The Student Room Group
Reply 1
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)

question s find the shaded area, leawing your answer in terms of pi.

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?
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
Reply 3
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?
Reply 4
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
Reply 5
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
Reply 6
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
Reply 7
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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