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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
- 8 years agoFavorite 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 44 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