Answer:
(a) t = ±2
(b) t ∈ {0, 1}
(c) In navigation terms: east by north. The slope is about 0.42 at that point.
Step-by-step explanation:
(a) dy/dx = 0 when dy/dt = 0
dy/dt = 3t^2 -12 = 0 = 3(t -2)(t +2)
The slope is zero at t = ±2.
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(b) dy/dx = (dy/dt)/(dx/dt) = <em>undefined</em> when dx/dt = 0
dx/dt = 6t^2 -6t = 6(t)(t -1) = 0
The slope is undefined for t ∈ {0, 1}.
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(c) At t=3, dy/dx = (dy/dt)/(dx/dt) = 3(3-2)(3+2)/(6(3)(3-1)) = 15/36 = 5/12
The general direction of movement is away from the origin along a line with a slope of 5/12, about 22.6° CCW from the +x direction.
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The first attachment shows the derivative and its zeros and asymptotes. It also shows some of the detail of the parametric curve near the origin.
The second attachment shows the parametric curve over the domain for which it is defined, along with the point where t=3.
Pictures show u have to solve and the answers
Note that 8.00 = (2.00)^3, that implies that a decay to 1/8.00 means that the Carbon-14 has passed 3 half-life. (1 to decay to 1/2, other to decay to 1/4, and other to decay to 1/8). And, three half-life is 3 * 5730 years = 17,920 years. So,<span> the answer is that the plant was alive 17,920 years.</span>