[bigbadb1319]

Exercise 4A:
Q2). a). Find the equation of the tangent to the curve with parametric equations x=3t-2sint, y=t² + tcost, at the point P, where t=pi/2

k i've done majority of this question, n got the gradient to be: pi/6
i can't work out the equation of the tangent, i know ur suppose to put in t=pi/2 in both x and y, but sumthings goin wrong!

Q3a). Find the equation of the normal to the curve with parametric equations x=e^t, y=e^t + e^-t, at point P, where t=2

again, i've done all of it, but at da back of the book it says da answer is x=1, but should it not be y=2, becuaue the gradient is 0, n when u put it in
y-y1=m(x-x1) it comes out as y=2?

Q3b). Find the equation of the normal to the curve with parametric equations x=1 - cos2t, y=sin2t at the point P, where t=pi/6

i got dy/dx to be: 2cos2t/2sin2t, is that the same as 1/tan2t? and i can't seem to work out the rest, don't know what i'm doing wrong....!

any help would be much appreiated!! thanx

[Malik]

If gradient is pi/6 then gradient of tangent is pi/6 .
You need corrosponding x and y values so sub t=pi/2 into your x and y equations.

x=(3/2)pi-2 y=pi²/4 m=gradient=pi/6
plug all xym into =>y-y1=m(x-x1)
y-(pi²/4)=pi/6[x-(1.5pi-2)]
simiplify this for your answer.

[bigbadb1319]

Originally Posted by Malik

If gradient is pi/6 then gradient of tangent is pi/6 .
You need corrosponding x and y values so sub t=pi/2 into your x and y equations.

x=(3/2)pi-2 y=pi²/4 m=gradient=pi/6
plug all xym into =>y-y1=m(x-x1)
y-(pi²/4)=pi/6[x-(1.5pi-2)]
simiplify this for your answer.

alrite kl, k found out where i was goin wrong, wasn't sqaring pi!, any idea for the other ones?