Year 13 Maths Help Thread

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    (Original post by RDKGames)
    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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    Did you get that the coordinates of M are  (a\cos^{2}{\phi},a\sin{\phi}\cos  {\phi}) ? If so, then the next part, you can do by verification.

    Edit: this is the Cartesian coordinates.

    Alternatively, if you do not seek the method of verification, you can do so by doing this:

    As  X = a\cos^{2}{\phi} we can use the fact that  \cos^{2}{\phi} = \frac{1}{2}(1+\cos{2\phi}) so we get that  \dfrac{2X}{a} = (1 + \cos{2\phi}) so we can obtain that  \cos^{2}{2\phi} = \Big( \dfrac{2X - a}{a}\Big)^{2}

    Using the similar approach, notice that  Y = \frac{a}{2}\sin{2\phi} Hence,  \sin^{2}{2\phi} = \Big( \dfrac{2Y}{a} \Big)^{2} .Using  \sin^{2}{2\phi} + \cos^{2}{2\phi} = 1 \Longrightarrow \Big( \dfrac{2X-a}{a} \Big)^{2} + \Big( \dfrac{2Y}{a} \Big)^{2} = 1 .

    Rearranging,  4X^{2} - 4aX + a^{2} + 4Y^{2} = a^{2} \Longrightarrow X^{2} + Y^{2} = aX Notice for any varying  \theta and  r we can say that  X = r\cos{\theta}, Y = r\sin{\theta} .

    Finally, notice that  X^{2} + Y^{2} = r^2 . Therefore  r^2 = ar\cos{\theta} .

    We can cancel the r, aslong as  C_{2} covers the point at the pole i.e at  \theta = \pm \frac{\pi}{2} .

    This will give us the required result.
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    (Original post by RDKGames)
    C: r=a\sqrt{sin(2\theta)}

    Someone explain to me what I am doing wrong here, the answer is 2a^2.
    Their answer is obviously nonsensical (analytically and geometrically) unless the curve is meant to be r = 2a\sqrt{\sin \theta}.
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    (Original post by AMarques)
    Did you get that the coordinates of M are  (a\cos^{2}{\phi},a\sin{\phi}\cos  {\phi}) ? If so, then the next part, you can do by verification.

    Edit: this is the Cartesian coordinates.

    Alternatively, if you do not seek the method of verification, you can do so by doing this:

    As  X = a\cos^{2}{\phi} we can use the fact that  \cos^{2}{\phi} = \frac{1}{2}(1+\cos{2\phi}) so we get that  \dfrac{2X}{a} = (1 + \cos{2\phi}) so we can obtain that  \cos^{2}{2\phi} = \Big( \dfrac{2X - a}{a}\Big)^{2}

    Using the similar approach, notice that  Y = \frac{a}{2}\sin{2\phi} Hence,  \sin^{2}{2\phi} = \Big( \dfrac{2Y}{a} \Big)^{2} .Using  \sin^{2}{2\phi} + \cos^{2}{2\phi} = 1 \Longrightarrow \Big( \dfrac{2X-a}{a} \Big)^{2} + \Big( \dfrac{2Y}{a} \Big)^{2} = 1 .

    Rearranging,  4X^{2} - 4aX + a^{2} + 4Y^{2} = a^{2} \Longrightarrow X^{2} + Y^{2} = aX Notice for any varying  \theta and  r we can say that  X = r\cos{\theta}, Y = r\sin{\theta} .

    Finally, notice that  X^{2} + Y^{2} = r^2 . Therefore  r^2 = ar\cos{\theta} .

    We can cancel the r, aslong as  C_{2} covers the point at the pole i.e at  \theta = \pm \frac{\pi}{2} .

    This will give us the required result.
    Your working out makes sense and it gets the right answer, however I'm unsure how the get the co-ordinates of M. I'm not used to working with lines in polar form. I can see that the line's equation is \theta=\phi but I'm unsure what substitutions go where to get it. I can only see M being (acos\phi,asin\phi)
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    (Original post by RDKGames)
    Your working out makes sense and it gets the right answer, however I'm unsure how the get the co-ordinates of M. I'm not used to working with lines in polar form. I can see that the line's equation is \theta=\phi but I'm unsure what substitutions go where to get it.
    I basically turned both equations into Cartesian form, and then subbed one into the other and made x the subject. By doing this, I get three x-coordinates (including the pole which we don't count as either A or B), and hence three y-coordinates.

    For the polar equation  \theta = \phi this is the same as  y = \tan{\phi}x in Cartesian form. For  C_{1} we get  x^{2} + y^{2} = a(x \pm \sqrt{x^{2} + y^{2}}) the + sign for a > 0, and - for a < 0.

    We then sub  \tan{\phi}x for  y in the second equation. By doing this, we can find both A and B, and then M will just be the average of the x/y coordinates.

    Out of curiosity how did you find the distance AB without finding A and B explicitly?
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    (Original post by AMarques)
    I basically turned both equations into Cartesian form, and then subbed one into the other and made x the subject. By doing this, I get three x-coordinates (including the pole which we don't count as either A or B), and hence three y-coordinates.

    For the polar equation  \theta = \phi this is the same as  y = \tan{\phi}x in Cartesian form. For  C_{1} we get  x^{2} + y^{2} = a(x \pm \sqrt{x^{2} + y^{2}}) the + sign for a > 0, and - for a < 0.

    We then sub  \tan{\phi}x for  y in the second equation. By doing this, we can find both A and B, and then M will just be the average of the x/y coordinates.

    Out of curiosity how did you find the distance AB without finding A and B explicitly?
    To find the distance, I firstly knew that \theta=\phi and used that substitution into r=a(1+cos\theta) and that would give me intersection point A as a function of \phi.
    For point B (and I'm not entirely sure how to explain this properly), I simply imagined that line at angle \phi above the initial line OL and the angle of \pi-\phi clockwise, below OL, would give the second point, therefore B would be r_B=a(1+cos[-(\pi-\phi)])=a(1-cos\phi).

    Since r represents the length, I simply added r_A and r_B: AB=a(1+cos\phi)+a(1-cos\phi) thus giving 2a which is a constant in dependent of \phi that I required.
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    (Original post by RDKGames)
    Your working out makes sense and it gets the right answer, however I'm unsure how the get the co-ordinates of M. I'm not used to working with lines in polar form. I can see that the line's equation is \theta=\phi but I'm unsure what substitutions go where to get it. I can only see M being (acos\phi,asin\phi)
    At point A the coordinates are  (a\cos \phi (1+\cos \phi ), a\sin \phi (1+\cos \phi )) and the coordinates at B are  (a\cos (\phi - \pi )(1+\cos (\phi - \pi ),a\sin (\phi -\pi )(1+\cos (\phi -\pi ))) . You can simplify for B.
    Then the coordinates of the midpoint at point M are  (X, Y) =(\frac{1}{2} (x_A + x_b) , \frac{1}{2} (y_A + y_B )) .
    You can convert from parametric to Cartesian equation in  X and  Y and then back to polar form.
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    (Original post by RDKGames)
    To find the distance, I firstly knew that \theta=\phi and used that substitution into r=a(1+cos\theta) and that would give me intersection point A as a function of \phi.
    For point B (and I'm not entirely sure how to explain this properly), I simply imagined that line at angle \phi above the initial line OL and the angle of \pi-\phi clockwise, below OL, would give the second point, therefore B would be r_B=a(1+cos[-(\pi-\phi)])=a(1-cos\phi).

    Since r represents the length, I simply added r_A and r_B: AB=a(1+cos\phi)+a(1-cos\phi) thus giving 2a which is a constant in dependent of \phi that I required.
    Oh that is quite a nice way of doing it. However, not sure how using polar coordinates would work to find M. Does what I said previously make sense?

    Edit: The post above explains how to do it
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    (Original post by AMarques)
    Oh that is quite a nice way of doing it. However, not sure how using polar coordinates would work to find M. Does what I said previously make sense?

    Edit: The post above explains how to do it
    Yes it does make sense, thank you.

    I've only started polar stuff yesterday so I do prefer turning everything into Cartesian as of now so I'll keep this method handy, along with my geometric observation as shown. Is there just a method to notice the intersections for A and B straight away or would I necessarily have to do some substitutions before deriving them?
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    (Original post by RDKGames)
    Yes it does make sense, thank you.

    I've only started polar stuff yesterday so I do prefer turning everything into Cartesian as of now so I'll keep this method handy, along with my geometric observation as shown. Is there just a method to notice the intersections for A and B straight away or would I necessarily have to do some substitutions before deriving them?
    Using polar equations substitutions is normally the easiest method as demonstrated above, however there is no harm in doing Cartesian. When you say "notice the intersections" what do you mean? I normally sketch the given curves and it becomes apparent where they occur. In this case you could just substitute  \theta = \phi and  \theta = \phi - \pi , but in more common polar coordinates questions you'll get, you will normally get equations of the form  r = f(\theta) (of course they can make it more complicated). In this case, you just substitute for  r and find the corresponding  \theta and so on...

    Glad I was helpful, let me know how the polar coordinates stuff goes.
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    (Original post by AMarques;[url="tel:66724118")
    66724118[/url]]Using polar equations substitutions is normally the easiest method as demonstrated above, however there is no harm in doing Cartesian. When you say "notice the intersections" what do you mean? I normally sketch the given curves and it becomes apparent where they occur. In this case you could just substitute  \theta = \phi and  \theta = \phi - \pi , but in more common polar coordinates questions you'll get, you will normally get equations of the form  r = f(\theta) (of course they can make it more complicated). In this case, you just substitute for  r and find the corresponding  \theta and so on...

    Glad I was helpful, let me know how the polar coordinates stuff goes.
    Ah yeah, thanks. I simply meant whether there is a fast way to get the points of interaction just by inspection. I'll post more polar stuff here which I may not understand so I can give you a tag if you're able to help further on this topic
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    (Original post by RDKGames)
    Ah yeah, thanks. I simply meant whether there is a fast way to get the points of interaction just by inspection. I'll post more polar stuff here which I may not understand so I can give you a tag if you're able to help further on this topic
    Anytime!
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    Towards the end when it says "assuming that I has been found, the differential equation becomes..."

    Where do they pluck that from? If I is as they define it there as e to the integral of P(x), where do they make that substitution? I am failing to understand how they get that line

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    (Original post by RDKGames)
    Towards the end when it says "assuming that I has been found, the differential equation becomes..."

    Where do they pluck that from? If I is as they define it there as e to the integral of P(x), where do they make that substitution? I am failing to understand how they get that line

    Name:  ImageUploadedByStudent Room1470220492.982574.jpg
Views: 52
Size:  141.2 KB


    Posted from TSR Mobile
    The second and third line of working out on the sheet should make you notice how they substituted it. As they have found I to be defined as  e^{\int P.dx} , then the result in line 3 and 4 of their working holds, so we can just substitute  I\dfrac{dy}{dx} + IPy for  \dfrac{d}{dx}(Iy) in line 2 of their working.
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    (Original post by RDKGames)
    Towards the end when it says "assuming that I has been found, the differential equation becomes..."

    Where do they pluck that from? If I is as they define it there as e to the integral of P(x), where do they make that substitution? I am failing to understand how they get that line



    Posted from TSR Mobile
    No, you have \frac{dy}{dx} + p(x) y  =q(x) multiply this equation through by I(x) to get I(x) \frac{dy}{dx} + I(x)p(x) y = I(x)q(x). You would really like this to be an 'exact' differential equation, i.e: of the form \frac{d}{dx}(R(x)y) = I(x)q(x) because then you can integrate both sides and be done.

    Indeed, we want to choose an I such that \frac{d}{dx}(I(x)y) is the LHS of our differential equation. Differentiate that using the product rule to get I(x) \frac{dy}{dx} + I'(x)y.

    Once we have that, we're done since we get \frac{d}{dx}(Iy) = I(x)q(x) \Rightarrow Iy = \int I(x)q(x) \, \mathrm{d}x and then divide both sides by I(x).

    Notice how this almost looks like our original differential equation multiplied through by I(x)? Infact, the only thing that's different is that the coefficient of our y term is I'(x) here and I(x) p(x) there. Since we want to choose I(x) such that \frac{d}{dx}(Iy) = I(x) \frac{dy}{dx} + I(x)p(x)y and hence allowing us to solve our differential equation, we need to make those two coefficients equal.

    i.e: \displaystyle I(x) = I'(x) p(x) \Rightarrow I(x) = \exp \left(\int p(x) \, \mathrm{d}x\right).

    That is the motivating thing to do here is to take a DE, multiply it through by a certain function that turns it into an exact differential equation; to find this function, it must satisfy the above.
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    (Original post by AMarques)
    ...
    (Original post by Zacken)
    ....

    Ah thank you very much for the explanations. After spending much time and getting confused due to all the P's, Q's and I's, I managed to understand this finally. Had to prove to myself that IP=I' which made the substitution clear as you've mentioned.
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    (Original post by RDKGames)
    Ah thank you very much for the explanations. After spending much time and getting confused due to all the P's, Q's and I's, I managed to understand this finally. Had to prove to myself that IP=I' which made the substitution clear as you've mentioned.
    You're not proving that though, you're asserting it.
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    The answer for 2.b) is 196 but i got 256

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    (Original post by kiiten)
    The answer for 2.b) is 196 but i got 256

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    256 is correct.
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    (Original post by B_9710)
    256 is correct.
    Thanks - my teacher must have got it wrong then
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    (Original post by kiiten)
    The answer for 2.b) is 196 but i got 256

    Posted from TSR Mobile
    It is indeed 256. Also it's the area under the curve from -5 to 2, and above it from 2 to 3 (reference to the sketch).
 
 
 
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