A particle moves along a horizontal line such that its position s(t) = (t^2e^3t/3)for time t. Find when the particle is at rest.
To be at rest, wouldn't the velocity have to be zero ?
And isn't velocity the first derivative of your s(t) ?
so...
take the first derivative, set it equal to zero and solve for t.
hint: use the product rule.
To find when the particle is at rest, we need to find the times at which its velocity is zero. We can do this by finding the derivative of the position function, s(t), and setting it equal to zero.
Let's start by finding the derivative of s(t):
v(t) = d/dt (s(t))
= d/dt ((t^2 * e^(3t))/3)
To simplify the differentiation, we can apply the product rule:
v(t) = (d/dt (t^2)) * (e^(3t))/3 + (t^2) * (d/dt (e^(3t)))/3
Differentiating each term gives us:
v(t) = 2t * (e^(3t))/3 + (t^2) * (3 * e^(3t))/3
= (2t * e^(3t)) / 3 + (t^2 * e^(3t))
Now, let's set v(t) equal to zero and solve for t:
0 = (2t * e^(3t)) / 3 + (t^2 * e^(3t))
In order to solve this equation, we can factor out an e^(3t):
0 = (t * e^(3t)) * (2/3 + t)
Now we have a product of two terms that is equal to zero. This means that at least one of the factors must be equal to zero:
t * e^(3t) = 0 or 2/3 + t = 0
The first equation, t * e^(3t) = 0, implies that either t = 0 or e^(3t) = 0. However, e^(3t) is never zero, so the only solution is t = 0.
The second equation, 2/3 + t = 0, gives us t = -2/3.
Therefore, the particle is at rest at t = 0 and t = -2/3.
To find when the particle is at rest, we need to find the values of time (t) for which the particle's velocity is equal to zero.
Given that the particle's position function is s(t) = (t^2e^3t / 3), we need to differentiate this function to find the velocity function.
Taking the derivative of s(t) with respect to t, we use the product rule and chain rule:
s'(t) = (2t * e^3t / 3) + (t^2 * 3e^3t)
Simplifying this, we get:
s'(t) = (2t * e^3t) / 3 + (3t^2 * e^3t)
Now, we set s'(t) equal to zero and solve for t:
(2t * e^3t) / 3 + (3t^2 * e^3t) = 0
Factoring out e^3t as a common factor:
e^3t (2t / 3 + 3t^2) = 0
This equation will be equal to zero if either e^3t = 0 or (2t / 3 + 3t^2) = 0.
However, it is not possible for e^3t to equal zero since e^x is always positive for any value of x. Therefore, we can only focus on the second part of the equation.
Setting (2t / 3 + 3t^2) equal to zero and solving for t:
2t / 3 + 3t^2 = 0
Dividing both sides by t:
2 / 3 + 3t = 0
Simplifying further:
2 + 9t = 0
Subtracting 2 from both sides:
9t = -2
Dividing both sides by 9:
t = -2 / 9
Therefore, the particle is at rest when t = -2 / 9.