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# P3 trig Q watch

1. Line c has equation x=5cos a y=4 sin a (a=theta) 0<=a<2pi

The curve look like below.

Gradient of tangent = -4/5 at P
equation of tangent = y-2(,/2) = -4/5 (x-5/,/2))
co-ordinates or R = (5,/2 , 0)

well the answer is 10-25pi but i cant get it dont even know what im doing wrong.
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2. (Original post by lexazver203)
Line c has equation x=5cos a y=4 sin a (a=theta) 0<=a<2pi

The curve look like below.

Gradient of tangent = -4/5 at P
equation of tangent = y-2(,/2) = -4/5 (x-5/,/2))
co-ordinates or R = (5,/2 , 0)

well the answer is 10-25pi but i cant get it dont even know what im doing wrong.
What's the value of theta for P?
3. (Original post by meepmeep)
What's the value of theta for P?
dy/da = 4cosa
dx/da = -5sina
dy/dx = -4cosa/5sina = -4/5

therefore cosa = sina
sina/cosa = 1
tana = 1
a = pi/4 or 5pi/4

now you an finish it.

MB
4. (Original post by musicboy)
dy/da = 4cosa
dx/da = -5sina
dy/dx = -4cosa/5sina = -4/5

therefore cosa = sina
sina/cosa = 1
tana = 1
a = pi/4 or 5pi/4

now you an finish it.

MB
didnt quite follow that how can u make sina=cosa?
5. (Original post by lexazver203)
didnt quite follow that how can u make sina=cosa?
he didnt.. look again, they're not the same thing

remember when you're parametrically differentiating:

dy/dx = dy/da * da/dx
due to the chain rule
6. (Original post by kimoni)
he didnt.. look again, they're not the same thing

remember when you're parametrically differentiating:

dy/dx = dy/da * da/dx
due to the chain rule

yeah i know the chain rule but it still doesnt make much sense
7. No he was right...

dy/da = 4cosa
dx/da = -5sina

.'. da/dx = -1/5sina

dy/dx = -4cosa/5sina = -4/5 <--here you just compare coefficients

therefore cosa = sina
sina/cosa = 1
tana = 1
a = pi/4 or 5pi/4
8. (Original post by jonas123)
No he was right...

dy/da = 4cosa
dx/da = -5sina

.'. da/dx = -1/5sina

dy/dx = -4cosa/5sina = -4/5 <--here you just compare coefficients

therefore cosa = sina
sina/cosa = 1
tana = 1
a = pi/4 or 5pi/4
i actually understood it but i still cant get the right answer forthe area of the thing

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