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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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