A particle moves along the x-axis so that at time "t" its position is given by s(t)=(t+3)(t-1)^3 ,t>0
For what values of "t" is the velocity of the particle increasing?
a) 0<t<1
b) t>1
c) t>0
d) the velocity is never increasing
You know that the velocity of s(t) is the first derivative of s(t)=(t+3)(t-1)^3 for t > 0
s ' (t) = (t+3)(3)(t-1)^2 + (t-1)^3
= (t-1)^2 (3t + 9 + t-1)
= (t-1)^2 (4t + 8)
you also know that if the function increases, (t-1)^2 (4t + 8) must be positive
so the factor (t-1)^2 is always positive except when t = 1
and since t > 0, 4t+8 will be always positive
so for t > 0, s(t) will be increasing for all values of t, except t = 1
check: https://www.wolframalpha.com/input/?i=plot+s(t)%3D(t%2B3)(t-1)%5E3
your choices:
a) 0<t<1 ----- correct but does not include all values of t
b) t>1 -------- correct but does not include all values of t
c) t>0 -------- true except when t = 1
when t = 1, the function is "stationary"
d) clearly false
s = t^4 - 6 t^2 + 8 t - 3 (check my multipication)
v = ds/dt = 4 t^3 - 12 t + 8
a = d^2 s/dt^2 = 12 t^2 -12 = 12 (t^2-1)
where is that acceleration positive?
where t^2 >1
but t is only positive so
t > 1
Okay I understand how you got the answer, thank you! But which answer choice would I pick? Thanks in advance.
Actually I get it, thanks again
To determine the values of 't' for which the velocity of the particle is increasing, we need to find the derivative of the position function with respect to time. The velocity function is the derivative of the position function.
Given:
s(t) = (t+3)(t-1)^3
Let's find the derivative of s(t) with respect to t:
s'(t) = d/dt [(t+3)(t-1)^3]
To find this derivative, we can use the product rule. The product rule for differentiation states that if we have two functions u(t) and v(t), then the derivative of their product can be found as follows:
(d/dt) [u(t) * v(t)] = u'(t) * v(t) + u(t) * v'(t)
Applying the product rule to our expression:
s'(t) = (d/dt) (t+3) * (t-1)^3 + (t+3) * (d/dt) (t-1)^3
To find each individual derivative, we can further simplify.
1. Derivative of (t+3):
The derivative of t+3 with respect to t is simply 1.
2. Derivative of (t-1)^3:
To find the derivative of (t-1)^3, we can use the chain rule. The chain rule states that if we have a composite function, then the derivative can be found by multiplying the derivative of the outer function with the derivative of the inner function.
Let u = t-1
So, (t-1)^3 = u^3
The derivative of u^3 with respect to u is 3u^2.
But since our function is in terms of t, we need to multiply by the derivative of u(t) as well.
(d/dt) (t-1)^3 = 3(t-1)^2 * (d/dt) (t-1)
The derivative of (t-1) with respect to t is simply 1.
Substituting these values back into our expression for s'(t):
s'(t) = 1 * (t-1)^3 + (t+3) * 3(t-1)^2 * 1
Simplifying further:
s'(t) = (t-1)^3 + 3(t+3)(t-1)^2
Now to answer the question, for what values of 't' is the velocity of the particle increasing, we need to find when the velocity function s'(t) is positive.
Therefore, we need to solve the inequality:
s'(t) > 0
By analyzing the sign of the expression (t-1)^3 + 3(t+3)(t-1)^2, we can determine the intervals of 't' for which the velocity is increasing.
Since the expression (t-1)^3 + 3(t+3)(t-1)^2 is positive for all t > 1, we can conclude that the velocity of the particle is increasing for t > 1.
Therefore, the answer is option b) t > 1.