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    Gernald's Avatar
    Gernald Posts: 901, Reputation: 93
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    #1

    Jan 19, 2009, 01:04 PM
    finding the slope of the tangent
    I came into my math class a few days late and now I'm trying to do the homework and totally hate myself.

    I was able to do the first half of the problem:
    The point P(1,1/2) lies on the curve y=x/1+x
    if q is the point (x,x/1+x) then does the slope of the secent line PQ for the following values of x. And then it has a bunch of x values among them 0.5 0.9 and 0.99... no problem with those though.

    Part B is what has me confused it asks to use the previous results to guess the value of the slope of the tangent line to the curve p(1,1/2)

    and then I need to use the slope to find an equation for the tangent line.

    How do I do part B? I keep getting 1/6, but my book says it's 1/4.

    :eek:
    Capuchin's Avatar
    Capuchin Posts: 5,255, Reputation: 656
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    #2

    Jan 19, 2009, 03:07 PM

    First, you need to be more clear with your notation, x/1+x is not the same as x/(1+x), remember BODMAS (or BIDMAS, or however you were taught it). Try to be clearer in future.

    For the part you are confused with, your teacher is trying to get you to realise something very important, so I'm not going to spell it out for you:

    If you draw a curve (actually do it right now), such as a parabola y = x^2, then take a point on that curve, say P(2,4). Now take different points Q(x,x^2) and work out the slope of the line PQ.

    Start with x=0, then move x closer to 2, so x=1, x=1.8, x=1.99 etc. (draw these points and the line PQ in each case on your graph).

    Also draw th tangent to the curve at point P.

    Can you now see what your teacher is trying to get you to realise with how this relates to the slope of the tangent at point P?

    If you need further help, please ask. :)
    Gernald's Avatar
    Gernald Posts: 901, Reputation: 93
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    #3

    Jan 20, 2009, 02:15 PM

    Thanks, sorry about mistyping it, I did end up with the right graph when I tried it myself, just wrote it wrong here.
    I figured it out after a while. It's so obvious after doing one, it's just getting that one right.
    Thanks for your help!
    Capuchin's Avatar
    Capuchin Posts: 5,255, Reputation: 656
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    #4

    Jan 20, 2009, 04:41 PM

    I'm glad you figured it out. :)

    I'll give the observation here for others to read: As Q approaches P, then the slope of the line PQ approaches the slope of the tangent to the curve at P.

    In the limit that the length of PQ is 0, (P ~= Q), then these slopes are the same.

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