C4 Differentiation

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dx/dt = 2sin2t
dy/dt = 2cos2t

gradient of the normal = -dx/dy = -tan2t
at P : gradient = -tanpi/3
y = sinpi/3
x = 1- cospi/3

y - sinpi/3 = -tanpi/3 (x-(1-cospi/3))

Reply 2

JeremyC

Substituting

\large{ t = \frac{\pi}{6} }

into the equations for x and y gives:

\large{ y = \sin \frac{2 \pi}{6} = \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \, \, \, x = 1 - \cos \frac{2 \pi}{6} = 1 - \cos \frac{\pi}{3} = 1 - \frac{1}{2} = \frac{1}{2} }

You should know that

\large{ \frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx}}

(*)

Thus, differentiating y and x with respect to t we get:

\large{\frac{dy}{dt} = 2 \cos 2t \, \, \frac{dx}{dt} = 2 \sin 2t }

Now using (*)

\large{\frac{dy}{dx} = \frac{ 2 \cos 2t}{2 \sin 2t} = \cot 2t}

Therefore, the gradient where

\large{ t = \frac{\pi}{6} }

\large{ \cot \frac{2 \pi}{6} = \cot \frac{\pi}{3} = \frac{1}{\sqrt{3}}}

Next, using

\large{ m_1 m_2 = -1}

, the gradient of the normal is

\large{ \frac{-1}{(\frac{1}{\sqrt{3}})} = -\sqrt{3}}

Finally, using

\large{y - y_1 = m(x - x_1) }

gives

\large{y - \frac{\sqrt{3}}{2} = -\sqrt{3}(x - \frac{1}{2})}

which rearranges to

\large{y = -\sqrt{3}x + \sqrt{3}}

EDIT: beaten to it :frown:

Reply 3

KAISER_MOLE

fatuous_philomath

EDIT: beaten to it :frown:

lol, welcome to the maths subforum during exam season :rolleyes:

Reply 4

JeremyC

Indeed.

Although procrastinating by typing posts requiring extensive use of LaTeX is far more enjoyable than revising for the A level exams I have next week, whether or not my post is useful, and at least I get to learn LaTeX a little better. :smile: