Curve y = cos πx

Discussion in 'Calculus' started by nycmathguy, Oct 10, 2021.

  1. nycmathguy

    nycmathguy

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    Exercises 1.4

    For part (a), I must evaluate the given trigonometric function at each x-value. Yes?
    Once this is done, how do I find the slope of the secant line PQ?

    I would like you to do part (d) using a different color per lines. Cool?

    20211009_195057.jpg
     
    nycmathguy, Oct 10, 2021
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  2. nycmathguy

    MathLover1

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    yes, evaluate the given trigonometric function at each x-value and find points
     
    MathLover1, Oct 10, 2021
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  3. nycmathguy

    nycmathguy

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    Sounds good. I will play with B question until all three parts.
     
    nycmathguy, Oct 10, 2021
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  4. nycmathguy

    nycmathguy

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    Part (a)

    When x = 0, y = 0.
    When x = 0.4, y = -0.4
    When x = 0.49, y = -0.49
    When x = 0.499, y = -0.499
    When x = 1, y = -1
    When x = 0.6, y = -0.6
    Whe x = 0.51, y = -0.51
    When x = 0.501, y = -0.501

    From the above, the following points are formed and their slopes:

    P(0.5, 0)
    Q(0, 0)

    Slope of PQ = 0

    P(0.5, 0)
    Q(0.4, -0.4)

    Slope of PQ = 4

    P(0.5, 0)
    Q(0.49, -0.49)

    Slope of PQ = 49

    P(0.5,0)
    Q(0.499, -0.499)

    Slope of PQ = 499

    P(0.5, 0)
    Q(1, -1)

    Slope of PQ = -2

    P(0.5, 0)
    Q(0.51, -0.51)

    Slope of PQ = -51

    P(0.5, 0)
    Q(0.501, -0.501)

    Slope of PQ = -501

    1. Is this right?

    2. How do I use the the points and slopes, considering the information is right, to answer part (b)?

    Forget parts (c) and (d) for now.
     
    nycmathguy, Oct 10, 2021
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  5. nycmathguy

    MathLover1

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    part a is correct

    b. find average slope for tangent line at P
     
    MathLover1, Oct 10, 2021
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  6. nycmathguy

    nycmathguy

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    Can you set up part (b) for me? Just set it up for me to do.

    Thanks.
     
    nycmathguy, Oct 11, 2021
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  7. nycmathguy

    MathLover1

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    use point P(0.5, 0)
    find average slope m (choose two slopes from part a
    use point slope formula y-y[1]=m(x-x[1]
     
    MathLover1, Oct 11, 2021
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  8. nycmathguy

    nycmathguy

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    1. Why do I need point P?

    2. Do you mean (slope 1 + slope 2)/2?

    3. Why use the point-slope formula here?

    Stuck here....
     
    nycmathguy, Oct 11, 2021
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  9. nycmathguy

    MathLover1

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    1. Why do I need point P?
    point slope formula y-y[1]=m(x-x[1]) -> so, you need a slope m (you have (slope 1 + slope 2)/2) and the coordinates of the point of tangency (x[1],y[1]) which is P(0.5, 0) to find equation of tangent

    2. Do you mean (slope 1 + slope 2)/2?
    yes
    3. Why use the point-slope formula here?
    explained in 1.

    Point-slope form = y − y 1 = m ( x − x 1 ) , where ( x 1 , y 1 ) is the point given and m is the slope given.
     
    MathLover1, Oct 11, 2021
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  10. nycmathguy

    nycmathguy

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    Let's do this:

    Complete this problem for me parts b through d. I will then use your reply to answer a similar question in the same exercise problem set. We can then move forward.
     
    nycmathguy, Oct 11, 2021
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  11. nycmathguy

    MathLover1

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    b.

    P(0.5, 0)
    Q(0, 0)

    Slope of PQ = 0

    P(0.5, 0)
    Q(0.4, -0.4)

    Slope of PQ = 4


    m=(0 + 4)/2=2

    c.

    P(0.5, 0)

    y-0=2(x-0.5)
    y=2(x-1/2)
    y=2x-1


    d. Sketch the curve, two of the secant lines, and the tangent line.

    y=cos(pi*x)
    y=2x-1-> tangent line
    the secant lines:
    A secant line is simply a linear equation and with two given points you can find the equation.
    The two points on the secant line are:
    Q(0.4, -0.4)
    Q(0.49, -0.49), and slope is PQ = -1
    Next, solve for the y-intercept:

    y=49x+b ....find b, substitute (0.4, -0.4)
    -0.4=-1*0.4+b
    b=0

    y=-x
    upload_2021-10-11_17-30-23.png
    View attachment 649
     
    Last edited: Oct 11, 2021
    MathLover1, Oct 11, 2021
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  12. nycmathguy

    nycmathguy

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    What can I say? You are amazing. It is my hope that you will be my online calculus tutor just as you have been for precalculus. Let's step off the calculus 1 boat for now. One course at a time is the best choice for someone like me who took precalculus in 1993. Long ago, right?
     
    nycmathguy, Oct 11, 2021
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  13. nycmathguy

    MathLover1

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    right
     
    MathLover1, Oct 12, 2021
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  14. nycmathguy

    nycmathguy

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    Calculus is not going anywhere. After precalculus, I will step into calculus 1 and take it from there. Good to know what lies ahead.
     
    nycmathguy, Oct 12, 2021
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