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FP1 Differentiation - Parabolas watch

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    I can't seem to figure out what I've done wrong in this question:

    Find the equation of the tangent to the curve:
    y^2 = 7x
    At the point (7, -7)

    Here's what I did:

    y = \sqrt {7} x^\frac {1}{2}

    \frac {dy}{dx} = \frac {\sqrt {7}}{2}\times \frac {1}{\sqrt{x}}

    \frac {dy}{dx} = \frac {\sqrt{7}}{2\sqrt {x}}

    When x = 7:

    \frac {dy}{dx} = \frac {\sqrt{7}}{2\sqrt {7}}

    \frac {dy}{dx} = \frac {1}{2}

    Using y - y_1 = m(x - x_1):

    y + 7 = \frac{1}{2}(x - 7)

    y + 7 = \frac{1}{2}x - \frac{7}{2}

    2y + 14 = x - 7

    x - 2y - 21 = 0

    But the answer is:

    x + 2y + 7 = 0
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    Do u know how to do implicit differentiation ?
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    I got dy/dx to be sqrt((7/2)x)^-(1/2)
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    (Original post by lyamlim97)
    Do u know how to do implicit differentiation ?
    Haven't done C4 yet. 😡
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    When u square root both sides. On the right side, its +-√7x
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    (Original post by The_Big_E)
    x
    You'd have been much better off using implicit differentiation but your error is square rooting both sides and taking the positive square root. As you can see, x=7 \Rightarrow y^2 = 49 \Rightarrow y = \pm \sqrt{49} but more so, you need to take the negative square root because you can see that you need y=-7 from the point given hence

    \displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{\sqrt{7}}{2\sqrt{x}}
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    (Original post by Zacken)
    You'd have been much better off using implicit differentiation but your error is square rooting both sides and taking the positive square root. As you can see, x=7 \Rightarrow y^2 = 49 \Rightarrow y = \pm \sqrt{49} but more so, you need to take the negative square root because you can see that you need y=-7 from the point given hence

    \displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{\sqrt{7}}{2\sqrt{x}}
    I'll try learn implicit differentiation, but how do you know when to use the negative or positive gradient function with this method?
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    (Original post by The_Big_E)
    I'll try learn implicit differentiation, but how do you know when to use the negative or positive gradient function with this method?
    Like I said, because you can see that from the point they've given you, x=7 means y=-7, note the minus. Hence you take the negative square root.
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    I'd go with implicit differentiation - even if you haven't learnt C4, often it's a much simpler way to tackle the coordinate geometry you meet in FP1. Look through the C4 book if you've got it or take a look at this video: http://www.examsolutions.net/maths-r...tutorial-1.php
    It's pretty simple.
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    This is a common problem : students doing FP1 before C4 and tackling problems that are best done using implicit differentiation,

    I would always recommed a sketch which should show whether to use the positive or negative square root.
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    (Original post by The_Big_E)
    I'll try learn implicit differentiation, but how do you know when to use the negative or positive gradient function with this method?
    Your first reaction when seeing something like

    y^2 = 7x

    should be to try sketching the corresponding curve and asking yourself "is it possible to write y = f(x)?" where f(x) is a function. In this case the answer is obviously "no" because for any point on the curve (except the vertex) there are 2 y-values for every x-value.

    The 'top half' of the curve can be described by the function

    y = \sqrt{7x}

    and the 'bottom half' can be described by the function

    y = -\sqrt{7x}

    In your case, you should have taken the 2nd option because the point you are given (7, -7) lies on that curve, but not on the 1st one!

    Implicit differentiation will get you to the answer very quickly. Alternatively, see my note to notnek below

    (Original post by notnek)
    This is a common problem : students doing FP1 before C4 and tackling problems that are best done using implicit differentiation,

    I would always recommed a sketch which should show whether to use the positive or negative square root.
    If you have a choice of techniques then implicit differentiation is certainly the neatest IMHO.

    But you can also do this question by pretending it's a C1 'tangent to a circle' question - let y = mx + c be the equation of the tangent. Because P(7, -7) lies on the tangent and curve we know that -7 = 7m + c or c = -7(m+1).

    Putting y = mx + c into y^2 = 7x gives us the quadratic

    m^2x^2 + (2cm - 7)x + c^2 = 0

    and since we require equal roots we need the discriminant to be 0 i.e.

    (2cm - 7)^2 = 4m^2c^2

    Expanding the LHS we can cancel the m^2c^2 term and set c = -7(m + 1) leaving us with the nice quadratic 4m^2 + 4m + 1 = 0 which leads to m = -1/2 and hence c = -7/2.

    Tidying up the fractions gives us 2y + x + 7 = 0 as required
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    it's easier to find dx/dy in this case. then dy/dx you just turn it upside down ?
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    (Original post by the bear)
    it's easier to find dx/dy in this case. then dy/dx you just turn it upside down ?
    That is actually quite a bit neater and nicer than the other methods proposed here (mine included).
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    (Original post by the bear)
    it's easier to find dx/dy in this case. then dy/dx you just turn it upside down ?
    I believe finding dx/dy is on the C3 spec, so chances are the OP hasn't done it. I do agree that it's the easiest method to pick up on, though - takes about a minute to learn
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    (Original post by aymanzayedmannan)
    I believe finding dx/dy is on the C3 spec, so chances are the OP hasn't done it. I do agree that it's the easiest method to pick up on, though - takes about a minute to learn
    Hi,
    I'll soon be studying FP1.
    Is there any other topics in FP1, that C3/C4 concepts can assist in?
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    (Original post by MathsAndCompSci)
    Hi,
    I'll soon be studying FP1.
    Is there any other topics in FP1, that C3/C4 concepts can assist in?
    Nope, I don't think so. It's only implicit differentiation. C3 iteration techniques (numerical methods in the book) are related to numerical methods in FP1 but the questions aren't similar.
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    (Original post by aymanzayedmannan)
    Nope, I don't think so. It's only implicit differentiation. C3 iteration techniques (numerical methods in the book) are related to numerical methods in FP1 but the questions aren't similar.
    I've got no idea what the spec is like, but trapezium rule thingy for integral estimates?
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    (Original post by Zacken)
    I've got no idea what the spec is like, but trapezium rule thingy for integral estimates?
    Trapezium rule appears only in C2 and C4 (and oddly has shown up in M3 as well...) usually. Integration isn't tested in FP1 but can be to a C2 level, which could include the trapezium rule. Very very unlikely though
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    (Original post by aymanzayedmannan)
    Trapezium rule appears only in C2 and C4 (and oddly has shown up in M3 as well...) usually. Integration isn't tested in FP1 but can be to a C2 level, which could include the trapezium rule. Very very unlikely though
    Ah, I must have seen it in a C4 paper then, thought I saw it pop up in my FP1 mock yesterday, must have been mistaken.
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    Okay thank you, learned implicit differentiation and it really made things easier. Thanks everyone
 
 
 
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