STEP III 2007 question 5 solution

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STEP III 2007 question 5 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP III 2007 question 5 solution

$y = \ln(x^2-1), \coth \theta = x, r = \sqrt{x^2-1}$

First find expressions for $\cosh \theta, \sinh \theta$ : $\coth^2-1 = \frac{\cosh^2-\sinh^2}{\sinh^2} = \frac{1}{\sinh^2}$ , so

$\displaystyle \sinh \theta = \frac{1}{\sqrt{\coth^2 \theta -1}}=\frac{1}{r}$ .

Then $\displaystyle \cosh \theta = \coth \theta \sinh \theta = \frac{x}{r}$

Now $\displaystyle \frac{\text{d} y}{\text{d} x} = \frac{2x}{x^2-1} = 2\frac{x}{r^2} = \frac{2(x/r)}{r} = \frac{2\cosh \theta}{r}$

Next, find $\displaystyle \frac{\text{d} r}{\text{d} x} = \frac{x}{\sqrt{x^2-1}} = \frac{x}{r} = \cosh \theta$ .

$\displaystyle \frac{\text{d} \theta}{dx}$ is a little harder, but we have $\cosh \theta = x \sinh \theta$ , so $\displaystyle \sinh \theta \frac{\text{d} \theta}{\text{d} x} = \sinh \theta + x \cosh \theta \frac{\text{d} \theta}{\text{d} x}$ and so $\displaystyle (1 - x \coth \theta)\frac{\text{d} \theta}{\text{d} x} = 1$ .

Since $x = \coth \theta$ we deduce $\frac{\text{d} \theta}{\text{d} x} = -\frac{1}{r^2} = -\frac{1}{r}\sinh \theta$

Now with the benefit of having looked at this question before, we can do the rest in one big jump:

$\displaystyle \frac{\text{d} }{\text{d} x} \frac{\cosh n\theta}{r^n} = n \sinh n\theta \frac{\text{d} \theta}{\text{d} x} \frac{1}{r^n} -n \frac{\cosh n\theta}{r^{n+1}}\frac{\text{d} r}{\text{d} x}$

$\displaystyle = -\frac{n}{r^{n+1}} (\sinh n\theta \sinh\theta + \cosh n\theta \cosh \theta)$

$\displaystyle = -\frac{n \cosh(n+1)\theta}{r^{n+1}}$

So $\displaystyle \frac{\text{d} ^2y}{\text{d} x^2} = 2 \frac{\text{d} }{\text{d} x} \frac{\cosh \theta}{r} = -\frac{2 \cosh 2\theta}{r^2}$

$\displaystyle \frac{\text{d} ^3y}{\text{d} x^3} = \frac{4 \cosh 3\theta}{r^3}$ and generally $\displaystyle \frac{\text{d} ^ny}{\text{d} x^n} = 2\frac{(-1)^{n-1}(n-1)!\cosh n\theta}{r^n}$ .