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Year 12 Maths Help Thread

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Original post by am99
ImageUploadedByStudent Room1462212501.956516.jpg
Can someone help me on part bi, am not sure how to integrate it
ImageUploadedByStudent Room1462212567.543703.jpg
What I tried was to integrate it without the x^4 first then put the x^4 there and take way the power but that is wrong.


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Expand the top: (2+x2)3=(2+x2)(2+x2)(2+x2)=(4+4x2+x4)(2+x2)=(8+⋯ )(2+x^2)^3 = (2+x^2)(2+x^2)(2+x^2) = (4 + 4x^2 + x^4)(2+x^2) = (8 + \cdots)

Then (2+x2)3x4=8x4+⋯+x2\frac{(2+x^2)^3}{x^4} = \frac{8}{x^4} + \cdots + x^2 and that's easy to integrate, isn't it?

Remember that ∫1xn dx=∫x−ndx=x−n+1−n+1+c\int \frac{1}{x^n} \, \mathrm{d}x = \int x^{-n} \mathrm{d}x = \frac{x^{-n +1}}{-n + 1} + c
Reply 4061
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am99
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Original post by Zacken
Expand the top: (2+x2)3=(2+x2)(2+x2)(2+x2)=(4+4x2+x4)(2+x2)=(8+⋯ )(2+x^2)^3 = (2+x^2)(2+x^2)(2+x^2) = (4 + 4x^2 + x^4)(2+x^2) = (8 + \cdots)

Then (2+x2)3x4=8x4+⋯+x2\frac{(2+x^2)^3}{x^4} = \frac{8}{x^4} + \cdots + x^2 and that's easy to integrate, isn't it?

Remember that ∫1xn dx=∫x−ndx=x−n+1−n+1+c\int \frac{1}{x^n} \, \mathrm{d}x = \int x^{-n} \mathrm{d}x = \frac{x^{-n +1}}{-n + 1} + c


Am sorry but I forgot how to integrate that :/, I think this is wrong ImageUploadedByStudent Room1462213922.765195.jpg


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Original post by am99
Am sorry but I forgot how to integrate that :/, I think this is wrong ImageUploadedByStudent Room1462213922.765195.jpg


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S, you have: ∫8+12x2+6x4+x6x4=∫8x4+12x2+6+x2\int \frac{8 + 12x^2 + 6x^4 + x^6}{x^4} = \int \frac{8}{x^4} + \frac{12}{x^2} + 6 + x^2

Now interage that using the fact that "add one to the power and divide by the new power".
FP3/2 question here. It's pretty lengthy but might be a solid question (I'm not sure).

The polar curve r(1−sin⁡θ)=4 r(1-\sin \theta )=4 is a parabola with Cartesian equation,
y=18x2−2 \displaystyle y=\frac{1}{8} x^2 -2 .

(a) Find the focus and directrix of the parabola.

(b) A curve, C \text{C} , has equation 2(x2+y2)+x=y+42 \sqrt{2(x^2+y^2)} + x=y+4\sqrt 2 .

It is further given that the curve C C is a conic section.Find in any order the following of C \text{C}
(i) The type of conic section.
(ii) The eccentricity.
(iii) The focus/foci and directix/directrices.
(iv) The line of symmetry.
(v) The polar equation.
(edited 7 years ago)
Reply 4064
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am99
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Original post by Zacken
S, you have: ∫8+12x2+6x4+x6x4=∫8x4+12x2+6+x2\int \frac{8 + 12x^2 + 6x^4 + x^6}{x^4} = \int \frac{8}{x^4} + \frac{12}{x^2} + 6 + x^2

Now interage that using the fact that "add one to the power and divide by the new power".


I feel so dumb for asking this but for the first term 8x, it would be 8x^2 and you divide x^4 by 2, which gives me x^2 so final answer for that term is 8x^2/x^2?


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Original post by am99
I feel so dumb for asking this but for the first term 8x, it would be 8x^2 and you divide x^4 by 2, which gives me x^2 so final answer for that term is 8x^2/x^2?


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Where are you getting 8x from? It's 8 + px^2 + ..., look at your answer to (a).
Reply 4066
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am99
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Original post by Zacken
Where are you getting 8x from? It's 8 + px^2 + ..., look at your answer to (a).


Omg I don't even know why I wrote 8x, I had 8 written down on part a -_- sorrry, let me try again now


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am99
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Original post by Zacken
Where are you getting 8x from? It's 8 + px^2 + ..., look at your answer to (a).


I finally got it, thank you !!!' It was such a dumb mistake and I didn't even realise it


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Original post by am99
I finally got it, thank you !!!' It was such a dumb mistake and I didn't even realise it


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You're very welcome! :h:

Glad you got it. :biggrin:
Can someone help me with this I can get the gradient but that's it:
Point A lies on curve y=x1/2 , the tangent of the curve at A is parallel to line 3y-2y=1. Find the equation of the curve at A.

The gradient is 2/3x
Original post by TheGreatPumpkin
Can someone help me with this I can get the gradient but that's it:
Point A lies on curve y=x1/2 , the tangent of the curve at A is parallel to line 3y-2y=1. Find the equation of the curve at A.

The gradient is 2/3x

If the gradient of the line is 2/3 then the gradient of the tangent must also be 2/3 since it's parallel to the line.

If the tangent at A has gradient 2/3 then this means that dy/dx = 2/3 at A.

Solve dy/dx = 2/3 to find the x-coordinate of A.
Original post by notnek
If the gradient of the line is 2/3 then the gradient of the tangent must also be 2/3 since it's parallel to the line.

I understand that
Original post by notnek

Solve dy/dx = 2/3 to find the x-coordinate of A.


That I have no clue, what do you mean solve, there is no variable?
Original post by TheGreatPumpkin
I understand that


That I have no clue, what do you mean solve, there is no variable?


What is dy/dx?
Original post by Zacken
What is dy/dx?


The gradient? 2/3
Original post by TheGreatPumpkin
The gradient? 2/3


Yes, but it's also something else. You know that y=x1/2y = x^{1/2}, so what's dy/dx? Equate the expression you get to 2/3 and solve for x.
Original post by Zacken
Yes, but it's also something else. You know that y=x1/2y = x^{1/2}, so what's dy/dx? Equate the expression you get to 2/3 and solve for x.


So I integrate dy/dx and simultaneously solve the equation?
Original post by TheGreatPumpkin
So I integrate dy/dx and simultaneously solve the equation?


No. We have: y=x1/2y = x^{1/2}. We differentiate this to get dydx=12x1/2\frac{dy}{dx} = \frac{1}{2x^{1/2}}.

But we know that dy/dx = 2/3, so: 12x1/2=23\frac{1}{2x^{1/2}} = \frac{2}{3}, now solve this.
Original post by Zacken
No. We have: y=x1/2y = x^{1/2}. We differentiate this to get dydx=12x1/2\frac{dy}{dx} = \frac{1}{2x^{1/2}}.

But we know that dy/dx = 2/3, so: 12x1/2=23\frac{1}{2x^{1/2}} = \frac{2}{3}, now solve this.


x=9/16, right?
Original post by TheGreatPumpkin
x=9/16, right?


Yeah.
Original post by Zacken
Yeah.


Then I just substitute it in to find y

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