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Year 12 further maths help!

I'm really struggling with this 2018 WJEC paper 😭 and it's due tommorow
1736264759311207888626891014744.jpg
Reply 2
Original post by Maximum-tragedy
I'm really struggling with this 2018 WJEC paper 😭 and it's due tommorow

which one / what have you tried?
Original post by mqb2766
which one / what have you tried?


4 - I've differentiated to get the velocity but I have no idea how to try to get the value of t let alone the rest of the question
Reply 4
Original post by Maximum-tragedy
4 - I've differentiated to get the velocity but I have no idea how to try to get the value of t let alone the rest of the question

If its (instantaneously) at rest, then each component must be zero, which is just working out when a few trig terms are simultaneously zero? The momentum part should be straightfoward as should the force (acceleration/rate of change of momentum)?
(edited 1 month ago)
Original post by mqb2766
If its (instantaneously) at rest, then each component must be zero, which is just working out when a few trig terms are simultaneously zero? The momentum part should be straightfoward as should the force (acceleration/rate of change of momentum)?


I'm not going to lie - what you just said went right over my head. So could you just make cos(t) 0? And then mess round with the sine to get the answer?
Reply 6
Original post by Maximum-tragedy
I'm not going to lie - what you just said went right over my head. So could you just make cos(t) 0? And then mess round with the sine to get the answer?

You have something like
v(t) = cos(t) i + sin(2t) j + cos(t) k
so you want each of the i, j, k coefficients to be zero so
cos(t) = 0
sin(2t) = 0
cos(t) = 0
A simple sketch of the trig functions or when is cos(t) = 0 and which of those solutions correspond to sin(2t) = 0.
Original post by mqb2766
You have something like
v(t) = cos(t) i + sin(2t) j + cos(t) k
so you want each of the i, j, k coefficients to be zero so
cos(t) = 0
sin(2t) = 0
cos(t) = 0
A simple sketch of the trig functions or when is cos(t) = 0 and which of those solutions correspond to sin(2t) = 0.

17362703887038345991749878593164.jpg Thank you! I think I got it
Reply 8
Original post by Maximum-tragedy
17362703887038345991749878593164.jpg Thank you! I think I got it

Looks about right. Minor quibble (which you should have come across) is when you say
cos^(-1)(0) = pi/2
in this case youre lucky as the first value also gives sin(2t) = 0. More generally it would be better to write something like
cos(t) = 0
t = pi/2 + k*pi
so explicitly write out all the possible solutions for cos, then verify which is the first one that satisfies sin(2t)=0, so k=0.

If you just want to refer to the principal solution, it would be better to say
arccos(0) = pi/2
as arccos is the inverse of cos on the restricted domain 0 to pi.

Also, you can get force by differentiating momentum (as youve already stated that). Theres no need to go back to velocity and then multiply by mass.
(edited 1 month ago)
Original post by mqb2766
Looks about right. Minor quibble (which you should have come across) is when you say
cos^(-1)(0) = pi/2
in this case youre lucky as the first value also gives sin(2t) = 0. More generally it would be better to write something like
cos(t) = 0
t = pi/2 + k*pi
so explicitly write out all the possible solutions for cos, then verify which is the first one that satisfies sin(2t)=0, so k=0.
If you just want to refer to the principal solution, it would be better to say
arccos(0) = pi/2
as arccos is the inverse of cos on the restricted domain 0 to pi.
Also, you can get force by differentiating momentum (as youve already stated that). Theres no need to go back to velocity and then multiply by mass.


Ok thank you!
17362728436173693492427531972941.jpg Would anyone know how to do this either? I don't understand how to reduce the top from that to something more understandable
Original post by Maximum-tragedy
17362728436173693492427531972941.jpg Would anyone know how to do this either? I don't understand how to reduce the top from that to something more understandable

If youre asked for partial fractions, then that requires the original fraction is proper. Do you know what that means, What to do it first?
Original post by mqb2766
If youre asked for partial fractions, then that requires the original fraction is proper. Do you know what that means, What to do it first?


Nope! 🙃
Do you have a textbook? If so, have a read of the section, if not, get one.

But simply your numerator is x^4 and denominator is x^3 (highest power) so as 4>=3, its improper. It would be proper if it was something quadratic/cubic. So if you do the division first youd end up with something like
ax + b + (cx^2 + dx + e)/((x^2+7)(x-1))
So the (remainder) fraction is now quadratic/cubic which is proper and you can do partial fractions on that.
(edited 1 month ago)

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