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Trig parametric differentiation problem C4

TheKevinFang

Hi there, I can't do any of the trig problems on question 9. Urgent help and/or solutions needed!

Reply 1

Zacken

Original post by TheKevinFang

Hi there, I can't do any of the trig problems on question 9. Urgent help and/or solutions needed!

Urgent thoughts and/or workings needed!

Reply 2

B_9710

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

Reply 3

HapaxOromenon2

Original post by TheKevinFang

Hi there, I can't do any of the trig problems on question 9. Urgent help and/or solutions needed!

I'll do an example for you. For the first part of Q9, since x=t, dx/dt = 1, so dt/dx=1/1 = 1, and since y=t^2, dy/dt = 2t.
Thus by the Chain Rule, dy/dx = dy/dt * dt/dx = 2t * 1 = 2t, so when t = -1, dy/dx = -2.
Also you can calculate that since t = -1, x = -1 and y = 1.
Thus the equation of the tangent is y-1=-2(x-(-1)) -> y-1=-2(x+1) = -2x-2 -> y=-2x-1.