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Homework Help: All my Calculus Troubles

  1. Jun 20, 2005 #1
    Well, seeing as I will be working on calculus for a long time, and I figure at the rate I've been going I'll get stuck a lot and need help, I'll just make one thread for all my calculus troubles.

    I just read the chapter on Tangent Lines and Slopes. I did a problem, double-checked the math, then turned to the back of the book and found out the answer was completely incorrect. I retraced again, and found no problem, but the book is a bit confusing so I might have missed something.

    The problem reads: Find a formula that gives the slope at any point P (x,y) on the given curve. [itex]y=4-x^2[/itex] [itex]P:(1,-1)[/itex]

    My work is as follows:

    1)I begin by making a secant line from point P to a point Q, which I arbitrarily place. The coordinates are [itex]Q:(-1+\Delta x, 4-(-1+\Delta x)^2[/itex]

    2)To find my slope: [tex]\frac{\Delta y}{\Delta x} = \frac{4-(-1+\Delta x)^2}{\Delta x}[/tex]

    3)This becomes: [tex]\frac{4-1+2\Delta x-\Delta x^2}{\Delta x}[/tex]

    4)Finally, I get [itex]5-\Delta x=m[/itex] Getting rid of the [itex]\Delta x[/itex] my slope is [itex]m=5[/itex]

    The back of the book says the answer is -2x.

    What happened?
  2. jcsd
  3. Jun 20, 2005 #2
    I'm new here, but I might be able to help.

    Since you read the section on tangent lines, did you read about:

    [tex]f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}[/tex]?

    Try using this formula on your original equation [tex]f(x)=4-x^2[/tex] and that will give you the tangent line at any point.
  4. Jun 20, 2005 #3
    The section on limits and derivatives comes right after this one.
  5. Jun 20, 2005 #4


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    Homework Helper

    Hmmm, I don't understand what you are doing here...
    P(1, -1) is not on the curve.
    Let [itex]P(\varepsilon, 4 - \varepsilon ^ 2)[/itex]be a point on a curve.
    Let [itex]Q(\varepsilon + \Delta x, 4 - (\varepsilon + \Delta x)^ 2)[/itex]be another point on a curve.
    You have as [itex]\Delta x \rightarrow 0 => Q \rightarrow P[/itex]
    Can you find [tex]\frac{Q_y - P_y}{Q_x - P_x}[/tex] as [tex]\Delta x \rightarrow 0[/tex]?
    Er, I think limits should be learn before tangent line.
    Viet Dao,
    Last edited: Jun 20, 2005
  6. Jun 20, 2005 #5
    Should I skip past this chapter for now, learn limits, and then return to it?

    You mentioned P(1,-1) is not on the curve — and you're right. It's (-1,3). The math I did uses (-1,3) though, I just mistyped it here.

    Can you make that itexed slope formulation tex-ed? It's too small, can't read it.

    Should I substitute 0 for [tex]\Delta x[/tex]?
    Last edited: Jun 20, 2005
  7. Jun 20, 2005 #6
    i dont know how far along you are, but all you need to do is find the derivative of the given function:

    [tex]y=4-x^2[/tex] the derivative of this is -2x.

    the derivative of a function is the slope of the tangent line at a certain point.
    for the point (1,1) the equation of the tangent line is :

    [tex]y-1=-2(x-1)[/tex] :surprised
  8. Jun 20, 2005 #7


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    Homework Helper

    I think the OP is just beginning with calculus and is not allowed to "just differentiate".

    So, going back to first principles, the slope of a tangent to a curve at a point is defined by [tex]\frac{dy}{dx} = \lim_{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}[/tex]

    We have [tex]y = 4 - x^2[/tex]. ---eqn(1)


    [tex]y + \Delta y = 4 - (x + \Delta x)^2 = 4 - x^2 - 2x\Delta x - (\Delta x)^2[/tex] ---eqn(2)

    using the binomial expansion.

    Take eqn(2) - eqn(1),

    we have [tex]\Delta y = -2x\Delta x - (\Delta x)^2[/tex]

    Divide that by [tex]\Delta x[/tex],

    [tex]\frac{\Delta y}{\Delta x} = -2x - \Delta x[/tex]

    And taking the limit as [tex]\Delta x[/tex] goes to zero, we have :

    [tex]\frac{dy}{dx} = -2x[/tex]

    as we would expect. This is the equation that gives the slope of the tangent to the curve at any point (x,y)

    If you want the equation of a line that has that slope and that passes through the point (1, -1), the equation of that line is simply,

    [tex]y - (-1) = -2X(x - 1)[/tex]

    or [tex]y = -(2X)x + 2X - 1[/tex]

    Note that that line will NOT actually be a tangent to any point on the curve, it will only be parallel to the tangent to the curve at the chosen point [itex](X,Y)[/itex].

    If you want the equation of the actual tangent line to the curve at a point [itex](X,Y)[/itex], then it would be given by

    [tex]y - Y = -2X(x - X)[/tex]

    [tex]y - (4 - X^2) = -2X(x - X)[/tex]

    which rearranges to

    [tex]y = -(2X)x + X^2 + 4[/tex]
    Last edited: Jun 21, 2005
  9. Jun 20, 2005 #8
    It seems as if you are all suggesting that I use derivatives and/or limits. I believe I will skip this chapter, learn how to do limits and derivatives, and then return to it at a later time. Wish me luck with derivatives and limits. I will need it.
  10. Jun 21, 2005 #9
    choose an arbitrary x
    then the point P = (x, 4 - x^2) is on the curve
    now move to another x a short distance Δx away: x + Δx
    then the point Q = (x + Δx, 4 - (x + Δx)^2 ) is also on the curve
    the slope of the line connecting P and Q is:

    [tex] \mbox{slope \overline{PQ} } \ = \ \frac {\mbox{ (y value of Q) } \ - \ \mbox{ (y value of P) } } {\mbox{ (x value of Q) } \ - \ \mbox{ (x value of P) } } \ = \ \displaystyle \frac { ( 4 - (x + \Delta x)^2) \ - \ ( 4 - x^2 ) } { (x + \Delta x) \ - \ (x)} \ = \ \displaystyle \frac { - 2x\Delta x \ - \ (\Delta x)^2 } { \Delta x} \ = \ -2x \, - \, \Delta x [/tex]

    now what happens to Δx when point P approaches point Q?
    if you answered that Δx → 0, then what is the slope of the tangent to the curve?
    Last edited: Jun 21, 2005
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