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# P3 Parametrics question. watch

1. (Original post by mathematician)
hey u guys theres no need to keep diferentiating, thts for the first part!
the second part just asks u to find where the line of y=2x-4, crosses the curve c,
sub parametric equations into the function, and your left with
2t^3-t^2=4

t^2(2t-1)=4
t^2=4
t= +/- 2 - im assuming on the question there were constraints, becos there are actually two points on which the line crosses, subbing the values of t back in and u get your co ordinates 8 and 4
remember curve c is a circle or an ellipse, so the line would have to pass thru twice
2. (Original post by mathematician)
remember curve c is a circle or an ellipse, so the line would have to pass thru twice
surely only when you've got sin t and cos t?
I plotted it on graph (see previous post) and got a Y shape
3. (Original post by kimoni)
surely only when you've got sin t and cos t?
I plotted it on graph (see previous post) and got a Y shape
what are the axis of your graph?
tht shud help u to where uve gone wrong.
4. (Original post by mathematician)
what are the axis of your graph?
tht shud help u to where uve gone wrong.
youve got 3 variables remember, x,t,y,
re aragning u get the curve c as y=x^2/3
being crossed by y=2x-4
5. Mathemtician in this case it is not an elipse i dont think it is when it described in terms of sins and cos it's an elipse. When it is an elipse question tho whhen it tells you where the line cuts an elipse at one point how would u find out where it would cut it 2nd time???
6. Actually 'V' shape is more accurate, with both lines divergent from x=0
7. This is what I got
Attached Images

8. (Original post by mathematician)
t^2(2t-1)=4
t^2=4
t= +/- 2 - im assuming on the question there were constraints, becos there are actually two points on which the line crosses, subbing the values of t back in and u get your co ordinates 8 and 4
These lines don't work - try putting t = 2 into the first line - it doesn't equal 4! You need to rearrange to 2t^3 - t^2 - 4 = 0 to be able to solve for t. There are no integer solutions.
9. Er Mathsey!

t^2(2t-1)=4 <----------------- this step here wheres the (2t-1) go?
t^2=4
t= +/- 2 - im assuming on the question there were constraints, becos there are actually two points on which the line crosses, subbing the values of t back in and u get your co ordinates 8 and 4

this question is not cool
10. (Original post by Bezza)
These lines don't work - try putting t = 2 into the first line - it doesn't equal 4! You need to rearrange to 2t^3 - t^2 - 4 = 0 to be able to solve for t. There are no integer solutions.
wot were the values true for with respect to t or x, it must have said?
11. Sub x = t^3 and y = t^2 into the equation of l and multiply through by (-1) to get 2t^3 - 3t^2 - 4 = 0 By the factor theorem, (t-2) is a factor, taking this out gives (t - 2)(2t^2 + t + 2) = 0 The only real solution is t = 2, point B is (8,4)
12. (Original post by XYTER)
Er Mathsey!

t^2(2t-1)=4 <----------------- this step here wheres the (2t-1) go?
t^2=4
t= +/- 2 - im assuming on the question there were constraints, becos there are actually two points on which the line crosses, subbing the values of t back in and u get your co ordinates 8 and 4

this question is not cool
i didnt write the question to the end there is also part (c) where it says - prove that the line l only cuts C at the point B
13. (Original post by XYTER)
Er Mathsey!

t^2(2t-1)=4 <----------------- this step here wheres the (2t-1) go?
t^2=4
t= +/- 2 - im assuming on the question there were constraints, becos there are actually two points on which the line crosses, subbing the values of t back in and u get your co ordinates 8 and 4

this question is not cool
tht is the way the examiner wud expect u to do it, ok
in the question it wud say, that t>0 or x>0 woteva from this info, t has to b +/-2
14. (Original post by Bezza)
Sub x = t^3 and y = t^2 into the equation of l to get 2t^3 - 3t^2 - 4 = 0 By the factor theorem, (t-2) is a factor, taking this out gives (t - 2)(2t^2 + t + 2) = 0 The only real solution is t = 2, point B is (8,4)
only thing is that it is +4 not -4
15. (Original post by lexazver203)
i didnt write the question to the end there is also part (c) where it says - prove that the line l only cuts C at the point B
look there is clrly confusion with this question, is it possible for every aspect of the question to b written down so i show u
16. (Original post by mathematician)
look there is clrly confusion with this question, is it possible for every aspect of the question to b written down so i show u

Ok full question with no mistakes this time.

The curve C has parametric equations x=t^3 y=t^2 t>0

(a) Find an equation of the tangent to C at A(1,1)

Given that hte line l with equation 3y-2x+4=0 cuts the curve C at B,

(b) find the coordinates of B,
(c) prove that the line l only cuts C at the point B.
17. (Original post by mathematician)
tht is the way the examiner wud expect u to do it, ok
in the question it wud say, that t>0 or x>0 woteva from this info, t has to b +/-2
The examiner wouldn't expect it cos it doesn't make sense. You have the wrong equation for a start - it should be 2t^3 - 3t^2 = 4. You said:
2t^3-t^2=4

t^2(2t-1)=4
t^2=4
t= +/- 2
Put t = 2 into the LHS of the first line: 2*8-4 = 16-4 = 12 which doesn't equal 4. so t=2 is not a solution of this equation. You can't just get rid of the (2t-1) like you have done
18. That was i was saying and has anyone actually read the prev post sayin he wrote the questions down incorrectly?
19. (Original post by lexazver203)
Ok full question with no mistakes this time.

The curve C has parametric equations x=t^3 y=t^2 t>0

(a) Find an equation of the tangent to C at A(1,1)

Given that hte line l with equation 3y-2x+4=0 cuts the curve C at B,

(b) find the coordinates of B,
(c) prove that the line l only cuts C at the point B.
ok there u have it, thts why we discard 2t-1, becos t>0
20. 3t^2 - 2t^3 +4 =0
t^2(3-2t)=-4
therofer t =2 then do the rest U SELF

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