C4 parametric equation help!
Please I need help completing these questions I know my part c isnwrong which will affect my part d hence why I haven't attempted all of part d. I seriously don't know what I've done wrong with part b that has lead to the mistake in c) any help please
Above attachment are the questions and part c)
These are the answers, all help is appreciated thanks very much 😬
Posted from TSR Mobile
Above attachment are the questions and part c)
These are the answers, all help is appreciated thanks very much 😬
Posted from TSR Mobile
Original post by Hudl
Please I need help completing these questions I know my part c isnwrong which will affect my part d hence why I haven't attempted all of part d. I seriously don't know what I've done wrong with part b that has lead to the mistake in c) any help please
Above attachment are the questions and part c)
These are the answers, all help is appreciated thanks very much 😬
Posted from TSR Mobile
Above attachment are the questions and part c)
These are the answers, all help is appreciated thanks very much 😬
Posted from TSR Mobile
wow
I did this question with one of my students this morning (June 2001 paper)
expand sin2t
then either recognition or substitution u = sint
Original post by TeeEm
wow
I did this question with one of my students this morning (June 2001 paper)
expand sin2t
then either recognition or substitution u = sint
I did this question with one of my students this morning (June 2001 paper)
expand sin2t
then either recognition or substitution u = sint
Is this for part b) or c)
For part b) I clearly have it in the form they mean, meaning its correct but then that isn't the answer apparently.
If it is the answer, I used the formula in the c3 booklet where -2Sin(A+B)/2 x Sin (A-B)/2
but just realised I did it wrong still whilst typing this as. its -27 as the coefficient so if I want it to work I would have to divide my integral by 27/2 meaning my cosA-cosb will have a coefficient of 27/2.
But still I'm stuck, can you explain part b) possibly and we'll take it from there? thanks a lot
Original post by Hudl
Is this for part b) or c)
For part b) I clearly have it in the form they mean, meaning its correct but then that isn't the answer apparently.
If it is the answer, I used the formula in the c3 booklet where -2Sin(A+B)/2 x Sin (A-B)/2
but just realised I did it wrong still whilst typing this as. its -27 as the coefficient so if I want it to work I would have to divide my integral by 27/2 meaning my cosA-cosb will have a coefficient of 27/2.
But still I'm stuck, can you explain part b) possibly and we'll take it from there? thanks a lot
For part b) I clearly have it in the form they mean, meaning its correct but then that isn't the answer apparently.
If it is the answer, I used the formula in the c3 booklet where -2Sin(A+B)/2 x Sin (A-B)/2
but just realised I did it wrong still whilst typing this as. its -27 as the coefficient so if I want it to work I would have to divide my integral by 27/2 meaning my cosA-cosb will have a coefficient of 27/2.
But still I'm stuck, can you explain part b) possibly and we'll take it from there? thanks a lot
what is the problem?
I understood that it was part (c)
Original post by TeeEm
what is the problem?
I understood that it was part (c)
I understood that it was part (c)
Sorry if I'm not being clear. To begin with part b) is the problem as that seems to be apparently wrong so what do I do? or is it right, because my friend told me its wrong and he may be wrong
Original post by Hudl
Sorry if I'm not being clear. To begin with part b) is the problem as that seems to be apparently wrong so what do I do? or is it right, because my friend told me its wrong and he may be wrong
I will check (b)
Original post by Hudl
Sorry if I'm not being clear. To begin with part b) is the problem as that seems to be apparently wrong so what do I do? or is it right, because my friend told me its wrong and he may be wrong
limits are the wrong way round in part b
that the only mistake
A = +27
Original post by TeeEm
limits are the wrong way round in part b
that the only mistake
A = +27
that the only mistake
A = +27
+27? oh I got -27 from doing integral of y dx/dt
y = 9sin 2t x = 3cos t
dx/dt = -3sin t as cos differentiates to -sin
therefore y dx/dt = 9sin 2t * -3sin t = -27sin2tsint? dt
Original post by Hudl
+27? oh I got -27 from doing integral of y dx/dt
y = 9sin 2t x = 3cos t
dx/dt = -3sin t as cos differentiates to -sin
therefore y dx/dt = 9sin 2t * -3sin t = -27sin2tsint? dt
y = 9sin 2t x = 3cos t
dx/dt = -3sin t as cos differentiates to -sin
therefore y dx/dt = 9sin 2t * -3sin t = -27sin2tsint? dt
READ THE LAST POST
the limits are the wrong way round
Original post by TeeEm
READ THE LAST POST
the limits are the wrong way round
the limits are the wrong way round
Ohhhhh, I see the limits for A=-27 should be the other way round.
So is that a rule because I haven't learnt that if you change the limites so it was meant to be 0 and pi/2 for my A=-27 with 0 being at the top
but instead its pi/2 0 with pi/2 being at the top. When you change the limits like this is it a rule that you change sign?
I dont get why a change in sign will be needed please, thanks
Original post by Hudl
Ohhhhh, I see the limits for A=-27 should be the other way round.
So is that a rule because I haven't learnt that if you change the limites so it was meant to be 0 and pi/2 for my A=-27 with 0 being at the top
but instead its pi/2 0 with pi/2 being at the top. When you change the limits like this is it a rule that you change sign?
I dont get why a change in sign will be needed please, thanks
So is that a rule because I haven't learnt that if you change the limites so it was meant to be 0 and pi/2 for my A=-27 with 0 being at the top
but instead its pi/2 0 with pi/2 being at the top. When you change the limits like this is it a rule that you change sign?
I dont get why a change in sign will be needed please, thanks
there is no magic
you integrate in cartesian left to right
at x=0 use the x=... equation and you get pi/2
when y=0 or (x=3, can you see why?) you get t=0
Original post by TeeEm
there is no magic
you integrate in cartesian left to right
at x=0 use the x=... equation and you get pi/2
when y=0 or (x=3, can you see why?) you get t=0
you integrate in cartesian left to right
at x=0 use the x=... equation and you get pi/2
when y=0 or (x=3, can you see why?) you get t=0
Hahahah there is no magic, sorry I get that but then my integral at the top in the pic below can be written in the form of the integral at the bottom in the pic below.
I'm asking is tht a general rule, for you to change signs if you want to change limits?
Posted from TSR Mobile
(edited 9 years ago)
Original post by hudl
hahahah there is no magic, sorry i get that but then my integral at the top in the pic below can be written in the form of the integral at the bottom in the pic below.
I'm asking is tht a general rule, for you to change signs if you want to change limits?
posted from tsr mobile
I'm asking is tht a general rule, for you to change signs if you want to change limits?
posted from tsr mobile
absolutely
Original post by TeeEm
absolutely
Beautiful, thanks a lot, appreciate it wish I could rep you but seems you've helped me to many times before!
Original post by Hudl
Beautiful, thanks a lot, appreciate it wish I could rep you but seems you've helped me to many times before!
not too worry.
it was my pleasure
Quick Reply
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