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 x^2 gives 2b - \frac{1}{2}\cdot \frac{4}{3}= 0 \Rightarrow b=\frac{1}{3}

    x^3 gives 2c = 0 \Rightarrow c=0

    x^4 gives 2d - \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.
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    Someone explain to me how to find the polar equation of a circle with centre (a,\frac{\pi}{2}). I can see that the angle will make the circle go into a form of x^2+(y-a)^2=a^2 but I don't understand how to get the 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+y^2=a^2 \rightarrow r=2acos(\theta) and I understand where this comes from; so for x^2+(y-a)^2=a^2 it would be an anticlockwise rotation of \frac{\pi}{2} radians therefore r=2acos(\theta) \mapsto r=2acos(\theta-\frac{\pi}{2})=2asin(\theta)?
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    (Original post by RDKGames)
    Someone explain to me how to find the polar equation of a circle with centre (a,\frac{\pi}{2}). I can see that the angle will make the circle go into a form of x^2+(y-a)^2=a^2 but I don't understand how to get the 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+y^2=a^2 \rightarrow r=2acos(\theta) and I understand where this comes from; so for x^2+(y-a)^2=a^2 it would be an anticlockwise rotation of \frac{\pi}{2} radians therefore r=2acos(\theta) \mapsto r=2acos(\theta-\frac{\pi}{2})=2asin(\theta)?
    You could just use the fact that  x=r\cos \theta and  y=r\sin \theta and expand and simplify.
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    (Original post by B_9710)
    You could just use the fact that  x=r\cos \theta and  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,\frac{\pi}{2}) centre, would I necessarily have to convert it into a Cartesian form before I sub anything in?
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    (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,\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  \theta = \pi /2 you k ow that it lies on the y axis on the Cartesian coordinate system. So it has equation  x^2+(y-a)^2 = r^2 . But without anymore information you can't do anymore with it.
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    (Original post by B_9710)
    Well as it lies on the line  \theta = \pi /2 you k ow that it lies on the y axis on the Cartesian coordinate system. So it has equation  x^2+(y-a)^2 = r^2 . But without anymore information you can't do anymore with it.
    Thanks.

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

    If 0<\theta<\frac{\pi}{2} (lets take only the first quadrant here), would the cartesian equation be (x-a)^2+(y-b)^2=c^2 where the substitutions (apart from x and y) are a=ccos\theta, b=csin\theta?
    No because here you're using 'c' as the radius of the circle. If you think about it geometrically,  c\cos \theta is not going to give the correct value of a.
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    (Original post by B_9710)
    No because here you're using 'c' as the radius of the circle. If you think about it geometrically,  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 (\lambda,\alpha).
    This means the Cartesian equation will be (x-a)^2+(y-b)^2=\lambda^2
    We can define general x and y as x=rcos\theta and y=rsin\theta

    If the centre is at (\lambda,\alpha) then that means the radius of the circle will be \lambda as it goes through the origin, and if I construct a right-angled triangle with \lambda as hypotenuse and \alpha as the angle, I get the horizontal and vertical distances to be a=\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=2\lambda cos(\theta-\alpha). Is this correct working for any general circle?
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    Question:

    r_1=1-sin\theta ; -\pi<\theta\leq \pi
    r_2=\frac{1}{2}

    How can I go about finding the total area region which lies inside r_1 and r_2? I have found their points of intersection to be at (\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.
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    (Original post by RDKGames)
    Finding polar equation of a circle with centre (\lambda,\alpha).

    If the centre is at (\lambda,\alpha) then that means the radius of the circle will be \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 \lambda) or have I mis-interpreted something?
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    (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 \lambda) or have I mis-interpreted something?
    Circle isn't going through the origin as I've stated.
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    (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 (\lambda, \alpha) and radius a the equation is

    \displaystyle 

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

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

    r_1=1-sin\theta ; -\pi<\theta\leq \pi
    r_2=\frac{1}{2}

    How can I go about finding the total area region which lies inside r_1 and r_2? I have found their points of intersection to be at (\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:

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    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 \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.
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    (Original post by Zacken)
    Why are you assuming that if you're 'generalising'?
    In full generality, the equation of a circle with centre (\lambda, \alpha) and radius a the equation is

    \displaystyle 

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

    which yields your result if you assume \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=c^2 given that I still want a circle with centre (\lambda,\alpha)? I want to derive that general result for all cases.
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    (Original post by Zacken)
    The area bounded by the black line, the red curve and the y-axis is given by \frac{1}{2} \displaystyle \int_{\pi/6}^{\pi/2} (1-\sin \theta)^2 \, \mathrm{d} \theta.
    Makes sense, thanks.
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    A way to find out whether a number k 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 k 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 n?
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    (Original post by Palette)
    A way to find out whether a number k 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 k 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 n?
    Write n = 10a + b then n \equiv 10(a-2b) \pmod{7} so n \equiv 0 \pmod{7} \iff a - 2b \equiv 0 \pmod{7} and we're done.

    (note that  0 \leq b \leq 9 and a \geq 0)
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    I might not be super good with math but i can help if you want
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    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.

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    C: r=a\sqrt{sin(2\theta)}

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

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

    =\frac{1}{2}\displaystyle\int^{\  frac{\pi}{2}}_0 (a\sqrt{sin(2\theta)})^2 .d\theta

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

    =\frac{a^2}{2}\left[ -\frac{1}{2}cos(2\theta) \right]_0^\frac{\pi}{2}

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

    =\frac{a^2}{2}

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