91Ó°ÊÓ

Expand \(\left(c-\frac{1}{c}\right)^{5}\) using the binomial series. $$ \begin{aligned} \left(c-\frac{1}{c}\right)^{5}=& c^{5}+5 c^{4}\left(-\frac{1}{c}\right) \\ &+\frac{(5)(4)}{(2)(1)} c^{3}\left(-\frac{1}{c}\right)^{2} \\ &+\frac{(5)(4)(3)}{(3)(2)(1)} c^{2}\left(-\frac{1}{c}\right)^{3} \\ &+\frac{(5)(4)(3)(2)}{(4)(3)(2)(1)} c\left(-\frac{1}{c}\right)^{4} \\ &+\frac{(5)(4)(3)(2)(1)}{(5)(4)(3)(2)(1)}\left(-\frac{1}{c}\right)^{5} \end{aligned} $$ i.e. \(\left(c-\frac{1}{c}\right)^{5}=c^{5}-5 c^{3}+10 c-\frac{10}{c}+\frac{5}{c^{3}}-\frac{1}{c^{5}}\)

Find the middle term of \(\left(2 p-\frac{1}{2 q}\right)^{10}\) In the expansion of \((a+x)^{10}\) there are \(10+1\), i.e. 11 terms. Hence the middle term is the sixth. Using the general expression for the \(r\) th term where \(a=2 p, x=-\frac{1}{2 q}\), \(n=10\) and \(r-1=5\) gives $$ \begin{aligned} &\frac{(10)(9)(8)(7)(6)}{(5)(4)(3)(2)(1)}(2 p)^{10-5}\left(-\frac{1}{2 q}\right)^{5} \\ &=252\left(32 p^{5}\right)\left(-\frac{1}{32 q^{5}}\right) \end{aligned} $$ Hence the middle term of \(\left(2 p-\frac{1}{2 q}\right)^{10}\) is \(-252 \frac{p^{5}}{q^{5}}\)

Evaluate (1.002) \(^{9}\) using the binomial theorem correct to (a) 3 decimal places and (b) 7 significant figures. $$ \begin{aligned} &(1+x)^{n}=1+n x+\frac{n(n-1)}{2 !} x^{2} \\ &+\frac{n(n-1)(n-2)}{3 !} x^{3}+\cdots \\ &(1.002)^{9}=(1+0.002)^{9} \end{aligned} $$ Substituting \(x=0.002\) and \(n=9\) in the general expansion for \((1+x)^{n}\) gives: $$ \begin{gathered} (1+0.002)^{9}=1+9(0.002)+\frac{(9)(8)}{(2)(1)}(0.002)^{2} \\ +\frac{(9)(8)(7)}{(3)(2)(1)}(0.002)^{3}+\cdots \\ =1+0.018+0.000144 \\ +0.000000672+\cdots \\ =1.018144672 \ldots \end{gathered} $$ Hence \((1.002)^{9}=1.018\), correct to 3 decimal places \(=1.018145\), correct to 7 significant figures.

Determine the value of \((3.039)^{4}\), correct to 6 significant figures using the binomial theorem. (3.039) \(^{4}\) may be written in the form \((1+x)^{n}\) as: $$ \begin{aligned} (3.039)^{4}=&(3+0.039)^{4} \\ =&\left[3\left(1+\frac{0.039}{3}\right)\right]^{4} \\ =& 3^{4}(1+0.013)^{4} \\ (1+0.013)^{4}=1+4(0.013) \\ \qquad+\frac{(4)(3)}{(2)(1)}(0.013)^{2} \\ +\frac{(4)(3)(2)}{(3)(2)(1)}(0.013)^{3}+\cdots \\ =1+0.052+0.001014 \\ & \quad+0.000008788+\cdots \\ =& 1.0530228 \end{aligned} $$ correct to 8 significant figures Hence \((3.039)^{4}=3^{4}(1.0530228)\) \(=85.2948\), correct to 6 significant figures.

The second moment of area of a rectangle through its centroid is given by \(\frac{b l^{3}}{12}\) Determine the approximate change in the second moment of area if \(b\) is increased by \(3.5 \%\) and \(l\) is reduced by \(2.5 \%\) New values of \(b\) and \(l\) are \((1+0.035) b\) and \((1-0.025) l\) respectively. New second moment of area $$ \begin{aligned} &=\frac{1}{12}[(1+0.035) b][(1-0.025) l]^{3} \\ &=\frac{b l^{3}}{12}(1+0.035)(1-0.025)^{3} \\ &\approx \frac{b l^{3}}{12}(1+0.035)(1-0.075), \text { neglecting } \end{aligned} $$ powers of small terms $$ \approx \frac{b l^{3}}{12}(1+0.035-0.075), \text { neglecting } $$ products of small terms $$ \approx \frac{b l^{3}}{12}(1-0.040) \text { or }(0.96) \frac{b l^{3}}{12}, \text { i.e. } 96 \% $$ of the original second moment of area Hence the second moment of area is reduced by approximately \(4 \%\)

The resonant frequency of a vibratung shatt is given by: \(f=\frac{1}{2 \pi} \sqrt{\frac{k}{I}}\), where \(k\) is the stiffness and \(I\) is the inertia of the shaft. Use the binomial theorem to determine the approximate percentage error in determining the frequency using the measured values of \(k\) and \(I\) when the measured value of \(k\) is \(4 \%\) too large and the measured value of \(I\) is \(2 \%\) too small. Let \(f, k\) and \(I\) be the true values of frequency, stiffness and inertia respectively. Since the measured value of stiffness, \(k_{1}\), is \(4 \%\) too large, then $$ k_{1}=\frac{104}{100} k=(1+0.04) k $$ The measured value of inertia, \(I_{1}\), is \(2 \%\) too small, hence $$ I_{1}=\frac{98}{100} I=(1-0.02) I $$ The measured value of frequency, $$ \begin{aligned} f_{1} &=\frac{1}{2 \pi} \sqrt{\frac{k_{1}}{I_{1}}}=\frac{1}{2 \pi} k_{1}^{\frac{1}{2}} I_{1}^{-\frac{1}{2}} \\ &=\frac{1}{2 \pi}[(1+0.04) k]^{\frac{1}{2}}[(1-0.02) I]^{-\frac{1}{2}} \\ &=\frac{1}{2 \pi}(1+0.04)^{\frac{1}{2}} k^{\frac{1}{2}}(1-0.02)^{-\frac{1}{2}} I^{-\frac{1}{2}} \\ &=\frac{1}{2 \pi} k^{\frac{1}{2}} I^{-\frac{1}{2}}(1+0.04)^{\frac{1}{2}}(1-0.02)^{-\frac{1}{2}} \end{aligned} $$ $$ \text { i.e. } f_{1}=f(1+0.04)^{\frac{1}{2}}(1-0.02)^{-\frac{1}{2}} $$ $$ \begin{aligned} &\approx f\left[1+\left(\frac{1}{2}\right)(0.04)\right]\left[1+\left(-\frac{1}{2}\right)(-0.02)\right] \\\ &\approx f(1+0.02)(1+0.01) \end{aligned} $$ Neglecting the products of small terms, $$ f_{1} \approx(1+0.02+0.01) f \approx 1.03 f $$ Thus the percentage error in \(f\) based on the measured values of \(k\) and \(I\) is approximately \([(1.03)(100)-100]\), i.e. \(3 \%\) too large.