STEP III 2007 question 4 solution

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STEP III 2007 question 4 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP III 2007 question 4 solution

$\displaystyle x = a\cos t + a\ln \tan \frac{t}{2}$

$\displaystyle y = a\sin t$

$\displaystyle \frac{\text{d} y}{\text{d} x} = \frac{\text{d} y}{\text{d} t}.\frac{dt}{dx}$

$\displaystyle \frac{\text{d} y}{\text{d} t} = a\cos t$

$\displaystyle \frac{\text{d} x}{\text{d} t} = - a\sin t + \frac{a\sec^2\frac{t}{2}}{2\tan \frac{t}{2}} = -a\sin t + \frac{\frac{a}{\cos^2\frac{t}{2}}}{\frac{2\sin \frac{t}{2}}{\cos\frac{t}{2}}}$ $= -a\sin t + \frac{a}{2\sin \frac{t}{2}\cos\frac{t}{2}} = -a\sin t + \frac{a}{\sin t}= \frac{a(1-\sin ^2t)}{\sin t} = \frac{a\cos^2t}{\sin t}$

$\displaystyle \frac{\text{d} t}{\text{d} x} = \frac{\sin t}{a\cos^2t}$

$\displaystyle \frac{\text{d} y}{\text{d} x} = a\cos t.\frac{\sin t}{a\cos^2t}$

$\displaystyle \frac{\text{d} y}{\text{d} x} = \tan t$

As required.

Tangent to curve at P is

$\displaystyle y - a\sin t = \tan t (x - a\cos t - a\ln \tan \frac{t}{2})$

y = 0

$\displaystyle - a\sin t = x\tan t + \tan t (-a\cos t - a\ln \tan \frac{t}{2})$

$\displaystyle \tan t x = -a\sin t - \tan t(-a\cos t - a\ln \tan \frac{t}{2})$

$\displaystyle x = -a\cos t + a\cos t + a\ln \tan \frac{t}{2}$

$\displaystyle x = a\ln \tan \frac{t}{2}$

$\displaystyle PQ^2 = (a\sin t)^2 + (a\ln \tan \frac{t}{2} - (a\cos t + a\ln \tan \frac{t}{2}))^2$

$\displaystyle PQ^2 = a^2\sin ^2t + (-a\cos t)^2$

$\displaystyle PQ^2 = a^2$

$\displaystyle PQ = a$

As required.

Next part.

$\displaystyle y = a\sin t$

$\displaystyle \frac{\text{d} y}{\text{d} t} = a\cos t$

$\displaystyle \frac{\text{d}^2y}{\text{d} t^2} = -a\sin t$

$\displaystyle x = a\cos t + a\ln \tan \frac{t}{2}$

$\displaystyle \frac{\text{d} x}{\text{d} t} = -a\sin t + \frac{a}{\sin t}$

$\displaystyle \frac{\text{d}^2x}{\text{d} t^2} = - a\cos t - a\cot t \cosect$

Right so

$\displaystyle \rho = \frac{ [(-a\sin t + \frac{a}{\sin t})^2 + a^2\cos^2t]^{\frac{3}{2}}}{|[(-a\sin t + \frac{a}{\sin t}).(-a\sin t) -a\cos t.(-a\cos t - a\cot t\cosect)]|}$

$\displaystyle \rho = \frac{[a^2\sin ^2t + \frac{a^2}{\sin ^2t} - 2a^2 + a^2\cos^2t]^{|\frac{3}{2}}}{|a^2\sin ^2t-a^2 + a^2\cos^2t + a^2 \cot ^2t|}$

$\displaystyle \rho = \frac{[\frac{a^2-a^2\sin ^2t}{\sin ^2t}]^{\frac{3}{2}}}{a^2\cot ^2t}$

$\displaystyle \rho = \frac{[a^2\cot ^2t]^{\frac{3}{2}}}{a^2\cot ^2t}$

$\displaystyle \rho = \frac{a^3\cot ^3t}{a^2\cot ^2t}$

$\displaystyle \rho = a\cot t$

As required.

Normal to P

$\displaystyle y - a\sin t = -\frac{1}{\tan t} ( x - a\cos t - a\ln \tan \frac{t}{2})$

You really need to look at a diagram here. If CQ is parallel to y axis the equation $x = a\ln \tan \frac{t}{2}$ should intersect the normal to P at C. (The prove of this comes to the fact that we can confirm the distance between P and C is rho which will be explained)

Let $\displaystyle x = a\ln \tan \frac{t}{2}$

$\displaystyle y - a\sin t = -\frac{\cost}{\sin t}(-a\cos t)$

$\displaystyle y = \frac{a\cos^2t}{\sin t} + a\sin t$

$\displaystyle y = \frac{a\cos^2t+a\sin ^2t}{\sin t}$

$\displaystyle y = \frac{a}{\sin t}$

Now We are told that the distance between P and C is rho = a cot t

$\displaystyle \text{Distance}^2 = (a\cos t)^2 + (\frac{a}{\sin t} - a\sin t)^2$

$\displaystyle \text{Distance}^2 = a^2\cos^2t + [(\frac{a-a\sin ^2t}{\sin t}]^2$

$\displaystyle \text{Distance}^2 = a^2\cos^2t + [\frac{a\cos^2t}{\sin t}]^2$

$\displaystyle \text{Distance}^2 = a^2\cos^2t + \frac{a^2\cos^4t}{\sin ^2t}$

$\displaystyle \text{Distance}^2 = \frac{a^2\cos^2t\sin ^2t + a^2\cos^4t}{\sin ^2t}$

$\displaystyle \text{Distance}^2 = \frac{a^2\cos^2t(\sin ^2t + \cos^2t)}{\sin ^2t}$

$\displaystyle \text{Distance}^2 = \frac{a^2\cos^2t}{\sin ^2t}$

$\displaystyle \text{Distance} = a\cot t = \rho$

So if the coordinates of C are $\displaystyle (a\ln\tan \frac{t}{2},\frac{a}{\sin t})$ and i.e it intersects with $\displaystyle x = a\ln \tan \frac{t}{2}$

The distance between P and C is rho. Which we are told it is in the question. This means that the coordinates of C are $\displaystyle (a\ln\tan \frac{t}{2},\frac{a}{\sin t})$ and intersects with $x = a\ln \tan \frac{t}{2}$ at C. Therefore CQ must be parallel to the y axis.

Solution by insparato.

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