Calculus FRQ Help?

Question

A particle moves along the y-axis so that its position at time t is given by y(t)= t²tan(1/t) for t>1.

a. Show and explain that the velocity of the particle at time t is given by v(t)= 2ttan(1/t) - sec²(1/t) for t>1.

b. At time t= 4/π, is the particle moving toward or away from the origin? Give a reason for your answer.

c. The velocity of the particle at time t can be written as v(t)= 2tan(1/t) / 1/t - sec²(1/t) for t>1. Find the limit of v(t) as t approaches infinity. Show the work that led you to your answer.

Answer 1

a)

Velocity is the first derivative of position.

Using the product and chain rules:

y'(t) = v(t) = 2ttan(1/t) + t²sec²(1/t)(-1/t²) = 2ttan(1/t)- sec²(1/t)

b)

v(4/π) = (8/π)tan(π/4) -sec²(π/4) = (8/π)(1)-(√2)² ≈2.5465-2=0.5465

It is moving away since v(4/π) is positive.

c)

lim 2tan(1/t) / 1/t - sec²(1/t) = L

t→∞

The first part above is a 0/0 situation, so need to use L'Hospital's rule. The second part is 1.

For the first part,

lim 2sec²(1/t)(-1/t²)/(-1/t²) = 2 lim sec²(1/t) = 2(1) = 2

t→∞ t→∞

Putting both parts together

L = 2-1 = 1