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.
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Below is the solution:
<span>distance = 475
</span>time = 9.5 hrs
<span>since distance and time varies directly
</span>
therefore in 9.5 hrs train travels 475 miles
=> In 1 hr train travels 475 / 9.5 miles = 50 miles
<span>and in 4 hrs train travells 50*4 i.e 200 miles </span>
Answer:
angle Z=20°
side xy≈15.36
hypotenuse≈19.49
Step-by-step explanation:
Find angle z adding the other two angles and subtracting that by 180 to get 20.
(12)tan(20) gets you side xy which is 15.36
12^2+15.36^2=xz^2
Answer:
C
Step-by-step explanation:
When doing 4 divided by 2 you get 2. Looking at the sign it is saying 3 is greater than the number we solve for. Since we solved and got 2 we can double check and ask is 3 greater than 2? Yes it is so that would make the statement correct.
Hope this Helps :)