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# C4 Differentiation watch

1. Need help on this question please

2. 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))
3. Substituting into the equations for x and y gives:

You should know that (*)

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

Now using (*)

Next, using , the gradient of the normal is

Finally, using gives

which rearranges to

EDIT: beaten to it
4. (Original post by fatuous_philomath)

EDIT: beaten to it
lol, welcome to the maths subforum during exam season
5. 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.
6. Can you please explain how you differentiate x and y in this question. I can't seem to get the answer you are getting
7. x = 1-cos2t
dx/dt = 0 + - (-1)(2)(sin2t)

y = sin2t
dy/dt = (2)cos2t

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Updated: June 14, 2006
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