Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

? asked in Science & MathematicsMathematics · 8 years ago

Calculus 2 Parametric equations?

parametric equation : x= squareroot(t) , y= squareroot(t-1)

point : t=2

1)dy/dx = squareroot (t/(t-1))

2) d^2y/dx^2 = ????????????

3)slope = squareroot (2)

4)concavity = ???????

any one? thanks in advance!

2 Answers

Relevance
  • Favorite Answer

    x = sqrt(t)

    y = sqrt(t - 1)

    x^2 = t

    y^2 = t - 1

    2x * dx = dt

    dx/dt = 1/(2x)

    dx/dt = 1/(2 * sqrt(t))

    y^2 = t - 1

    2y * dy = dt

    dy/dt = 1/(2y)

    dy/dt = 1/(2 * sqrt(t - 1))

    (dy/dx) =>

    (dy/dt) / (dx/dt) =>

    (1/(2sqrt(t))) / (1/(2sqrt(t - 1))) =>

    2 * sqrt(t - 1) / (2 * sqrt(t)) =>

    sqrt(t - 1) / sqrt(t) =>

    sqrt((t - 1) / t) =>

    sqrt(1 - 1/t)

    d2y/dx^2

    dy/dx = sqrt((t - 1) / t)

    dy/dx = sqrt(t - 1) / sqrt(t)

    dy/dx = y / x

    d2y/dx^2 = (x * dy/dx - x) / x^2

    d2y/dx^2 = (x * (y/x) - x) / x^2

    d2y/dx^2 = (y - x) / x^2

    d2y/dx^2 = (sqrt(t - 1) - sqrt(t)) / t

    slope

    dy/dx = y/x = sqrt(t - 1) / sqrt(t)

    t = 2

    sqrt(2 - 1) / sqrt(2) = sqrt(1) / sqrt(2) = sqrt(2)/2

    concavity

    (sqrt(t - 1) - sqrt(t)) / t =>

    (sqrt(1) - sqrt(2)) / 2 =>

    (1 - sqrt(2)) / 2

  • ?
    Lv 4
    4 years ago

    x(t) = x? + (x? - x?) t y(t) = y? + (y? - y?) t the position (x?, y?) is initial position and (x?, y?) is position even as t = a million x(t) = 2 + (-5-2) t = 2 - 7t y(t) = 10 + (-3-10) t = 10 - 13t -------------------- speed: v(t) = ?((dx/dt)² + (dy/dt)²) v(t) = ?((-7)² + (-13)²) v(t) = ?218 v(t) ? 14.7648 instruments/sec

Still have questions? Get your answers by asking now.