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Math Help - slope of a tangent to a curve

  1. #1
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    slope of a tangent to a curve

    I could use some help on this one. Calculate the slope of a line tangent to the curve of each of the functions y = f(x) for the given point P.

    Y = x^2 ; P is (2,4)
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by OzzMan View Post
    I could use some help on this one. Calculate the slope of a line tangent to the curve of each of the functions y = f(x) for the given point P.

    Y = x^2 ; P is (2,4)
    First find the derivative f'(x)=2x...then you want to find the slope of the tangent line at the point (2,4) and the derivative evaluated at a point x has the same slope as the tangent line at that point so the slope of the tangent is f'(2)=2\cdot{2}=4
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  3. #3
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    Uh I'm new to calculus so if you could kind of walk me through it a little bit more that be great.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    ok

    Quote Originally Posted by OzzMan View Post
    Uh I'm new to calculus so if you could kind of walk me through it a little bit more that be great.
    If you haven't learned differntiation rules then we will go with the quotient difference...ok the slope of a curve is given by f'(x)=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}..so to get the devative of f(x)=x^2 before you know differentiation rules is to go through this thus the derivative of x^2 is \lim_{h \to 0}\frac{(x+h)^2-x^2}{h}=\lim_{h \to 0}\frac{x^2+2xh+h^2-x^2}{h}=\lim_{h \to 0}\frac{2xh+h^2}{h} =\lim_{h \to 0}\frac{2xh}{h}+\lim_{h \to 0}\frac{h^2}{h}=2x+\lim_{h \to 0}h=2x...so the slope at the point x is 2\cdot{x}...so the slope of the curve of x^2 at x=2 is 2\cdot{2}=4...so the slope of the tangent line at the point x=2 has the same slope as the curve so its slope is 4
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  5. #5
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    wow lol. and im only in chapter 2. i got 7 more chapters to go tonight and tomorrow.
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    here let me make it a little easier

    Quote Originally Posted by OzzMan View Post
    wow lol. and im only in chapter 2. i got 7 more chapters to go tonight and tomorrow.
    all that work just to get x^2...but I'll tell you a little trick if f(x)=x^{n}...then the derivative of f(x) or f'(x) is f'(x)=nx^{n-1}...so to get the derivative of x...just multiply the coefficient by the exponent and reduce the exponent by one..so it would be f'(x)=2*x^{2-1}=2x^1=2x
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  7. #7
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    I'm a little confused though where you got those equations from. In my book they seem to be a little different. Damn I'm lost.
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    ?

    Quote Originally Posted by OzzMan View Post
    I'm a little confused though where you got those equations from. In my book they seem to be a little different. Damn I'm lost.

    WHich equations?
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  9. #9
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    The 4th post in this thread
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    If you mean

    Quote Originally Posted by OzzMan View Post
    The 4th post in this thread
    the limit stuff...hmm...thats something you just have to learn...here I think this will really help you its a great site..good luck! if you have any more questions just ask


    Visual Calculus
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  11. #11
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    Well what I mean is that it just appeared a little differently in my book. It's all good though.
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  12. #12
    MHF Contributor Mathstud28's Avatar
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    Ok then

    Quote Originally Posted by OzzMan View Post
    Well what I mean is that it just appeared a little differently in my book. It's all good though.
    Well good luck!
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