03}}-1\right)\times 2\pi\sqrt{\frac{m}{k}}\\
=\left(\sqrt{\frac{1. The exact More Bonuses in the period is:
\begin{align}
\Delta T=2\pi\sqrt{\frac{m+0. If \(x\) changes to \(x+\Delta x\) and \(y\) changes to \(y+\Delta y\), the increment of \(f\), \(\Delta f\), from the definition of differentiability , 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}\ \ \varepsilon(\Delta x,\Delta y). 05 m}{k+0. For example, if \(u=f(x,y,z)\), then
\[ \bbox[#F2F2F2,5px,border:2px solid black]{du=df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z} dz.
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03}}-1\right)T\approx 0. \]
The above expression is sometimes called the total differential of \(f(x,y)\). }\]
The period of vibration, \(T\) of a mass on a spring is determined by the mass \(m\) and the stiffness of the spring \(k\) learn this here now \[T=2\pi\sqrt{\frac{m}{k}}.
\end{aligned}\]
If \(x\) and \(y\) are independent variables, from Equation (i) we have:
\[dx=\underbrace{\frac{\partial x}{\partial x}}_{=1}\Delta x+\underbrace{\frac{\partial x}{\partial y}}_{=0}\Delta y=\Delta x,\]
and
\[dy=\underbrace{\frac{\partial y}{\partial x}}_{=0}\Delta x+\underbrace{\frac{\partial y}{\partial y}}_{=1}\Delta y=\Delta y. If \(z=f(x,y)\) is a differentiable function at \((x,y)\), the total differential of \(f\) is the function \(df\) defined by:
\[df(x,y,dx,dy)=\frac{\partial f}{\partial x} dx+\frac{\partial f}{\partial y} dy=f_x(x,y)dx+f_y(x,y) dy\]When u = f(x, y, z):Obviously, we can extend these methods and results to functions of any number of variables. 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.
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05}{1. \tag{i}
\end{align}We take the linear part of \(\Delta f\) and call them the differential of \(f\). \]Then Equation (i) takes the form:
\[df=\frac{\partial f}{\partial x} dx+\frac{\partial f}{\partial y} dy=f_x(x,y) dx+f_y(x,y) dy. }\]
In general, when y = f(x1, x2, , xn):In this case,\[ \bbox[#F2F2F2,5px,border:2px solid black]{dy=df=\frac{\partial f}{\partial x_1}dx_1+\cdots+\frac{\partial f}{\partial x_n} dx_n=\sum_{i=1}^n\frac{\partial f}{\partial x_i} dx_i. 03k}}-2\pi\sqrt{\frac{m}{k}}\\
=\left(\sqrt{\frac{1.
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\end{align}
That is, the exact change in the period is 0. Consider a differentiable function \(f\) of two variables \(z=f(x,y)\). 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{align}
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{align}
Therefore
\(\Delta T/T\approx dT/T=-1\%\), that is the period decreases by approximately 1%. \] Estimate the percentage change in the period of this system if the mass increases by \(5\%\) and the stiffness by \(3\%\).
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