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# Edexcel C4 Maths HELP

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1. Hello,
If the gradient of curve is dy/dx = (-2/3)cot2t
Then how do I find the coordinates of the points where the tangent to the curve is parallel to the x-axis.
I've written the gradient as, (-2)/(3tan2t)
so surely if the tangent is parallel with the x axis, the numerator of this gradient must = 0?
Thanks
2. the equation of any line is y = mx + c

lines parallel to the x-axis have equations y = c which can be written in the form y = 0x + c.

therefore, since m = 0, dy/dx = 0. so dy = 0.
3. (Original post by Raees_Sharif)
the equation of any line is y = mx + c

lines parallel to the x-axis have equations y = c which can be written in the form y = 0x + c.

therefore, since m = 0, dy/dx = 0. so dy = 0.
thank you! I understand that bit but isn't dy -2/3 or -2?
obviously -2/3 doesn't equate to 0 though so what is dy?
4. (Original post by Grewsters)
Hello,
If the gradient of curve is dy/dx = (-2/3)cot2t
Then how do I find the coordinates of the points where the tangent to the curve is parallel to the x-axis.
I've written the gradient as, (-2)/(3tan2t)
so surely if the tangent is parallel with the x axis, the numerator of this gradient must = 0?
Thanks
Exactly, the numerator has to be 0, so -2 cot 2t = 0.

But cot 2t = cos 2t / sin 2t.

So -2 cos 2t = 0.

So cos 2t = 0.

Solve the equation cos x = 0 where x = 2t.
5. If dy/dx=0 then cot2t=cos2t/sin2t=0
So cos2t=0.
6. (Original post by Zacken)
Exactly, the numerator has to be 0, so -2 cot 2t = 0.

But cot 2t = cos 2t / sin 2t.

So -2 cos 2t = 0.

So cos 2t = 0.

Solve the equation cos x = 0 where x = 2t.
ahhh yeah that makes sense now, thank you
7. (Original post by Grewsters)
Hello,
If the gradient of curve is dy/dx = (-2/3)cot2t
Then how do I find the coordinates of the points where the tangent to the curve is parallel to the x-axis.
I've written the gradient as, (-2)/(3tan2t)
so surely if the tangent is parallel with the x axis, the numerator of this gradient must = 0?
Thanks
(Original post by Zacken)
x
What past paper was this? I remember doing it quite a while back.
8. (Original post by Zacken)
Exactly, the numerator has to be 0, so -2 cot 2t = 0.

But cot 2t = cos 2t / sin 2t.

So -2 cos 2t = 0.

So cos 2t = 0.

Solve the equation cos x = 0 where x = 2t.
Yeah and if it says parallel to the y axis then dy/dx = o where the numerator is '0'
but the denominator is = 0 right?
9. (Original post by XxKingSniprxX)
Yeah and if it says parallel to the y axis then dy/dx = o where the numerator is '0'
but the denominator is = 0 right?
What?
10. (Original post by Zacken)
What?
read op question. "how do I find the coordinates of the points where the tangent to the curve is parallel to the x-axis. "

I said if its parallel to the y axis the numerator is '0' and the denominator is = 0
right?

where in OP case you just treat dy/dx = 0 as normal.
11. (Original post by XxKingSniprxX)
read op question. "how do I find the coordinates of the points where the tangent to the curve is parallel to the x-axis. "

I said if its parallel to the y axis the numerator is '0' and the denominator is = 0
right?

where in OP case you just treat dy/dx = 0 as normal.
tf you mean by 'the numerator is '0''?
12. (Original post by Zacken)
tf you mean by 'the numerator is '0''?
yeah so you would just set the denominator to zero
and solve if the question was asking for parallel to y axis.
13. (Original post by XxKingSniprxX)
yeah so you would just set the denominator to zero
and solve if the question was asking for parallel to y axis.
yeah, that isn't what you were saying previously though
14. (Original post by XxKingSniprxX)
What past paper was this? I remember doing it quite a while back.
C3 Solomon C. I hate how there are no video walkthroughs for solomon papers so I come here for all my issues with them hahah

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