really need help with derivatives

Question

Particle P moves along the x-axis so that its position at time t>0 is given by xP(t)= (e^(2−t)−2t) / (e^(2−t)+3t). A second particle, particle Q , also moves along the x -axis so that its position at time t is given by xQ(t)=t^3−3t^2+5 .

a) Show that the velocity of particle P at time t is given by vP(t)= −5te^(2−t)−5e^(2−t) / (e^(2−t)+3t)^2 .

b) At time t=2 , particle Q is at rest. At time t=2 , is particle P moving toward particle Q or away from particle Q ? Explain your reasoning.

c) The acceleration of particle Q is given by aQ(t) . Find the value of aQ(2) .

d) Describe the position of particle P and the position of particle Q as t approaches infinity. How did you get to your answers.

Answer 1

a) Use the quotient rule and note that the derivative position with respect to time is the velocity

vP(t) = xP'(t) = ((e^(2-t)+3t) * d/dt(e^(2−t)−2t)) - (e^(2−t)−2t) * d/dt(e^(2-t)+3t))/((e^(2-t)+3t)^2)

= ((e^(2-t)+3t) * (-e^(2-t)-2) - (e^(2−t)−2t) * (-e^(2-t)+3))/((e^(2-t)+3t)^2)

= −5te^(2−t)−5e^(2−t) / (e^(2−t)+3t)^2

as desired

b) First find the positions of particles P and Q at time t=2 which is given by

xP(2) = (e^(0)−4) / (e^(0)+6) = -3/7

xQ(2) = 8 - 12 + 5 = 1

Consider vP(2) = (-5*2*e^0 - 5*e^0)/((e^0 + 6)^2) = (-15)/(49) < 0

and thus particle P is moving in the negative direction and since it is already to the left of particle Q, it is moving away from particle Q

c) We first calculate, noting that the derivative of velocity is acceleration

vQ(t) = xQ'(t) = d/dt(t^3−3t^2+5) = 3t^2 - 6t

aQ(t) = vQ'(t) = d/dt(3t^2 - 6t) = 6t - 6

and plug in t=2 to get aQ(2) = 12 - 6 = 6

d) We first consider particle P. We see that as t goes to infinity, the quantity

lim_(t -> infty) (e^(2−t)−2t) / (e^(2−t)+3t) = -2/3 since lim_(t -> infty) e^(2-t) = 0

Next, we consider particle Q, and see that the first term, t^3, grows faster than every other term, and thus

lim_(t -> infty) t^3 - 3t^2 + 5 = infty