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Anna [14]
2 years ago
7

At a skills competition, a target is being lifted into the air by a cable at a constant speed. An archer standing on the ground

launches an arrow toward the target. The system of equations below models the height, in feet, of the target and the arrow t seconds after it was fired. Which statement most likely describes the situation modeled by this system?
The arrow is fired with an initial upward velocity of 2 ft/s.
The arrow is fired with an initial upward velocity of 4 ft/s.
The arrow is fired with an initial upward velocity of 8 ft/s.
The arrow is fired with an initial

Mathematics
1 answer:
xz_007 [3.2K]2 years ago
3 0
The arrow is fired with an initial upward velocity of 32ft/s
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A child wanders slowly down a circular staircase from the top of a tower. With x,y,zx,y,z in feet and the origin at the base of
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Answer:

a) The tower is 90 feet tall

b) She reaches the bottom at t = 18 minutes.

c) Her speed at time t is 5 \sqrt[]{5} ft/minute

d) Her acceleration at time t is 10 ft/minute^2

Step-by-step explanation:

Consider the path described by the child as going down the tower to have the following parametrization \gamma(t) = (10\cos t, 10 \sin t, 90-5t)

a) Assuming that the child is at the top of the tower when she starts going down, we have that at the initial time (t=0) we will have the value of the height of the tower. That is z = 90-5*0 = 90 ft.

b) The child reaches the bottom as soon as z =0. We want to find the value of t that does that. Then we have 0 = 90-5t, which gives us t = 18 minutes.

c) Given the parametrization we are given, the velocity of the child at time t is given by \frac{d\gamma}{dt}= (\frac{d}{dt}(10\cos t), \frac{d}{dt} (10 \sin t ), \frac{d}{dt}(90-5t)) = (-10 \sin t, 10 \cos t, -5). The speed is defined as the norm of the velocity vector,

so, the speed at time t is given by v = \sqrt[]{(-10 \sin t)^2+(10 \cos t)^2+(-5)^2} = \sqrt[]{100(\sin^2 t + \cos^2 t)+25} = \sqrt[]{125}= 5 \sqrt[]{5}

d) ON the same fashion we want to know the norm of the second derivative of \gamma.

We have that \gamma ^{''}(t) =(-10\cost t, -10 \sin t , 0) so the acceleration is given by \sqrt[]{100(\cos^2 t+ \sin^2 t )} = 10 

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