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Normals and tangent question

  1. Oct 26, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the equation of the tangent and normals of [tex] y=x+\frac{1}{\sqrt{x}}[/tex] at x = 4.

    2. Relevant equations
    [tex]
    \lim_{h \rightarrow 0}\frac{f(x_{0} + h) - f(x_{0})}{h} = m
    [/tex]

    Also:
    [tex]
    slope of normal = \frac{-1}{slope of the tangent}[/tex]


    3. The attempt at a solution

    [tex]m=\lim_{h \rightarrow 0}\frac{((4+h)+1^{\frac{1}{4}+h})-4+1^\frac{1}{4}}{h}
    [/tex]

    I'm not sure how to sub in for a function that has more than 1 x terms I guess, and I don't think I did it right. I understand that the tangent is the slope at a given point expressed as a limit where h approaches 0. Generally, I think I'd sub in the desired x and remove the h from the bottom so we're not dividing by 0. Also, since this function has a square root, I might need to multiply by the conjugate (not sure about this, I vaguely remember this part about limits--still studying).

    The correct answer according to the book is:

    tangent: 15x -16y = -12
    normal: 32x + 30y = 263

    Not sure how to get these solutions.
     
    Last edited: Oct 26, 2009
  2. jcsd
  3. Oct 26, 2009 #2
    Do you have to do it from first principles? Don't you have some rules for differentiation (which is the same as finding a slope)?

    If you don't invent some! If df/dx = c and dg/dx = d what is d(f+g)/dx?
    What is the derivative of x? x^{-1/2}? (x^n for any fixed number n?).
    This should be enough to solve your problem easily.
     
  4. Oct 26, 2009 #3
    If I use rules of differentiation, I think I can do this:

    [tex] y = x + \frac{1}{\sqrt{x}}
    [/tex]

    [tex]
    = x + (x^{1/2})^{-1}

    [/tex]

    Then I can do this:

    [tex]
    f\prime= 1 + \frac{1}{2}x^{-\frac{1}{2}}\times-1(x^{1/2})^{-2}

    [/tex]

    I applied chain rule and product rule to get that. Probably still wrong, but either way, I think I need to be able to do it by expressing it as a limit to get full points on a test, so yeah I need first principles. Also, I don't get how the equation becomes some nice equation that's up there (15x -16y = -12)
     
    Last edited: Oct 26, 2009
  5. Oct 26, 2009 #4

    lanedance

    User Avatar
    Homework Helper

    you should be able to split the limits pretty easy, in general say f(x) = g(x) + h(x)

    you should be able to split the limits reasonably easy, if the functions are both dfferntiable
    [tex] f'(x) = lim_{d \rightarrow 0}( \frac{f(x+d)-f(x)}{(x+d)-x})
    = lim_{d \rightarrow 0}( \frac{g(x+d)+h(x+d)-g(x)-h(x)}{(x+d)-x}[/tex]
    [tex]
    = lim_{d \rightarrow 0}( \frac{g(x+d)- g(x)}{(x+d)-x}) + lim_{d \rightarrow 0}( \frac{h(x+d)-h(x)}{(x+d)-x})
    [/tex]
    [tex]
    = g'(x) + h'(x)
    [/tex]
    the are some subtleties involved, but i think it probably fine to assume above

    now your differentiation is not quite correct, you only really need to use the power rule here, knowing from exponential rules that (x^a)^b = x^(ab)

    [tex] \frac{d}{dx}\frac{1}{\sqrt{x}} = \frac{d}{dx} x^{-1/2} = (-1/2)x^{-3/2}
    [/tex]

    if you want to do it form first principles, its a bit more involved, look at each limit separately as indicated
    for the sqrt you have:
    [tex]lim_{d \rightarrow 0}( \frac{ \frac{1}{\sqrt{x+d}} - \frac{1}{\sqrt{x}} }{(x+d)-x})
    [/tex]

    now multilpy through to get a common denominator, then rationalise and you should be on your way...
     
    Last edited: Oct 26, 2009
  6. Oct 26, 2009 #5
    Okay, if that's the derivative, then how do we get

    tangent: 15x -16y = -12
    normal: 32x + 30y = 263

    as a final answer?
     
  7. Oct 26, 2009 #6

    Mark44

    Staff: Mentor

    You're really doing the hard way here. You don't need the chain rule. Rewrite your function as y = x + x-1/2

    Now take the derivative using the power rule.
    First off, there is no mention of f, so it should be there. Use y' or dy/dx instead. Second, the expression to the right of the 1 term is extremely convoluted. I'm pretty sure it's incorrect, but I haven't checked it, and I'm not very inclined to do that for what should be a very simple expression.
     
  8. Oct 26, 2009 #7

    lanedance

    User Avatar
    Homework Helper

    once you get the derivative at x0 then the tangent is the line with slope
    [tex] m = f'(x_0) [/tex]

    through the point
    [tex] (x_0,f(x_0)) [/tex]

    so if we let the equation of the line be given by y(x), it must satisfy
    [tex] y(x) - f(x_0) = m(x - x_0) [/tex]

    the normal will be through the same point with a different gradient, to give it a 90 degree angle with the tangent
     
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