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    (Original post by thefatone)
    .... i don't really get what to do???

    i'm not getting the bit where i do something with the gradient and find the co-ordinates of the point of intersection.
    It's not clear which part you're stuck at. Can you point to the exact part of my last post where you get stuck / something is unclear?

    Are you able to solve the question that I wrote in italics?

    Initially I gave you this other method for you to look at and ignore if you like. But I think it's important for your understanding that you are able to follow this method.
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    (Original post by notnek)
    It's not clear which part you're stuck at. Can you point to the exact part of my last post where you get stuck / something is unclear?

    Are you able to solve the question that I wrote in italics?

    Initially I gave you this other method for you to look at and ignore if you like. But I think it's important for your understanding that you are able to follow this method.
    i got k as -4???? but that can't be right

    yes from the bit in italics i did this

     y=\dfrac{1}{3} \times 9 +8



y=11



11=\dfrac{1}{3} x^2 +8



33=x^2 +8



x^2=25



x=5



11=15+k



k=-4
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    (Original post by thefatone)
    i got k as -4???? but that can't be right

    yes from the bit in italics i did this

     ....



11=\dfrac{1}{3} x^2 +8 \\ 33=x^2 +8

...
    11=\frac{1}{3} x^2 +8 \nRightarrow 33=x^2 +8
    You forgot to multiply 8 by 3.

    Edit: That was the inconsistency in your working but notnek has spotted your actual error below.
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    (Original post by thefatone)
    i got k as -4???? but that can't be right

    yes from the bit in italics i did this

     y=\dfrac{1}{3} \times 9 +8



y=11



11=\dfrac{1}{3} x^2 +8



33=x^2 +8



x^2=25



x=5



11=15+k



k=-4
    You've substituted x=3 into the curve equation. Why? You should be thinking about why you're doing each step.


    The gradient of the curve at P is 3. Do you understand this part? Please tell us and we can explain more if you don't.

    If the gradient of the curve at a point is 3 then this means that \frac{dy}{dx}=3 at the point.

    \displaystyle y=\frac{1}{3}x^2+8 \Rightarrow \frac{dy}{dx} = \frac{2}{3}x

    Again, tell us if you don't understand this.

    \displaystyle \frac{dy}{dx} = 3 \Rightarrow  \frac{2}{3}x = 3

    If you solve this you will have the x-coordinate of P.
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    (Original post by Kvothe the arcane)
    11=\frac{1}{3} x^2 +8 \nRightarrow 33=x^2 +8
    You forgot to multiply 8 by 3.

    Edit: That was the inconsistency in your working but notnek has spotted your actual error below.
    it happened earlier above too >.> i need to be more careful......

    (Original post by notnek)
    You've substituted x=3 into the curve equation. Why? You should be thinking about why you're doing each step.


    The gradient of the curve at P is 3. Do you understand this part? Please tell us and we can explain more if you don't.

    If the gradient of the curve at a point is 3 then this means that \frac{dy}{dx}=3 at the point.

    \displaystyle y=\frac{1}{3}x^2+8 \Rightarrow \frac{dy}{dx} = \frac{2}{3}x

    Again, tell us if you don't understand this.

    \displaystyle \frac{dy}{dx} = 3 \Rightarrow  \frac{2}{3}x = 3

    If you solve this you will have the x-coordinate of P.
    I do understand what that means, at point P on the curve C the gradient is 3

    why do i equate gradients??? what does that do?
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    (Original post by thefatone)
    it happened earlier above too >.> i need to be more careful......



    I do understand what that means, at point P on the curve C the gradient is 3

    why do i equate gradients??? what does that do?
    Can you tell me where I have equated gradients?

    It's easier for me to understand if you say the specific part of my post that confuses you.
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    (Original post by notnek)
    Can you tell me where I have equated gradients?

    It's easier for me to understand if you say the specific part of my post that confuses you.
    where you put the gradient of

     y=3x+k

and



y=\dfrac{1}{3} x^2 +8

    together....
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    (Original post by thefatone)
    where you put the gradient of

     y=3x+k

and



y=\dfrac{1}{3} x^2 +8

    together....
    The gradient of the parabola is constantly changing but as the straight line is a tangent you know that it hits the curve at one point, P. At that point, the gradient of the parabola is equal to the gradient of the straight line. This allows us to work out the coordinate of the point of intersection and by extension, k.
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    (Original post by thefatone)
    where you put the gradient of

     y=3x+k

and



y=\dfrac{1}{3} x^2 +8

    together....
    Please quote the part of my post where this happened.

    I'm finding it hard to pinpoint what you're unsure about.

    I'd appreciate if someone else has a go at explaining this - it's often good to hear two explanations
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    (Original post by Kvothe the arcane)
    The gradient of the parabola is constantly changing but as the straight line is a tangent you know that it hits the curve at one point, P. At that point, the gradient of the parabola is equal to the gradient of the straight line. This allows us to work out the coordinate of the point of intersection and by extension, k.
    i see so you've used gradients instead of equalling the whole bloody equation like me xD

    (Original post by notnek)
    Please quote the part of my post where this happened.

    I'm finding it hard to pinpoint what you're unsure about.

    I'd appreciate if someone else has a go at explaining this - it's often good to hear two explanations
    ok y=3x+k
    \dfrac{\mathrm d y}{\mathrm d x}=3


    y=\dfrac{1}{3} x^2 +8

    \dfrac{\mathrm d y}{\mathrm d x}=\dfrac{2}{3}x
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    (Original post by notnek)
    Please quote the part of my post where this happened.
    Well you said that \dfrac{d}{dx} \left(\dfrac{1}{3}(x^2+8) \right)=\dfrac{d}{dx}(3x+k) \therefore \dfrac{2x}{3}=3 which is what I suppose he's referring to as you equating the gradients.
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    There are two conditions for a line to be tangent to a curve.

    1. It needs to touch the curve at one point.

    2. The line needs to have the same gradient of the curve at the point of intersection or touching.

    So if the line y = 3x + c is tangent to the curve then it must be tangent at some point where the gradient of the curve has the same value as the gradient of the line.

    Now we know that \frac{2x}{3} is the gradient of the curve at a general point 'x'.

    We want to know at what point does the curve have a gradient that is the same as the gradient of the line.

    We know the gradient of the line is 3.

    Hence the point at which the gradient has a gradient of 3 is the point satisfying \frac{2x}{3} = 3 \iff x = \frac{9}{2}.

    This is the x-coordinate of the point of tangency, you want to find the y coordinate as well. This is easy - plug it into the formula that you have to find y for any given x, that is: y = \frac{x^2}{3} + 8 gives you the y coordinate for any x coordinate.

    But this point of tangency has to lie on the line as well...duuuh - this should make sense to you.

    Hence, if you plug this point into y = 3x + c you will get a number = 3*a number + c and you can then solve for c.
 
 
 
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