Consider a differentiable function \(f\) of two variables \(z=f(x,y)\). If \(x\) changes to \(x+\Delta x\) and \(y\) changes to \(y+\Delta y\), the increment of \(f\), \(\Delta f\), from Definition 3.10.1 , can be written as:
\begin{align}
\Delta f= f(x+&\Delta x,y+\Delta y)-f(x,y)\\
& =f_x(x,y)\Delta x+f_y(x,y)\Delta y+\sqrt{(\Delta x)^2+(\Delta y)^2}\cdot \varepsilon(\Delta x,\Delta y).\tag{i}
\end{align}

We take the linear part of \(\Delta f\) and call them the differential of \(f\). The differential part of \(f\) is denoted by \(df\) or \(dz\):
\[\begin{aligned} \label{Eq:df0} dz=df=\frac{\partial f}{\partial x}\Delta x+\frac{\partial f}{\partial y}\Delta y=f_x(x,y)\Delta x+f_y(x,y)\Delta y.\end{aligned}\]
Note that \(df\) is a function of four variables: \(x, y,\Delta x\), and \(\Delta y\). To emphasize on that, we may write it as \(df(x,y,\Delta x,\Delta y)\).

If \(x\) and \(y\) are independent variables, from Eq. 3.1 we have:
\[dx=\underbrace{\frac{\partial x}{\partial x}}_{=1}\Delta x+\underbrace{\frac{\partial x}{\partial y}}_{=0}\Delta y=\Delta x,\quad\text{and}\quad dy=\underbrace{\frac{\partial y}{\partial x}}_{=0}\Delta x+\underbrace{\frac{\partial y}{\partial y}}_{=1}\Delta y=\Delta y.\]

Then (3.1) takes the form:
\[df=\frac{\partial f}{\partial x} dx+\frac{\partial f}{\partial y} dy=f_x(x,y) dx+f_y(x,y) dy.\]
The above expression is sometimes called the total differential of \(f(x,y)\).

Definition 1. If \(z=f(x,y)\) is a differentiable function at \((x,y)\), the total differential of \(f\) is the function \(df\) defined by:
\[df(x,y,dx,dy)=\frac{\partial f}{\partial x} dx+\frac{\partial f}{\partial y} dy=f_x(x,y)dx+f_y(x,y) dy\]

Obviously, we can extend these methods and results to functions of any number of variables. For example, if \(u=f(x,y,z)\), then \[du=df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z} dz.\]

Example 1

The period of vibration, \(T\) of a mass on a spring is determined by the mass \(m\) and the stiffness of the spring \(k\) as: \[T=2\pi\sqrt{\frac{m}{k}}.\] Estimate the percentage change in the period of this system if the mass increases by \(5\%\) and the stiffness by \(3\%\).

Solution

We can use the total differential of \(T\) and say \(\frac{\Delta T}{T}\approx \frac{dT}{T}\):
\[\begin{aligned} dT&=\frac{\partial T}{\partial m} dm+\frac{\partial T}{\partial k} dk\\ &=2\pi\frac{1}{2\sqrt{m k}} dm-\frac{2\pi}{2} \sqrt{m} k^{-3/2} dk,\quad (\text{because } T=2\pi m^{1/2} k^{-1/2})\end{aligned}\]
Here \(m\) and \(k\) are independent variables, so:
\[dm=\Delta m=\frac{5}{100} m,\quad \text{and}\quad dk=\Delta k=\frac{3}{100} k\]
If we plug these expressions for \(dm\) and \(dk\) in \(dT\), we obtain:
\[\begin{aligned} dT=\underbrace{\pi\frac{1}{\sqrt{mk}}\times \frac{5}{100} m}_{\frac{5\pi}{100}\sqrt{\frac{m}{k}}}-\underbrace{\pi \sqrt{m} k^{-3/2} \times \frac{3}{100}k}_{\frac{3\pi}{100}\sqrt{\frac{m}{k}}}=-\frac{2\pi}{100}\sqrt{\frac{m}{k}}=-\frac{1}{100}\times \underbrace{2\pi\sqrt{\frac{m}{k}}}_{T}\end{aligned}\]
Therefore
\(\Delta T/T\approx dT/T=-1\%\), that is the period decreases by approximately 1%.

The exact change in the period is:
\[\Delta T=2\pi\sqrt{\frac{m+0.05 m}{k+0.03k}}-2\pi\sqrt{\frac{m}{k}}=\left(\sqrt{\frac{1.05}{1.03}}-1\right)\times 2\pi\sqrt{\frac{m}{k}}\]

\[=\left(\sqrt{\frac{1.05}{1.03}}-1\right)T\approx 0.009662 T.\]
That is, the exact change in the period is 0.9662%.

 

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