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Math Help - Rates of Change and Tangent Lines

  1. #1
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    Rates of Change and Tangent Lines

    Find
    A the slope of the curve
    B the equation of the tangent
    C. the equation of the normal
    C the equation of the normal draw graph of the curve, tangent line, and normal line in the same square viewing window.

    y= 1/(x-1) at x=2.

    I have some understand of the slope of a curve. The more complicated formula, because I don't know where to find the simpler one (please help with that) below


    [tex]m= (f(a+h)- f(a))/h

    I understand that normal and tangent have opposite slopes. HEY MY MISSING GAME BOY!!! That's cool. It's been lost for 2 years.
    Anyway, they have reciprocal slopes.

    According the example for the tangent of a curve, plugging in 2 as 'a' and then I have trouble. Something like the whole top row gets put into the 'x' in the equation 1/(x-1). How does that work?


    Thank yoU!
    Par
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Truthbetold View Post
    Find
    A the slope of the curve
    B the equation of the tangent
    C. the equation of the normal
    C the equation of the normal draw graph of the curve, tangent line, and normal line in the same square viewing window.

    y= 1/(x-1) at x=2.

    I have some understand of the slope of a curve. The more complicated formula, because I don't know where to find the simpler one (please help with that) below


    [tex]m= (f(a+h)- f(a))/h

    I understand that normal and tangent have opposite slopes. HEY MY MISSING GAME BOY!!! That's cool. It's been lost for 2 years.
    Anyway, they have reciprocal slopes.

    According the example for the tangent of a curve, plugging in 2 as 'a' and then I have trouble. Something like the whole top row gets put into the 'x' in the equation 1/(x-1). How does that work?


    Thank yoU!
    Par
    do you have to use the limit definition to find the derivative?

    anyway, find the derivative and plug in x = 2, that will give you the slope of the tangent line (note that the slope of the normal line is the negative reciprocal of this slope).

    when done, find the point on the graph where x = 2. (plug in x = 2 into the original equation and solve for y). call this point (x_1,y_1)

    then use the point slope form to find the equation of the tangent line:

    y - y_1 = m(x - x_1)

    where m is the slope.

    do one line for the tangent and another for the normal line
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  3. #3
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    do you have to use the limit definition to find the derivative?
    Yes. My fault.

    Lim as h--> 0 (f(a+h)- f(a))/h

    You lost me there at "find the derivative" How do you do that?
    I don't even know what a derivative is yet. (next chapter)
    I understand after that.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Truthbetold View Post
    Yes. My fault.

    Lim as h--> 0 (f(a+h)- f(a))/h

    You lost me there at "find the derivative" How do you do that?
    I don't even know what a derivative is yet. (next chapter)
    I understand after that.
    well, what you have there is the definition of the derivative. we call it f'(a). so,

    f'(a) = \lim_{h \to 0} \frac {f(a + h) - f(a)}h

    if this limit exists, it gives the derivative of a function f(x) at a point a (in the context that you want, the derivative gives the formula for the slope, so the derivative at a, gives the slope of the tangent line at the point a)

    now continue
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  5. #5
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    well, what you have there is the definition of the derivative. we call it f'(a). so,



    if this limit exists, it gives the derivative of a function f(x) at a point a (in the context that you want, the derivative gives the formula for the slope, so the derivative at a, gives the slope of the tangent line at the point a)

    now continue
    Okay. I understand the definition of a derivative and what you said.
    When plugging (a+h) as x, does it come to be
    (1/(a+h) -1 - 1/(a) -1 )/ h?
    That's the part I'm particularly confused about. How to plug it in.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Truthbetold View Post
    Okay. I understand the definition of a derivative and what you said.
    When plugging (a+h) as x, does it come to be
    (1/(a+h) -1 - 1/(a) -1 )/ h?
    That's the part I'm particularly confused about. How to plug it in.
    that's it! but you can plug in 2 for a to begin with, so you find the slope right away.

    f'(2) = \lim_{h \to 0} \frac {\frac 1{1 + h} - 1}h

    now combine the fractions in the numerator and simplify so you can cancel the h in the denominator and take the limit
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