Answer:
8) x ∈ {π/6, π/3, 2π/3, 5π/6}
9a) v = -2(sin(x) +cos(2x)); a = 4sin(2x) -2cos(x)
9b) v(3) ≈ -2.2 cm/s; a(3) ≈ 0.9 cm/s²
9c) negative direction (to the left)
Explanation:
You want the x-values in the interval [0, π] where b(x) = sin(4x) will have tangents parallel to 4x +2y -12 = 0. Given a particle's position is s(t) = 2cos(t) -sin(2t), you want equations for velocity and acceleration, and their values at t=3, and you want the direction of travel at t=20. Units are cm and seconds.
8) Tangent points
The equation for the given line can be written as ...
y = -2x +6
This has a slope of -2, so we want the x-values where b'(x) = -2.
The derivative is ...
b'(x) = 4cos(4x) = -2
cos(4x) = -1/2 . . . . . . . . . . divide by 4
The inverse cosine function tells us ...
4x = (2n+1)π ± π/3 . . . . . . . for integer n
x = nπ/2 + {π/6, π/3} . . . . . for n=0, 1
x ∈ {π/6, π/3, 2π/3, 5π/6}
9) Particle
a) Derivatives
The derivative of s(t) = 2cos(t) -sin(2t) is the velocity:
v(t) = s'(t) = -2sin(t) -2cos(2t)
And the derivative of v(t) is the acceleration:
a(t) = v'(t) = -2cos(t) +4sin(2t)
b) Values
The velocity and acceleration at t = 3 are ...
v(3) = -2(sin(3)+cos(2·3)) ≈ -2.2 . . . cm/s
a(3) = -2cos(3) +4sin(2·3) ≈ 0.9 . . . cm/s²
At 3 seconds, the velocity is about -2.2 cm/s, and the acceleration is about 0.9 cm/s².
c) Direction
The velocity at t=20 is about ...
v(20) = -2(sin(20) +cos(2·20)) ≈ -0.5
This has a negative sign, so the particle is moving in the negative direction.
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Additional comment
In the attachment, the function Y1(x) is defined as 2cos(x) - sin(2x). This calculator can do both symbolic and numerical derivatives.
We note that d(sin(u)) = cos(u)·du, and d(cos(u)) = -sin(u)·du, so we do not need the double-angle identity.
For evaluating the trig functions, the calculator is in radians mode.
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