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    (Original post by nuodai)
    Which equation? The original one, or the differentiated one? Either way it seems like a fairly ridiculous equation to have to solve. You'd have to move the \dfrac{-5}{a^6} to the other side, square both sides and multiply through by a^{12} to start with... and that would be near impossible to solve without iteration or Wolfram|Alpha.
    How about I scan in the question I'm doing?
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    (Original post by Hippysnake)
    How about I scan in the question I'm doing?
    Good idea
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    Can you give the full question?
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    (Original post by nuodai)
    Good idea
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    (Original post by Hippysnake)
    Right. Let's say that they walk x km down the side of the field and y km across it, and T is the amount of hours it takes them, where T = t_x + t_y. Then say that \dfrac{x}{t_x} = 5 \Rightarrow t_x = \dfrac{x}{5} and \dfrac{y}{t_y} = 3 \Rightarrow t_y = \dfrac{y}{3}. We can also work out by drawing a diagram or whatever that y = \sqrt{0.5^2 + (0.5 - x)^2} = \sqrt{x^2 + x - 0.5} (so far so good), so T = \dfrac{x}{5} + \dfrac{\sqrt{x^2 + x - 0.5}}{3}; what you need to do is find \dfrac{\mbox{d}T}{\mbox{d}x} (because you're minimising T), set it equal to zero and solve for x. This is effectively what you said, but you must have gone wrong somewhere or other because I have no idea where x^6 (= a^6) came from.

    When you differentiate this, you get \dfrac{\mbox{d}T}{\mbox{d}x} = \dfrac{1}{5} + \dfrac{2x + 1}{6\sqrt{x^2 + x - 0.5}}, which is much easier to solve.
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    (Original post by nuodai)
    Right. Let's say that they walk x km down the side of the field and y km across it, and T is the amount of hours it takes them, where T = t_x + t_y. Then say that \dfrac{x}{t_x} = 5 \Rightarrow t_x = \dfrac{5}{x} and \dfrac{y}{t_y} = 3 \Rightarrow t_y = \dfrac{3}{y}. We can also work out by drawing a diagram or whatever that y = \sqrt{0.5^2 + (0.5 - x)^2} = \sqrt{x^2 + x - 0.5} (so far so good), so T = \dfrac{5}{x} + \dfrac{3}{\sqrt{x^2 + x - 0.5}}; what you need to do is find \dfrac{\mbox{d}T}{\mbox{d}x} (because you're minimising T), set it equal to zero and solve for x. This is effectively what you said, but you must have gone wrong somewhere or other because one of your fractions was upside down and I have no idea where x^6 (= a^6) came from.
    Let's call the first part of the journey 'A' before we cross. Would that not be
    a/5 because we walk at 5kmh-1?

    And the second part would be
    ((a^2-a+0.5)^1/2)All over 3 because we're walking at 3kmh-1?

    So I need to add A to B, differentiate it, make it equal to zero and find the roots? I've tried that, but no luck.
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    (Original post by Hippysnake)
    Let's call the first part of the journey 'A' before we cross. Would that not be
    a/5 because we walk at 5kmh-1?

    And the second part would be
    ((a^2-a+0.5)^1/2)All over 3 because we're walking at 3kmh-1?

    So I need to add A to B, differentiate it, make it equal to zero and find the roots? I've tried that, but no luck.
    Sorry I made an error first time round; I've edited it so that it's correct now.

    You should be solving \dfrac{1}{5} + \dfrac{2x + 1}{6\sqrt{x^2 + x - 0.5}} = 0. Multiply through by \sqrt{x^2 + x - 0.5}, move the \dfrac{1}{5} to the other side, square both sides and then it's fairly straightforward.
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    (Original post by Hippysnake)

    I have created an Excel Document which gives an answer of 0.14 metres along the side of the field. I realise this is not the method expected but it may be of interest.
    Attached Files
  1. File Type: xls Distance along side.xls (22.0 KB, 83 views)
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    (Original post by nuodai)
    Sorry I made an error first time round; I've edited it so that it's correct now.

    You should be solving \dfrac{1}{5} + \dfrac{2x + 1}{6\sqrt{x^2 + x - 0.5}} = 0. Multiply through by \sqrt{x^2 + x - 0.5}, move the \dfrac{1}{5} to the other side, square both sides and then it's fairly straightforward.
    Thanks, it's finally starting to come together now. I'm just not sure how I'd multiply through....is that allowed?
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    (Original post by Hippysnake)
    Thanks, it's finally starting to come together now. I'm just not sure how I'd multiply through....is that allowed?
    It's fairly basic arithmetic. I'd also multiply through by 5\times 6 to make it easier

    \newline

\begin{matrix}

\dfrac{1}{5} + \dfrac{2x + 1}{6\sqrt{x^2 + x - 0.5}} = 0 & \times \sqrt{x^2 + x - 0.5} \\

\Rightarrow \dfrac{1}{5}\sqrt{x^2 + x - 0.5} + \dfrac{2x + 1}{6} = 0 & \times (5\times 6) \\

\Rightarrow 6\sqrt{x^2 + x - 0.5} + 5(2x + 1) = 0 & \mbox{rearrange} \\

\Rightarrow 6\sqrt{x^2 + x - 0.5} = -5(2x + 1) & \mbox{square} \\

\Rightarrow 36(x^2 + x - 0.5) = 25(2x+1)^2 & \mbox{expand + solve} \\

\end{matrix}

    ...and so on.
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    (Original post by nuodai)
    It's fairly basic arithmetic. I'd also multiply through by 5\times 6 to make it easier

    \newline

\begin{matrix}

\dfrac{1}{5} + \dfrac{2x + 1}{6\sqrt{x^2 + x - 0.5}} = 0 & \times \sqrt{x^2 + x - 0.5} \\

\Rightarrow \dfrac{1}{5}\sqrt{x^2 + x - 0.5} + \dfrac{2x + 1}{6} = 0 & \times (5\times 6) \\

\Rightarrow 6\sqrt{x^2 + x - 0.5} + 5(2x + 1) = 0 & \mbox{rearrange} \\

\Rightarrow 6\sqrt{x^2 + x - 0.5} = -5(2x + 1) & \mbox{square} \\

\Rightarrow 36(x^2 + x - 0.5) = 25(2x+1)^2 & \mbox{expand + solve} \\

\end{matrix}

    ...and so on.
    Cor, I've now got 64A^2-64A-7 = 0.

    ....nowhere near right?

    Soz, but is it not

    x^2 -x +0.5?
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    Pretty sure you're both wrong.

    I make it 1/8 km along edge then cross the field.
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    (Original post by rnd)
    Pretty sure you're both wrong.

    I make it 1/8 km along edge then cross the field.
    Indeed, I just noticed that 64x^2 - 64x + 7 = (8x-1)(8x-7), which would give that answer, so I must have made a sign error somewhere.
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    (Original post by rnd)
    Pretty sure you're both wrong.

    I make it 1/8 km along edge then cross the field.
    which would be right..:eek3:

    how do you do it?
    EDIT:Got it, thanks guys, will rep.
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    It's simpler if you have the edge bit as 0.5-x. then the other distance is just (x^2+0.25)^0.5
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    (Original post by rnd)
    Pretty sure you're both wrong.

    I make it 1/8 km along edge then cross the field.
    I agree with 1/8 along the edge of the field.

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    Surely \frac{d}{dx}f(a) = f'(a)\frac{da}{dx}. Do you all instead mean \frac{d}{da}f(a)??
 
 
 
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