For the following exercises, determine the slope of the tangent line, then find the equation of the tangent line at the given value of the parameter. 66. x = 3 sin t, y = 3 cost, t = À 67. x = cos t, y = 8 sin t, t = 5 68. x = 2t, y = f, t = -1 69. x = 1+1, y=t-7, t = 1 70. x = Vt, y = 2t, t = 4

Answer and Step-by-step explanation:

(66) x = 3 sin t, y = 3cos t at t = π/4

dx/dt = 3cos t, dy/dt = - 3sin t

Slope of the tangent, m = dy/dx = (dy/dt)/(dx/dt) = -(sin t)/(cos t) = - tan t = - tan (π/4) = - 1

At this point

x = 3 sin (π/4) = (3√2)/2

y = 3 cos (π/4) = (3√2)/2

Equation of the tangent

(y - ((3√2)/2) = -1(x - (3√2)/2)

y + x = ((3√2)/2) - ((3√2)/2) = 0

y + x = 0

(67) x = cos t, y = 8 sin t, t = π/2

dx/dt = - sin t, dy/dt = 8 cos t

slope, m = dy/dx = (dy/dt)/(dx/dt) = - (8cos t)/(sin t) = -(8 cos (π/2))/(sin (π/2)) = 0

At this point,

x = cos π/2 = 0, y = sin π/2 = 1

Equation of the tangent,

y - 1 = 0(x - 0)

y = 1.

(68) x = 2t, y = t³, t = -1

dx/dt = 2, dy/dt = 3t²

slope, m = dy/dx = (dy/dt)/(dx/dt) = 3t²/2 = 3(-1)²/2 = 3/2

At this point,

x = 2(-1) = -2, y = (-1)³ = -1

Equation of the tangent,

y + 1 = (3/2)(x + 2)

2y - 3x = 4

(69) x = t + (1/t), y = t - (1/t), t = 1

dx/dt = 1 - (1/t²), dy/dt = 1 + (1/t²)

slope, m = dy/dx = (dy/dt)/(dx/dt) = (t² + 1)/(t - 1) = (1+1)/(1-1) = 2/0

At this point, t = 1

x = 2 = -2, y = 0

Equation of the tangent,

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

x - 2 = 0

x = 2

(70) x = √t, y = 2t, t = 4

dx/dt = -(1/2√t), dy/dt = 2

slope, m = dy/dx = (dy/dt)/(dx/dt) = -4√t = -4√4 = -8

At this point,

x = √4 = 2, y = 2×4 = 8

Equation of the tangent,

y - 8 = - 8(x - 2)

y + 8x = 16

the answer is y < -3 or   y > 3

-3y > 7 + 2

-3 y > 9

y < -3