f(a,b)}\) |
| Percentage | \(\displaystyle {\dfrac {\Delta f}{f(a,b)}\times 100\%}\) | \(\dfrac {df}{f(a,b)\times 100\%}\) |
Suppose the variables \(r\) and \(h\) change the initial values of \((r_0,h_0) = (1,5)\) by the amounts \(dr = 0.03\) and \(dh = - 0.1\). Estimate the resulting absolute, relative and percentage changes in the value of the function \(V = \pi r^2 h\).
Solution.
To estimate the absolute change in \(V\) we evaluate \(dV = V_r (r_0,h_0)dr + V_h(r_0,h_0)dh\) to get \(dV = 2\pi rh dr + \pi r^2 dh\) \begin {align*} dV & = 2\pi (5)(0.03) + \pi (-0.1)\\ & = 0.3\pi - 0.1\pi \\ & = 0.2\pi \end {align*}
\begin {align*} \text {Relative Change}\hspace {0.5cm}= \dfrac {dV}{V(1,5)} & = \frac {0.2\pi }{5\pi }\\ & = 0.04 \end {align*}
\begin {align*} \text {Percentage Change}\hspace {0.5cm} & = \dfrac {dV}{V(1,5} \times 100\%\\ & = 0.04\times 100\%\\ & = 4\%\\\\ \end {align*}
Questions on this section
Stuck on something here? Ask below and it stays attached to this topic.