Given that $x$ is small enough for terms involving ${{x}^{4}}$ and above to be ignored, use

Question

Given that $x$ is small enough for terms involving ${{x}^{4}}$ and above to be ignored, use the Maclaurin series for $\sin x$ and $\cos x$ to show that $\tan x\approx x+\frac{1}{3}{{x}^{3}}$.[3] A straight pole, $AB$, leans against a wall as shown in the diagram below. $AC$ is horizontal where $C$ is the base of the wall and $AC=2$ metres. $B$ is vertically above $C$ and angle $BAC={{85}^{\circ }}$. Use the Maclaurin series for $\tan x$ to find an approximate value for $BC$, giving your answer correct to $5$ decimal places.[2] Use a calculator to find the value of $BC$, giving your answer correct to $5$ decimal places, and hence show that the error in your approximation is about $0.0008\%$.[2] Given that $y=\tan \left( \frac{1}{x+a} \right)$, where $a$ is a positive constant, show that ${{\left( x+a \right)}^{2}}\frac{\mathrm{d}y}{\mathrm{d}x}+{{y}^{2}}+1=0$.[2] Using $a=\frac{4}{\pi }$, find the Maclaurin series for $\tan \left( \frac{1}{x+a} \right)$, up to and including the term in ${{x}^{2}}$, giving all coefficients in exact form.[4]

Accepted Answer

(a) $\tan x\approx x+\frac13x^3$, $BC\approx22.86028\text{ m}$, while the calculator value is $22.86010\text{ m}$, giving an error of approximately $0.0008\%$. (b) $(x+a)^2\frac{\mathrm{d}y}{\mathrm{d}x}+y^2+1=0$, and for $a=\frac4\pi$, $\tan\left(\frac{1}{x+\frac4\pi}\right)=1-\frac{\pi^2}{8}x+\frac{\pi^2}{32}\left(\pi+\frac{\pi^2}{4}\right)x^2+O(x^3)$.

2017 ACJC Promo Q10

2017 ACJC Promo Q10

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