I am starting calculus - question about tangent line of a slope

Discussion in 'Math' started by jaygatsby, Jan 18, 2012.

  1. jaygatsby

    Thread Starter New Member

    Nov 23, 2011
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    I am finding the slope of a curve at a point. I am given the function and the point. For instance, I find the slope of the curve for y=x^2-3. No problem. But when we get into cubic equations, I don't know what to do. Instead of an h+3, for instance, I end up with a quadratic equation with h (h being delta x). Not sure what to do about this (first day of calculus!)

    Thank you
     
  2. MrChips

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    Oct 2, 2009
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    Not sure what you mean by h+3.

    The slope is the 1st derivative, dy/dx. It doesn't matter if the curve is quadratic or cubic.
     
  3. 1chance

    Member

    Nov 26, 2011
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    Since this is the first day of class I think you should either be using the"difference quotient", (f(a+h)-f(a))/h or the limit process where you pick points on the function closer and closer to the point you want computing the slope each time to see what value you are approaching. I teach the limit process first and then the diierence quotient which is the formal definition of a derrivative. Pardon any typos as I am using my phone for this.
     
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  4. MrChips

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    You apply the limit as h tends to zero. Hence h = 0.
     
  5. jaygatsby

    Thread Starter New Member

    Nov 23, 2011
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    Thank you very much. It looks like we are using the difference quotient, then. At the end, when doing quadratic functions, I can factor out an h using the denominator h, and have a nice h+4 or something as the result. But when they are cubic, the result ends up being h^2+2h+4 or something like that...
     
  6. DerStrom8

    Well-Known Member

    Feb 20, 2011
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    I think you're trying to go about it in a much more difficult way than you have to. You know how to take the derivative of the original equation--that is the tangent line, as mentioned above. So all you have to do is take the derivative, and you have the slope equation. Plug in the X value to find the slope number. Then use the line formula (your professor may ask for a specific format, such as y=mx+b or (Y-Y1)=m(X-X1)). Anyway, let's assume it is the latter, which is the most common in calculus. So, since you have the slope, you plug that in for 'm' in the formula. You also have the point, so simply put the X and Y values in for X1 and Y1 and you have your equation.

    For example, let's take a graph of f(x)=X^2, at the point (1,2). Take the derivative, and you get f'(x)=2x. Plug in the x value (1) and you have a slope of 2.

    Now we go to our equation for a line, (Y-Y1)=m(X-X1). Plug in the slope and you get (Y-Y1)=2(X-X1). Then, take the x and y values from the point (1 and 2 respectively) and put them in as well:

    Y-2=2(X-1)

    You can re-write that in another form as Y=2(X-1)+2, or Y=2X. That is the equation of the tangent line at that point.

    Did you follow that at all, or did I get too long-winded? :D:p
     
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  7. MrChips

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    Oct 2, 2009
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    Expand it and write out the something complicated.
    Then divide by h.
    Next set h = 0.
    Why? Because we start out with h a finite value and then approach the limit where h -> 0.
     
  8. MrChips

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    Oct 2, 2009
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    Can you write out the expression with the constants B and C filled in so that I can check your work?

    What happens when you set h = 0 into this expression?
     
  9. DerStrom8

    Well-Known Member

    Feb 20, 2011
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    I believe you can, as long as you plug in the value of h (x, from the point). That will give you the slope number at that particular x value.
     
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  10. Papabravo

    Expert

    Feb 24, 2006
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    If you do it correctly then all the terms involving h will disappear and you are left with an expression that is the derivative of f(x) as a function of x. For example

    f(x) = x^{3}+2x^{2}+x+1
    f(x+h) = (x+h)^{3}+2(x+h)^{2}+(x+h)+1
    f(x+h) - f(x) = 3x^{2}h+3xh^{2}+h^{3}+4xh+2h^{2}+1
    f(x+h) -f(x))/h = 3x^{2}+3xh+h^{2}+4x+2h+1
     h \rightarrow 0 gives \Rightarrow 3x^{2}+4x+1

    Is that clear?
    This method works for any polynomial of any order. It is helpful to recall Pascal's triangle and the Binomial Theorem when doing these expansions.
     
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