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    (Original post by Better)
    Oh I get you so if I was to "rotate" if I can call it that the green line, I can see its longer.

    So in essence its just like a Mathematical truth or something? like Pythagoras?
    Yeah i think it's very much an intuitive thing, although no doubt there is a mathematical proof for it
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    (Original post by Better)
    Quick Question just for understanding sakes.

    I know the fact that the closet distance from lets say Line L to point P, would be the perpendicular distance, but WHY is this?

    Can anyone care to explain or provide a link. I know it's applicable in C4 Vectors, but it just came up in a Physics 5 Question I did and I realized I don't understand the logic behind why it is that way.


    THANK YOU
    I don't have much time right now, but I'll post a simple proof for this later, if you want.
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    (Original post by Better)
    Quick Question just for understanding sakes.

    I know the fact that the closet distance from lets say Line L to point P, would be the perpendicular distance, but WHY is this?

    Can anyone care to explain or provide a link. I know it's applicable in C4 Vectors, but it just came up in a Physics 5 Question I did and I realized I don't understand the logic behind why it is that way.


    THANK YOU
    Here you go:

    Using the Pythagorean theorem, the distance between any point (x,y) on Line L (equation y=mx+c ) and a point P with coordinates (a,b) would be:

    D^2 = (x-a)^2 + (y-b)^2

    Where D is the distance

    Differentiating both sides of the equation with respect to x, we get:

    2D \cdot \dfrac{dD}{dx} = 2(x-a) + 2(y-b) \cdot \dfrac{dy}{dx}

    D \cdot \dfrac{dD}{dx} = (x-a) + (y-b) \cdot \dfrac{dy}{dx}

    We know that at the minimum value of D, dD/dx=0. So we set dD/dx=0;

    0 = (x-a) + (y-b) \cdot \dfrac{dy}{dx}

    (y-b) \cdot \dfrac{dy}{dx} = -(x-a)

    \dfrac{dy}{dx} = -\dfrac{x-a}{y-b} = -\dfrac{1}{\left(\dfrac{y-b}{x-a} \right)}

    Now, we know that \dfrac{dy}{dx} is equal to the gradient m of the line L, and \dfrac{y-b}{x-a} is equal to the gradient m_{1} of the line joining any point on the Line L and point P.

    So,

    m = -\dfrac{1}{m_{1}}

    m \times m_{1} = -1

    As the product of the gradient of two perpendicular lines is -1, we can conclude that the closest distance from a Line L to a point P would be the perpendicular distance.
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    (Original post by justinawe)
    Here you go:

    Using the Pythagorean theorem, the distance between any point (x,y) on Line L (equation y=mx+c ) and a point P with coordinates (a,b) would be:

    D^2 = (x-a)^2 + (y-b)^2

    Where D is the distance

    Differentiating both sides of the equation with respect to x, we get:

    2D \cdot \dfrac{dD}{dx} = 2(x-a) + 2(y-b) \cdot \dfrac{dy}{dx}

    D \cdot \dfrac{dD}{dx} = (x-a) + (y-b) \cdot \dfrac{dy}{dx}

    We know that at the minimum value of D, dD/dx=0. So we set dD/dx=0;

    0 = (x-a) + (y-b) \cdot \dfrac{dy}{dx}

    (y-b) \cdot \dfrac{dy}{dx} = -(x-a)

    \dfrac{dy}{dx} = -\dfrac{x-a}{y-b} = -\dfrac{1}{\left(\dfrac{y-b}{x-a} \right)}

    Now, we know that \dfrac{dy}{dx} is equal to the gradient m of the line L, and \dfrac{y-b}{x-a} is equal to the gradient m_{1} of the line joining any point on the Line L and point P.

    So,

    m = -\dfrac{1}{m_{1}}

    m \times m_{1} = -1

    As the product of the gradient of two perpendicular lines is -1, we can conclude that the closest distance from a Line L to a point P would be the perpendicular distance.
    You are a god among mortals.

    Thank you very much, I will try my best to get my head around it hahaha!
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    (Original post by Better)
    You are a god among mortals.

    Thank you very much, I will try my best to get my head around it hahaha!
    A god among mortals who has to re-sit C4 due to screwing it up in January? Some god! :lol:

    No problem, glad I could help.
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    (Original post by justinawe)
    A god among mortals who has to re-sit C4 due to screwing it up in January? Some god! :lol:

    No problem, glad I could help.
    Haha you'll do great mate especially with your Maths skills! I am impressed
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    (Original post by Better)
    Haha you'll do great mate especially with your Maths skills! I am impressed
    I certainly hope so! I think jan was a one-off, I just got too fixated on a particular question, ended up wasting a whole lot of precious time, and proceeded to panic after that

    I definitely won't make the same mistake again though, and since I've got FP2/FP3 coming up in June as well, hopefully that will make C4 seem easier.
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    (Original post by justinawe)
    I certainly hope so! I think jan was a one-off, I just got too fixated on a particular question, ended up wasting a whole lot of precious time, and proceeded to panic after that

    I definitely won't make the same mistake again though, and since I've got FP2/FP3 coming up in June as well, hopefully that will make C4 seem easier.
    Yeah mate lets do this! I'll do the same, with maths I don't get so nervous, I just rush through the papers too fast, I make so many basic errors as I go through. In Physics I get sooo nervous, and I get fixated on questions; its horrible. But I'm working on that haha

    About to do a D1 Past Paper, wish me luck brah!
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    (Original post by Better)
    Yeah mate lets do this! I'll do the same, with maths I don't get so nervous, I just rush through the papers too fast, I make so many basic errors as I go through. In Physics I get sooo nervous, and I get fixated on questions; its horrible. But I'm working on that haha

    About to do a D1 Past Paper, wish me luck brah!
    Ah, good luck with that I've self-studied D1 for the upcoming exams, and it's so incredibly boring!
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    solve cos2x=1-cosx any ideas??? tried 2cos^2-1 but am stumped
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    (Original post by Gekko123)
    solve cos2x=1-cosx any ideas??? tried 2cos^2-1 but am stumped
    Express cos2x in terms of cos only, you should get a quadratic.

    Posted from TSR Mobile
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    can any one help please

    i have part 1 of the question it was expressing 13-2x/(2x-3)(x+1) in partial fractions and i got
    4/(2X-3)-3/(x+1)

    and then part b) says b) Given that y = 4 at x = 2, use your answer to part (a) to find the solution of the differential equation

    dy/dx = [y(13-2x)]/[(2x-3)(x+1)], x > 1.5

    Express your answer in the form y = f(x)
    what do i do

    i got lny=2ln(2x-3)-3ln(x+1) but i dont know if its right and what do i do with the x an y values?
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    (Original post by Converse girl)
    can any one help please

    i have part 1 of the question it was expressing 13-2x/(2x-3)(x+1) in partial fractions and i got
    4/(2X-3)-3/(x+1)

    and then part b) says b) Given that y = 4 at x = 2, use your answer to part (a) to find the solution of the differential equation

    dy/dx = [y(13-2x)]/[(2x-3)(x+1)], x > 1.5

    Express your answer in the form y = f(x)
    what do i do

    i got lny=2ln(2x-3)-3ln(x+1) but i dont know if its right and what do i do with the x an y values?
    I think that's right, I'm a bit too lazy to check though sorry

    Use your C2 log rules. What does a*ln(b) equal? What does ln(a)-ln(b) equal?
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    (Original post by justinawe)
    A god among mortals who has to re-sit C4 due to screwing it up in January? Some god! :lol:

    No problem, glad I could help.
    Nice proof. What did you get in C4 in Jan?
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    (Original post by justinawe)
    I think that's right, I'm a bit too lazy to check though sorry

    Use your C2 log rules. What does a*ln(b) equal? What does ln(a)-ln(b) equal?
    so is the y=4 at x=2 is just there then? yeah i have that part with the logs

    thanks mr Lazy
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    (Original post by reubenkinara)
    Nice proof. What did you get in C4 in Jan?
    70, by far the worst maths mark I've ever received
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    (Original post by Converse girl)
    so is the y=4 at x=2 is just there then? yeah i have that part with the logs

    thanks mr Lazy
    The y=4 at x=2 part is to find the constant of integration. You haven't forgotten your "+c" now, have you?
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    (Original post by justinawe)
    70, by far the worst maths mark I've ever received
    How did it average with C3?
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    (Original post by reubenkinara)
    How did it average with C3?
    I got 95 for C3. Would have been 100, but I somehow ran out of time at the end and so missed out the final bit of the last question, which was worth 4 raw marks :lol:

    Obviously I'll be retaking C4 for the A*. It's not that C4 is that much harder than C3 though, I just had an extremely bad day personally.
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    (Original post by justinawe)
    The y=4 at x=2 part is to find the constant of integration. You haven't forgotten your "+c" now, have you?
    nooooooo i didnt okay fine i did

    thanks makes all sense now

    i think i need to tattoo +c on myself so i wont forget the it :facepalm:
 
 
 
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