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

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step-by-step explanation:

answer: option a.

step-by-step explanation:

you know that the quadrilateral opqr ( a four sided figure) is inscribed in circle. this means that all its vertices are on the circle.

then, by definition, the opposite angles of the quadrilateral are supplementary, this means that the sum of their measures is 180 degrees.

you can observe in the figure that the angles ∠opq   and the angle ∠orq   are opposite, therefore:

then, can be used to d to calculate the measure of ∠opq