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Year 13 Maths Help Thread

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Original post by Zacken
Math12345's answer is rather nicer, but anyways, for completeness:

Coefficient of x2x^2 gives 2b−12⋅43=0⇒b=132b - \frac{1}{2}\cdot \frac{4}{3}= 0 \Rightarrow b=\frac{1}{3}

x3x^3 gives 2c=0⇒c=02c = 0 \Rightarrow c=0

x4x^4 gives 2d−4b3+215=0⇒d=7452d - \frac{4b}{3} + \frac{2}{15} = 0\Rightarrow d = \frac{7}{45}


Original post by Math12345
sin(2x)=2x-(4/3)x^3+... (You can fill in the missing third term)

(sin(2x))^-1 = (2x-(4/3)x^3))^-1 = (2x(1-(2/3)x^2))^-1 = 1/(2x) [1+(2/3)x^2 +...]

You can continue (I used the expansion (1-y)^-1 above).


Ah thanks guys, I got it now. While the binomial is something I can deal with now, I do find comparison of coefficients less error heavy when I'm doing it. :smile:
Someone explain to me how to find the polar equation of a circle with centre (a,π2)(a,\frac{\pi}{2}). I can see that the angle will make the circle go into a form of x2+(y−a)2=a2x^2+(y-a)^2=a^2 but I don't understand how to get the r=...r=... form, the book doesn't explain it very well, they deal with a right-angled triangle in the circle with hyponeneuse 2a but I'm more confused on where would the cos/sin come from and why.

Edit: Okay I thought of this but unsure how credible this is or if I can use this generally: (x−a)2+y2=a2→r=2acos(θ)(x-a)^2+y^2=a^2 \rightarrow r=2acos(\theta) and I understand where this comes from; so for x2+(y−a)2=a2x^2+(y-a)^2=a^2 it would be an anticlockwise rotation of π2\frac{\pi}{2} radians therefore r=2acos(θ)↦r=2acos(θ−π2)=2asin(θ)r=2acos(\theta) \mapsto r=2acos(\theta-\frac{\pi}{2})=2asin(\theta)?
(edited 7 years ago)
Original post by RDKGames
Someone explain to me how to find the polar equation of a circle with centre (a,π2)(a,\frac{\pi}{2}). I can see that the angle will make the circle go into a form of x2+(y−a)2=a2x^2+(y-a)^2=a^2 but I don't understand how to get the r=...r=... form, the book doesn't explain it very well, they deal with a right-angled triangle in the circle with hyponeneuse 2a but I'm more confused on where would the cos/sin come from and why.

Edit: Okay I thought of this but unsure how credible this is or if I can use this generally: (x−a)2+y2=a2→r=2acos(θ)(x-a)^2+y^2=a^2 \rightarrow r=2acos(\theta) and I understand where this comes from; so for x2+(y−a)2=a2x^2+(y-a)^2=a^2 it would be an anticlockwise rotation of π2\frac{\pi}{2} radians therefore r=2acos(θ)↦r=2acos(θ−π2)=2asin(θ)r=2acos(\theta) \mapsto r=2acos(\theta-\frac{\pi}{2})=2asin(\theta)?


You could just use the fact that x=rcos⁡θ x=r\cos \theta and y=rsin⁡θ y=r\sin \theta and expand and simplify.
Original post by B_9710
You could just use the fact that x=rcos⁡θ x=r\cos \theta and y=rsin⁡θ y=r\sin \theta and expand and simplify.


Ah that's neat, makes sense to use those but how would I approach it if all I'm given is the (a,π2)(a,\frac{\pi}{2}) centre, would I necessarily have to convert it into a Cartesian form before I sub anything in?
(edited 7 years ago)
Original post by RDKGames
Ah that's neat, makes sense to use those but how would I approach it if all I'm given is the (a,π2)(a,\frac{\pi}{2}) centre, would I necessarily have to convert it into a Cartesian form before I sub anything in?


Well as it lies on the line θ=π/2 \theta = \pi /2 you k ow that it lies on the y axis on the Cartesian coordinate system. So it has equation x2+(y−a)2=r2 x^2+(y-a)^2 = r^2 . But without anymore information you can't do anymore with it.
Original post by B_9710
Well as it lies on the line θ=π/2 \theta = \pi /2 you k ow that it lies on the y axis on the Cartesian coordinate system. So it has equation x2+(y−a)2=r2 x^2+(y-a)^2 = r^2 . But without anymore information you can't do anymore with it.


Thanks.

If 0<θ<π20<\theta<\frac{\pi}{2} (lets take only the first quadrant here), would the cartesian equation be (x−a)2+(y−b)2=c2(x-a)^2+(y-b)^2=c^2 where the substitutions (apart from x and y) are a=ccosθ,b=csinθa=ccos\theta, b=csin\theta?
(edited 7 years ago)
Original post by RDKGames
Thanks.

If 0<θ<π20<\theta<\frac{\pi}{2} (lets take only the first quadrant here), would the cartesian equation be (x−a)2+(y−b)2=c2(x-a)^2+(y-b)^2=c^2 where the substitutions (apart from x and y) are a=ccosθ,b=csinθa=ccos\theta, b=csin\theta?


No because here you're using 'c' as the radius of the circle. If you think about it geometrically, ccos⁡θ c\cos \theta is not going to give the correct value of a.
Original post by B_9710
No because here you're using 'c' as the radius of the circle. If you think about it geometrically, ccos⁡θ c\cos \theta is not going to give the correct value of a.


I'm attempting to generalise it and I am thinking of it geometrically. Let me lay out a new scenario and show you how I'm seeing this at the moment:

Finding polar equation of a circle with centre (λ,α)(\lambda,\alpha).
This means the Cartesian equation will be (x−a)2+(y−b)2=λ2(x-a)^2+(y-b)^2=\lambda^2
We can define general x and y as x=rcosθx=rcos\theta and y=rsinθy=rsin\theta

If the centre is at (λ,α)(\lambda,\alpha) then that means the radius of the circle will be λ\lambda as it goes through the origin, and if I construct a right-angled triangle with λ\lambda as hypotenuse and α\alpha as the angle, I get the horizontal and vertical distances to be a=λcosα,b=λsinαa=\lambda cos\alpha , b=\lambda sin\alpha respectively. So it makes sense to me to use those as substitutions.

If I plug everything in and rearrange to get the polar form I get r=2λcos(θ−α)r=2\lambda cos(\theta-\alpha). Is this correct working for any general circle?
Question:

r1=1−sinθr_1=1-sin\theta ; −π<θ≤π-\pi<\theta\leq \pi
r2=12r_2=\frac{1}{2}

How can I go about finding the total area region which lies inside r1r_1 and r2r_2? I have found their points of intersection to be at (12,π6),(12,5π6)(\frac{1}{2},\frac{\pi}{6}) , (\frac{1}{2},\frac{5\pi}{6}) but I'm unsure how to go about finding the enclosed area.
(edited 7 years ago)
Untitled.png

Original post by RDKGames
Finding polar equation of a circle with centre (λ,α)(\lambda,\alpha).

If the centre is at (λ,α)(\lambda,\alpha) then that means the radius of the circle will be λ\lambda as it goes through the origin,




Doesn't make sense, look at the picture attached. You can't be telling me the radius of the circle is the red line, can you? (which has length λ\lambda) or have I mis-interpreted something?
Original post by Zacken

Doesn't make sense, look at the picture attached. You can't be telling me the radius of the circle is the red line, can you? (which has length λ\lambda) or have I mis-interpreted something?


Circle isn't going through the origin as I've stated.
Original post by RDKGames
Circle isn't going through the origin as I've stated.


Why are you assuming that if you're 'generalising'?
In full generality, the equation of a circle with centre (λ,α)(\lambda, \alpha) and radius aa the equation is

Unparseable latex formula:

\displaystyle [br]\begin{equation*}r^2 -2\lambda r \cos (\theta - \alpha) + \lambda^2 = a^2 \end{equation*}



which yields your result if you assume λ=a\lambda = a.
Original post by RDKGames
Question:

r1=1−sinθr_1=1-sin\theta ; −π<θ≤π-\pi<\theta\leq \pi
r2=12r_2=\frac{1}{2}

How can I go about finding the total area region which lies inside r1r_1 and r2r_2? I have found their points of intersection to be at (12,π6),(12,5π6)(\frac{1}{2},\frac{\pi}{6}) , (\frac{1}{2},\frac{5\pi}{6}) but I'm unsure how to go about finding the enclosed area.


Answer:

Untitled.png

Restrict ourselves to the first quadrant because we the shape is symmetric and we can multiply by two to get the area in the second quadrant.

Anywho, the purple shaded area is just the sector of a circle whose radius and angle you know. GCSE stuff.

The area bounded by the black line, the red curve and the y-axis is given by 12∫π/6π/2(1−sin⁡θ)2 dθ\frac{1}{2} \displaystyle \int_{\pi/6}^{\pi/2} (1-\sin \theta)^2 \, \mathrm{d} \theta.

Then add, multiply by two to include the second quadrant as well, then add in a the area of the semi-circle.
Original post by Zacken
Why are you assuming that if you're 'generalising'?
In full generality, the equation of a circle with centre (λ,α)(\lambda, \alpha) and radius aa the equation is

Unparseable latex formula:

\displaystyle [br]\begin{equation*}r^2 -2\lambda r \cos (\theta - \alpha) + \lambda^2 = a^2 \end{equation*}



which yields your result if you assume λ=a\lambda = a.


Ah I didn't mention that, at the time I thought all circles in polar form go through the origin which clearly isn't the case but it did make my substitutions for a and b simple. What substitutions would i make for a and b in (x−a)2+(y−b)2=c2(x-a)^2+(y-b)^2=c^2 given that I still want a circle with centre (λ,α)(\lambda,\alpha)? I want to derive that general result for all cases.
Original post by Zacken

The area bounded by the black line, the red curve and the y-axis is given by 12∫π/6π/2(1−sin⁡θ)2 dθ\frac{1}{2} \displaystyle \int_{\pi/6}^{\pi/2} (1-\sin \theta)^2 \, \mathrm{d} \theta.


Makes sense, thanks.
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Avatar for Palette
Palette
OP
14
A way to find out whether a number kk is divisible by 7 is to double the last digit and subtract it from the remaining digits. If the result is divisible by 7, then kk is divisible by 7. For example, 273 is divisible by 7 because 27-(3*2)=21 which is divisible by 7.

How can I prove that this method works for any integer with nn?
Original post by Palette
A way to find out whether a number kk is divisible by 7 is to double the last digit and subtract it from the remaining digits. If the result is divisible by 7, then kk is divisible by 7. For example, 273 is divisible by 7 because 27-(3*2)=21 which is divisible by 7.

How can I prove that this method works for any integer with nn?


Write n=10a+bn = 10a + b then n≡10(a−2b)(mod7)n \equiv 10(a-2b) \pmod{7} so n≡0(mod7)  ⟺  a−2b≡0(mod7)n \equiv 0 \pmod{7} \iff a - 2b \equiv 0 \pmod{7} and we're done.

(note that 0≤b≤9 0 \leq b \leq 9 and a≥0a \geq 0)
(edited 7 years ago)
I might not be super good with math but i can help if you want :smile:
How would I go about b. ii? I found the constant distance to be 2a in prev part but I'm not sure where to begin for this one.

ImageUploadedByStudent Room1470131704.087734.jpg



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C: r=asin(2θ)r=a\sqrt{sin(2\theta)}

Find the area A of that part of the interior of C which lies in the region 0≤θ≤π20\leq\theta\leq\frac{\pi}{2}

A=∫0π212r2.dθA=\displaystyle\int^{\frac{\pi}{2}}_0 \frac{1}{2}r^2 .d\theta

=12∫0π2(asin(2θ))2.dθ=\frac{1}{2}\displaystyle\int^{\frac{\pi}{2}}_0 (a\sqrt{sin(2\theta)})^2 .d\theta

Unparseable latex formula:

=\frac{a^2}{2}\displaystyle\int^{\frac{\pi}{2}}_0 sin(2\theta)} .d\theta



=a22[−12cos(2θ)]0π2=\frac{a^2}{2}\left[ -\frac{1}{2}cos(2\theta) \right]_0^\frac{\pi}{2}

=a22(1)=\frac{a^2}{2}(1)

=a22=\frac{a^2}{2}

Someone explain to me what I am doing wrong here, the answer is 2a22a^2.

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