Application of differentiation

[rachael]

2 The displacement of a particle from a fixed point is given by x = t(6 − t)(t + 1), 0 ≤ t ≤ 6.
a Find the exact time when the particle is a maximum distance from the fixed point.
b Find the velocity, v, when t = 3.
c Find the exact time when the particle is at maximum velocity.
d Find the maximum velocity.

i need help on part c and d
could any please help me?
thank you

 

[abercrombiems02]

I'll give you a few hints...

a. Whenever we're at a maximum we're at a critical point, it must be satified that for any well-defined critical point, whethere it be a minumum, maximum, or corner (for a bounded domain) that the first derivative of our quantity of interest is zero. Thus if we are to find a maximum displacement over time we must take the derivative of displacement with respect to time and set the quantity equal to zero. The second condition for a maximum is that the second derivative of displacement with respect to time (twice) must be less than zero. If we apply this principle to this problem, we can find the time at which these events occur and then plug that back into the original x = equation so find the maximum dispalcement.

b. simply take dx/dt, then evaluate what's left when t = 3

c. to maximize velocity we do a procedure similar to the steps in part one, however here we are looking for the time at which acceleration is zero and that the time derivative of acceleration is less than zero.

d. evaluate the algebraic expression for velocity at the time you found in part c.

since you asked for c and d specifically i'll give you an example.

suppose and objects displacement is given by -(4/3)t^3 + 2t^2
the velocity is then -4t^2 + 4t
the acceleration (dv/dt) us then -8t + 4
the critical points are -8t + 4 = 0 which gives t = 0.5
the second derivative (d2v/dt2) is -8 -8 < 0 therefore this is a maximum

now we just plot t = 0.5 into our original equation for v

suppose however that the velocity was given by (4/3)t^3 - 2t^2
this ends up giving us a second derivative of 8 which is positive
this means the critical point is a minimum. Therefore the only possible maximum are the corners, therefore we would need to be given a bounded time interval to evaluate the corners at to determine the local maxima.

Finally notice that in your problem, your t^3 term has a negative sign, this means your velocity does have a maximum that lies at a finite time, and thus you won't have to worry about a certain time interval or the need to evaluate the function at a given set of corners. So just follow the above example with your own expression and you should be good

 

[Architeuthis Dux]

Part c:

To find the exact time when the particle is at maximum velocity, we need to find the point where the velocity function, v(t), is equal to zero. This is because the maximum velocity occurs at the point where the slope of the displacement function, x(t), is equal to zero.

Using the chain rule, we can find the velocity function by taking the derivative of the displacement function with respect to time:

v(t) = x'(t) = (6 - 2t)(t + 1) + t(6 - t)

Setting this equal to zero and solving for t, we get:

0 = (6 - 2t)(t + 1) + t(6 - t)
0 = 6 - 2t + 6t - 2t^2 + 6t - t^2
0 = 4t^2 - 8t + 6

Using the quadratic formula, we can solve for t:

t = [8 ± √(64 - 4(4)(6))] / (2(4))
t = [8 ± √(16)] / 8
t = [8 ± 4] / 8
t = 1 or t = 3/2

Since the time, t, must be between 0 and 6, the particle is at maximum velocity at t = 1. Therefore, the exact time when the particle is at maximum velocity is t = 1.

Part d:

To find the maximum velocity, we can substitute t = 1 into the velocity function we found in part c:

v(1) = (6 - 2(1))(1 + 1) + 1(6 - 1)
v(1) = (6 - 2)(2) + 5
v(1) = 8 + 5
v(1) = 13

Therefore, the maximum velocity of the particle is 13 units per second.