Question

Integration question

Answer 1

x=acos3t dx/dt=-3asin3t y=asint dy/dt=acost
dy/dx=dy/dt*dt/dx=acost/-3asin3t=cost/-3sin3t
y-asint=cost/-3sin3t(x-acos3t)
at t=pi/6
y-1/2a=-(3)^1/2/6(x)

Answer 2

in order to calculate the area of triangle we need to let y=0 and figure out the base of triangle it turned out to be equal to x=(3)^1/2/4. That's what i have managed to do till now.

Answer 3

r' = tangent vector

(0, a/2) is the anchor point of (-3a, a sqrt(3)/2 )

so the line y=-x sqrt(3)/2 + a/2 defines our triangle right?

Answer 4

how did you find -3a,a sqrt(3)/2

Answer 5

set t=pi/6 and plug it into the dx/dt and dy/dt

Answer 6

okay got it!

Answer 7

but why are we doing all of this, why cant we try to find the equation of curve and integrate it with limits a and b

Answer 8

sorry a and 0.

Answer 9

we can, i was just working out part a

Answer 10

i have no problems with part a) i showed my work above.

Answer 11

i was just catching up then

Answer 12

if our work agrees then we are on the right track hopefully

Answer 13

can we solve t in terms of x; and input into y? dunno if that simplifies anything but its a thought i have in my head

Answer 14

t = 3 arccos(x/a)

y = a sin(3 arccos(x/a))

prolly not a very pretty integration is it

can we convert to polar?

Answer 15

nope, would be too much work.

Answer 16

dy/dx=cost/-3sin3t
y-y1=m(x-x1))
y-asint=cost/-3sint(x-acos3t)

Answer 17

trying to recall parametric integration; dotproduct seems useful

Answer 18

\[A=\int_{0}^{pi/6}y\frac{dx}{dt}dt\]
is what im seeing in my notes

Answer 19

-3a^2 sin(t) sin(3t)

and an identity for the product of sines is:

1/2 (sin(3t+t) + sin(3t-t))

dbl chk

sin sin is from cosines ..

-cos(3t+t) = -cos(3t)cos(t) + sin(3t)sin(t)
cos(3t-t) = cos(3t)cos(t) + sin(3t)sin(t)
-----------------------------------
1/2(cos(2t) - cos(4t)) = sin(3t)sin(t)

Answer 20

i believe that should work it real nice ....

Answer 21

asint(-3asin3t)
-3a^2sin3t^2
i am stuck in integration whats above

Answer 22

sin(t) sin(3t) is NOT sin(3t^2)

Answer 23

which is why i eventually worked the cosine rule out to show that this product is equal to what i posted

Answer 24

sin(3t)sin(t) = 1/2 (cos(2t) - cos(4t))