C4 parametric equations help Watch

oliviaaa98
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Can someone please explain question 7ii on paper A. Why is it tan (90-theta) etc??

https://mei.org.uk/files/papers/June13/C4_2013_June.pdf
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RDKGames
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(Original post by oliviaaa98)
Can someone please explain question 7ii on paper A. Why is it tan (90-theta) etc??

https://mei.org.uk/files/papers/June13/C4_2013_June.pdf
The gradient at A is equal to \tan \alpha. Let me know if you do not understand why. And it's not hard to deduce that \alpha + \theta  + 90 = 180 on that diagram, therefore \alpha = 90-\theta. Hence gradient is \tan (90-\theta).

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oliviaaa98
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don't understand why the gradient at A is equal to \tan \alpha , or why I even need to know the gradient at A :confused: ,but i don't think i need to know that anyway, i figured out how to do this question, they just wrote that on the markscheme as replacement for dy/dx i think
(Original post by RDKGames)
The gradient at A is equal to \tan \alpha. Let me know if you do not understand why. And it's not hard to deduce that \alpha + \theta  + 90 = 180 on that diagram, therefore \alpha = 90-\theta. Hence gradient is \tan (90-\theta).

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RDKGames
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(Original post by oliviaaa98)
don't understand why the gradient at A is equal to \tan \alpha , or why I even need to know the gradient at A :confused: ,but i don't think i need to know that anyway, i figured out how to do this question, they just wrote that on the markscheme as replacement for dy/dx i think
I meant the gradient at C. The x coordinate at that point is the same as the x coordinate of A.

Anyway, in order to determine \theta, you need to realise that it's somehow linked to the tangent at C. From my diagram above, all I did was introduce the angle \alpha. The gradient of the tangent at C is indeed the same as \tan \alpha because of the triangle at the end of this post. (1) What's the gradient of the hypotenuse?? (2) By basic trigonometry, what's the value of \tan \alpha ?? Can you see how the two are one and the same? This very idea applies to your question, hence \tan \alpha is the gradient at C. And I have already established that this can be rewritten as \tan (90-\theta).

But what *is* the gradient at C?? For that, you need to determine the value of \dfrac{dy}{dx} at that point, which turns out to be 1.98 as in the mark scheme.

So... \tan(90-\theta) = 1.98 as required.

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oliviaaa98
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thanks, I also need help on a question on this paper: https://mei.org.uk/files/papers/c406ju_83nd.pdf

can you please explain the last part of question 6 part iii) on paper A. 'Hence show that BF=3a/2 and find OF in terms of a, giving your answer exactly'.
And also part iv
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RDKGames
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(Original post by oliviaaa98)
thanks, I also need help on a question on this paper: https://mei.org.uk/files/papers/c406ju_83nd.pdf

can you please explain the last part of question 6 part iii) on paper A. 'Hence show that BF=3a/2 and find OF in terms of a, giving your answer exactly'.
And also part iv
In order to find the length BF, you need to know the y coordinates of B and F. The x coordinate is the same, so you don't need to know it for this length. The y coordinate of F is obviously 0, so what's the y coordinate of B? Use the fact that B occurs along the curve when \theta = \dfrac{2}{3}\pi. Hence the length of BF is...?

Then for length OF, you need to know the x coordinates of O and F. The y coordinate is the same, so you don't need to know it for this length. The x coordinate of O is obviously 0, so what's the x coordinate of F? Note that it's the same as the x coordinate of B. Hence the length of OF is...?

For (iv) use similar logic. For length BC, you now want the x coordinates of B and C. The y coordinates are the same, so you dont need to know what it is. You know the x coord of B, so what's the x coord of C?? There are a few ways to approach it, so I'll leave it to your creativity to see how you get it.

Length AF is easier, because you just consider the right-angled triangle AFB. You know the length BF. You are told the angle there is 30 degrees. So apply basic trig for AF.
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oliviaaa98
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for iv) when working out AF why do you have to multiple root 3 by 3a/2?, is it just to make it in terms of a? (and 3a/2 is equal to 1 using the standard 60 30 triangle, so its the same as multiplying by one?)??
(Original post by RDKGames)
In order to find the length BF, you need to know the y coordinates of B and F. The x coordinate is the same, so you don't need to know it for this length. The y coordinate of F is obviously 0, so what's the y coordinate of B? Use the fact that B occurs along the curve when \theta = \dfrac{2}{3}\pi. Hence the length of BF is...?

Then for length OF, you need to know the x coordinates of O and F. The y coordinate is the same, so you don't need to know it for this length. The x coordinate of O is obviously 0, so what's the x coordinate of F? Note that it's the same as the x coordinate of B. Hence the length of OF is...?

For (iv) use similar logic. For length BC, you now want the x coordinates of B and C. The y coordinates are the same, so you dont need to know what it is. You know the x coord of B, so what's the x coord of C?? There are a few ways to approach it, so I'll leave it to your creativity to see how you get it.

Length AF is easier, because you just consider the right-angled triangle AFB. You know the length BF. You are told the angle there is 30 degrees. So apply basic trig for AF.
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RDKGames
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(Original post by oliviaaa98)
for iv) when working out AF why do you have to multiple root 3 by 3a/2?, is it just to make it in terms of a? (and 3a/2 is equal to 1 using the standard 60 30 triangle, so its the same as multiplying by one?)??
\tan 30 = \dfrac{\text{BF}}{\text{AF}} \implies \text{AF} = \dfrac{\text{BF}}{\tan 30} = \dfrac{\frac{3}{2}a}{\tan 30}

What's the value of tan 30?
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