A particle moves along the x-axis with velocity v(t)=e^(3sinx)-2. At time t=0, the position of the particle is 3.

Question

a) Find the position when the particle is furthest to left.

b) On what intervals is the particle speeding up?

Answer 1

Here are some things we know

Since negative time has no meaning, we know that t >= 0

If v(t) > 0, x is moving to the right.

If v(t) < 0, x is moving to the left.

e^3sin(t)-2 cannot be integrated analytically

Part (a)

Consider the interval [0,2π]. A graph of the function shows it is negative until e^3sin(t)-2=0, or until t=arcsin((ln(2))/3). An approximate value of the integral of e^3sin(t)-2 for the interval [0,arcsin((ln(2))/3)] can still be approximated by numerical methods. Using Simpson's Rule and n=4, x has moved approximately -0.20388 from the starting point of 3. Next, looking at the graph, it is observable that the positive area is far greater than the negative areas (by counting squares if not by inspection). That means that by the end of the interval, the position is well to the right. Subsequent cycles will increase the positive position more than decrease it, so the leftmost position of the particle is approximately 3 -.20388 = 2.7961.

Part (b)

We know that:

The particle is speeding up when its acceleration is positive.

e^3sin(t) is always positive

Acceleration is the derivative of velocity, v(t)

v'(t) = 3cos(t)e^3sin(t) = a(t)

Since e^3sin(t) is always positive, the sign of a(t) is solely a function of cos(t). Thus,

a(t) > 0 from 0 to π/2, 3π/2 to 5π/2, 7π/2 to 9π/2 ...

Thus, the particle is speeding up in [0,π/2] and all subsequent intervals [(2n+1)π/2,(2n+3)π/2],n=1,3,5,...