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    I'm stuck on part (b) question 11 in Excercise 8K. The question is as follows...

    " The Curve C has the eqn y = f(x), where

    f(x) = 3lnx + (1/x), x > 0

    (a) Calculate the x cordinate of P. -> I've correctly identified it to be 1/3
    The point point Q on C has the x co-ordinate 1.
    (b) Find and equation for the normal to C at Q"

    ... I'm actually quite stuck on what (b) is asking for.

    Clues on how I could answer this question would be greatly appreciated. Thank you.
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    (Original post by adellie)
    I'm stuck on part (b) question 11 in Excercise 8K. The question is as follows...

    " The Curve C has the eqn y = f(x), where

    f(x) = 3lnx + (1/x), x > 0

    (a) Calculate the x cordinate of P. -> I've correctly identified it to be 1/3
    The point point Q on C has the x co-ordinate 1.
    (b) Find and equation for the normal to C at Q"

    ... I'm actually quite stuck on what (b) is asking for.

    Clues on how I could answer this question would be greatly appreciated. Thank you.
    first differentiate the function

    Spoiler:
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    f(x) = 3lnx + x^-1


    you should be able to do that
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    (Original post by Chaoslord)
    first differentiate the function

    Spoiler:
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    f(x) = 3lnx + x^-1


    you should be able to do that
    ... I'd done that for part (a), I got: 3/x - x^(-2)
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    You need to differentiate f(x) to find gradient of tangent at that point. The normals gradient is the negative reciprical of this.
    Use the equation y=mx+c to do the rest.
    Hope this helps.
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    oops, just relaised, i've differentiated differently to you! argh, but the i differentiated did get the the correct answer for (a)... i'm confused!
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    (Original post by adellie)
    ... I'd done that for part (a), I got: 3/x - x^(-2)
    okay well dy/dx = gradient

    if you sub in an x value into this equation you will get the gradient at x

    now we want the normal, the normal is perpendicular to the tangent. if two lines are perpendicular the multiple of their gradients is -1

    and we can use the formula

    y - y1 = m(x - x1)
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    (Original post by adellie)
    oops, just relaised, i've differentiated differently to you! argh, but the i differentiated did get the the correct answer for (a)... i'm confused!
    no your right i was re writing the formula to make it simpler to differentiate
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    heres a fully worked solution

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    \frac{dy}{dx} = \frac{3}{x} -\frac{1}{x^2}

    subbing in x = 1, we can see the gradient of the tangent is 2 therefore the gradient of the normal is -1/2

    subbing in the 1 into the original equation we get y = 1

    using y - y1 = m(x - x1)

    we get y = -\frac{1}{2}x + \frac{3}{2}

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    (Original post by Chaoslord)
    okay well dy/dx = gradient

    if you sub in an x value into this equation you will get the gradient at x

    now we want the normal, the normal is perpendicular to the tangent. if two lines are perpendicular the multiple of their gradients is -1

    and we can use the formula

    y - y1 = m(x - x1)
    I've subbed 1 into dy/dx, this has given me 2

    would the eqn for the normal be: -(1/3)lnx + 1/x

    ... what would I now do? I recognise this seems like C2 work, but I'm so tired at the momment and I can't seem to think properly or correctly!
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    ^read above
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    thanks for the spolier, I've relaised i've done the normal eqn totally wrong! i get it now! thanks x
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    (Original post by adellie)
    thanks for the spolier, I've relaised i've done the normal eqn totally wrong! i get it now! thanks x
    happy to health
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    ... another question, its reagardin question 15. How could simplify: 1 / (2e(ln)e + e) where e in this example is being used instead of x?
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    (Original post by adellie)
    ... another question, its reagardin question 15. How could simplify: 1 / (2e(ln)e + e) where e in this example is being used instead of x?

    this question is messy

    \frac{1}{2eln(e) + e}

    thats what im assuming the formula is

    first you can right it like this

    (2eln(e) + e)^{-1}

    now the chain rule states, (gof)' = g'of \times f'

    we can take 2eln(e) + e as one function and x^{-1} as the other

    now you may not be able to differentiate 2eln(e)

    you have to use the product rule (g(x) \times f(x))' = g'(x)f(x) + f'(x)g(x) where 2e = g(x) and ln(e) = f(x)

    hope this helps
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    I don't think i need to differentiate that as i already had, I think. If it helps, the answer should be 1/3e. How would I get to that compared to what I already had?
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    (Original post by adellie)
    I don't think i need to differentiate that as i already had, I think. If it helps, the answer should be 1/3e. How would I get to that compared to what I already had?
    oh sorry i miss read the question

    \frac{1}{2eln(e) + e}

    \frac{1}{ln(e^{2e}) + e}

    hmm im stuck are you sure you differentiated correctly? o.O
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    (Original post by Chaoslord)
    oh sorry i miss read the question

    \frac{1}{2eln(e) + e}

    \frac{1}{ln(e^{2e}) + e}

    hmm im stuck are you sure you differentiated correctly? o.O
    This is the whole question:

    Given that x = (y^2) ln y , y > 0,

    (a) find dy/dx

    ... I did:

    let u = y^2 and v = lny
    du/dy = 2y and dv/dy = 1/y
    ,', dx/dy = y + 2ylny

    ... this was correct at the back

    (b) use your answer to part (a) to find in terms of e, the value of dy/dx at y = e

    dy/dx = 1 dx/dy

    => 1 / (2e(lne) + e)

    ... and i'm stuck here! it should come to 1/3e

    Nb. i'm sorry about the messiness of my eqns, i don't know how to use LaTeX
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    (Original post by adellie)
    This is the whole question:

    Given that x = (y^2) ln y , y > 0,

    (a) find dy/dx

    ... I did:

    let u = y^2 and v = lny
    du/dy = 2y and dv/dy = 1/y
    ,', dx/dy = y + 2ylny

    ... this was correct at the back

    (b) use your answer to part (a) to find in terms of e, the value of dy/dx at y = e

    dy/dx = 1 dx/dy

    => 1 / (2e(lne) + e)

    ... and i'm stuck here! it should come to 1/3e

    Nb. i'm sorry about the messiness of my eqns, i don't know how to use LaTeX
    e is a constant...

    \frac{1}{2eln(e) + e}

    i'll give you a clue lne =1 xD

    jesus that was easy xD i was gonna ask you if it was e the constant but i assumed you would have informed me.... xD
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    (Original post by Chaoslord)
    e is a constant...

    \frac{1}{2eln(e) + e}

    i'll give you a clue lne =1 xD

    jesus that was easy xD i was gonna ask you if it was e the constant but i assumed you would have informed me.... xD
    Thank you! I'd forgotten lne = 1!!! ...That does seem very easy now!
 
 
 
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