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Math Help - I don't get these calc problems! Please help!

  1. #1
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    I don't get these calc problems! Please help!

    1) Find the points on the graph of y=sec x, 0≤x≤2π, where the tangent is parallel to the line 3y-2x=4.
    I understand that the derivative is the slope and that the slope should be the same as the slope obtained from the line. I don't understand how to get the points from the slope.

    2) A curve is parametrized by the equations x=
    √(t) and y=(t-3). Find an equation of the line tangent to the curve at the point defined by t=9.
    Is there two answers for this one or just one? I got an outrageous answer, like y=12x+36 when I did it.

    3) Let f(x)=int x
    (a) Find NDER (f(x), 3).
    (b) Is your answer to part (a) a meaningful estimate of a derivative of f(x)? explain.
    I'm completely boggled by this one. Firstly, I don't know how to get my grapher (TI-84 silver) to graph the NDER function. Secondly, I am not sure I completely understand the problem.

    4) Sketch a possible graph of a continuous function f that has domain [-3,3], where f(-1)=1 and the graph of y=f '(x) is shown below.
    Well, I'm not sure how to draw the graph on this forum, but I'll describe it: close circle at (-3,3), linear line to open circle point at (1,-2) & y=(x+2) (a normal parabola) with open circle at (1,3) and a close circle at (3,3).I don't understand for what the question is asking.


    Any help at all would be greatly appreciated! Thanks so much!
    Last edited by h4hv4hd4si4n; November 24th 2008 at 01:19 PM.
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  2. #2
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    Hi,

    2) You have 2 values of t that gives y=9 : t=0 and t=6
    The 2 corresponding points are A(0,9) and B(√(6),9)

    Derivation of x gives 1/(2√(t)). It is infinite for t=0 and 1/(2√(6)) for t=6.
    Derivation of y gives 2(t-3). It gives -6 for t=0 and 6 for t=6.

    Therefore the tangent in A is parallel to x axis
    The tangent in B has a vector (1/(2√(6) , 6) that can give you the equation of the tangent (I let you finish the job)
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  3. #3
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    1) this problem is more trig than calculus ...

    y' = \sec{x}\tan{x} = \frac{\sin{x}}{\cos^2{x}} = \frac{2}{3}

    3\sin{x} = 2\cos^2{x}

    3\sin{x} = 2(1 - \sin^2{x})

    2\sin^2{x} + 3\sin{x} - 2 = 0

    (2\sin{x} - 1)(\sin{x} + 2) = 0

    \sin{x} = \frac{1}{2} ... x = \frac{\pi}{6} , x = \frac{5\pi}{6}

    \sin{x} = -2 ... no solution

    lesson learned ... know your trig well.
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  4. #4
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    hey running-gag, i'm so sorry, i meant t=9, and accidentally typed y=9. (This is what happens after my cranium meets the sidewalk on this snowy afternoon. sorry.)
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  5. #5
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    Quote Originally Posted by h4hv4hd4si4n View Post
    hey running-gag, i'm so sorry, i meant t=9, and accidentally typed y=9. (This is what happens after my cranium meets the sidewalk on this snowy afternoon. sorry.)
    OK no problem
    Derivation of x gives 1/(2√(t)). It gives 1/6 for t=9.
    Derivation of y gives 2(t-3). It gives 12 for t=9.

    The tangent has a vector (1/6 , 12) that gives an equation y=72x-180
    Therefore I do not agree with your result !
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  6. #6
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    Quote Originally Posted by running-gag View Post
    OK no problem
    Derivation of x gives 1/(2√(t)). It gives 1/6 for t=9.
    Derivation of y gives 2(t-3). It gives 12 for t=9.

    The tangent has a vector (1/6 , 12) that gives an equation y=72x-180
    Therefore I do not agree with your result !
    Why do you use the points from the derivative and not the points you would get if you substituted 9 in for t in the original equations for x and y?

    Also, by using what you said, I got y=72x for the equation. I don't know where -180 came from.
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  7. #7
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    If you substitute 9 in for t in the original equations for x and y you get the coordinates of the point of the curve from which you want to find the tangent. Here it is (3,36)

    To get the coordinates of the vector of the tangent, you have to calculate the derivative first and then substitute t=9
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  8. #8
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    Quote Originally Posted by h4hv4hd4si4n View Post
    Also, by using what you said, I got y=72x for the equation. I don't know where -180 came from.
    The tangent must pass through the point (3,36)
    When you have the coordinates of the vector, you get only the slope of the tangent
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  9. #9
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    Quote Originally Posted by running-gag View Post
    If you substitute 9 in for t in the original equations for x and y you get the coordinates of the point of the curve from which you want to find the tangent. Here it is (3,36)

    To get the coordinates of the vector of the tangent, you have to calculate the derivative first and then substitute t=9

    Thanks for your help! now I understand.
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  10. #10
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    btw you can get the full equation of your curve
    x=√(t) and y=(t-3) gives y=(x-3)
    y'(x)=4x(x-3)
    y'(3)=72 which is effectively the slope of the tangent

    http://imageshack-france.com/out.php/i253011_Graph.JPG
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  11. #11
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    Quote Originally Posted by running-gag View Post
    btw you can get the full equation of your curve
    x=√(t) and y=(t-3) gives y=(x-3)
    y'(x)=4x(x-3)
    y'(3)=72 which is effectively the slope of the tangent

    http://imageshack-france.com/out.php/i253011_Graph.JPG

    this method seems so much easier!
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  12. #12
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    Sure but it is not always feasible
    This was only an exercice ...
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