Suppose the position of a particle in motion at time t is given by the vector pa

Question

Suppose the position of a particle in motion at time t is given by the vector parametric equation r(t) = /,2(t - 3)2,3, 2t3 - 9t2>. Find the velocity of the particle at time t. v(t) = Find the speed of the particle at time t. Find the time(s) when the particle is stationary. If there is more than one correct answer, enter your answers as a comma separated list. t = A satellite in a circular orbit 900 kilometers above the surface of the Earth. What is the period of the orbit? You may take the radius of the Earth to be 6370 kilometers and acceleration at the surface of the earth to be 9.81 m/sec/sec Be mindful of units! Find the limits, if they exist, or type DNE for any which do not exist.

Explanation / Answer

r(t) = (2(t-3)^2,3,2t^3-9t^2)

Velocity at time t = dr/dt = 4(t-3) i + 0 j + 6t*(t - 3) k

Speed at time t is the magnitude of velocity at t = (t-3)*sqrt(16+36t^2)

We see only at t=3s all components of particle velocity = 0

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For the satellite acc. due to gravity at orbit = centrepetal acceleration

acc. due to gravity at orbit = g' =g on ground * R^2/(R+h)^2 where R = 6370km h=900km

g' = 7.53m/s2 = V^2/(R+h) => V = 7.4km/s

Time period = 2*pi*(R+h) / V = 6169.7s

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Along x axis y=0 thus limit = 4

Along y axis x=0 thus function itself =0 everywhere and so limit = 0. Value of 0/2y^2 at y= 0 is indeterminate but we are intrested in the limit only. Since at all other points than y=0 along y axis function =0 limit =0

Along y=mx substitute and cancel x^2 thus limit = 4/(1+2m^2)

The limit DNE because its value depends on the direction of approach.

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