calculus ?

Find the distance traveled by a particle with position (x, y) as t varies in the given time interval.

x = cos^2(t), y = cos(t), 0 ≤ t ≤ 3𝜋

Update:

I know i have to use the formula 

sqrt((dx/dt)^2+(dy/dt)^2)

2 Answers

Relevance
  • x = cos(t)^2

    dx/dt = 2 * cos(t) * (-sin(t)) = -2sin(t)cos(t)

    y = cos(t)

    dy/dt = -sin(t)

    sqrt((dx/dt)^2 + (dy/dt)^2) * dt =>

    sqrt(4 * sin(t)^2 * cos(t)^2 + sin(t)^2) * dt =>

    sqrt(sin(t)^2 * (4cos(t)^2 + 1)) * dt =>

    sin(t) * sqrt(4 * cos(t)^2 + 1) * dt

    u = cos(t)

    du = -sin(t) * dt

    sqrt(4u^2 + 1) * (-du) =>

    -sqrt(1 + 4u^2) * du

    tan(m) = 2u

    sec(m)^2 * dm = 2 * du

    -sqrt(1 + tan(m)^2) * (1/2) * sec(m)^2 * dm =>

    (-1/2) * sqrt(sec(m)^2) * sec(m)^2 * dm =>

    (-1/2) * sec(m) * sec(m)^2 * dm

    int(sec(m)^3 * dm)

    a = sec(m)

    da = sec(m) * tan(m) * dm

    db = sec(m)^2 * dm

    b = tan(m)

    int(sec(m)^3 * dm) = sec(m) * tan(m) - int(sec(m) * tan(m)^2 * dm)

    int(sec(m)^3 * dm) = sec(m) * tan(m) - int(sec(m) * (sec(m)^2 - 1) * dm)

    int(sec(m)^3 * dm) = sec(m) * tan(m) - int(sec(m)^3 * dm) + int(sec(m) * dm)

    2 * int(sec(m)^3 * dm) = sec(m) * tan(m) + int(sec(m) * dm)

    2 * int(sec(m)^3 * dm) = sec(m) * tan(m) + ln|sec(m) + tan(m)|

    int(sec(m)^3 * dm) = (1/2) * (sec(m) * tan(m) + ln|sec(m) + tan(m)|)

    (-1/4) * (sec(m) * tan(m) + ln|sec(m) + tan(m)|) =>

    (-1/4) * (2u * sqrt(1 + 4u^2) + ln|2u + sqrt(1 + 4u^2)|) =>

    (-1/4) * (2 * cos(t) * sqrt(1 + 4 * cos(t)^2) + ln|2 * cos(t) + sqrt(1 + 4 * cos(t)^2)|)

    From t = 0 to t = 3pi

    cos(0) = 1

    cos(3pi) = -1

    (-1/4) * (2 * (-1) * sqrt(1 + 4 * 1) + ln|2 * (-1) + sqrt(1 + 4 * 1)|) + (1/4) * (2 * 1 * sqrt(1 + 4 * 1) + ln|2 * 1 + sqrt(1 + 4 * 1)|) =>

    (1/4) * (2 * sqrt(5) + ln|2 + sqrt(5)|) - (1/4) * (-2 * sqrt(5) + ln|-2 + sqrt(5)|) =>

    (1/4) * (2 * sqrt(5) + ln|2 + sqrt(5)|) + (1/4) * 2 * sqrt(5) - (1/4) * ln|sqrt(5) - 2|) =>

    (1/4) * (2 * sqrt(5) + 2 * sqrt(5)) + (1/4) * ln|(sqrt(5) + 2) / (sqrt(5) - 2)| =>

    (1/4) * 4 * sqrt(5) + (1/4) * ln|(sqrt(5) + 2)^2 / (5 - 4)| =>

    sqrt(5) + ln|(5 + 4 * sqrt(5) + 4) / 1| =>

    sqrt(5) + ln|9 + 4 * sqrt(5)| =>

    sqrt(5) + ln(9 + 4 * sqrt(5))

  • rotchm
    Lv 7
    2 months ago

    Hint: What formula or concept did you see for "the distance", or the "length of the curve" ?

    State it here and we will show you how to apply it. 

    (Dont go by Captain's answer; he made many mistakes)

    Ok, yes one of the formulas is to integrate sqrt((dx/dt)^2+(dy/dt)^2)  dt.

    So, evaluate dx/dt and dy/dt. What do you get for these?

    Then, plug into sqrt((dx/dt)^2+(dy/dt)^2). What does it give?

    And if you simplify it in various fashions, what expressions you get?

    The next difficulty will be in integrating it. But do the above first.

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