Given the position function, s of t equals t cubed divided by 3 minus 12 times t squared divided by 2 plus 36 times t, between t = 0 and t = 15, where s is given in feet and t is measured in seconds, find the interval in seconds where the particle is moving to the right.
Steve, you idiot. There were no typos you completely ignorant buffoon. This person copied and pasted the problem so don't suspect they made a typo and don't assume they made a mistake you cocky peon.
To determine the interval in seconds where the particle is moving to the right, we need to find the intervals where the velocity function is positive.
The velocity function (v) can be obtained by taking the derivative of the position function (s) with respect to time (t). Let's calculate the velocity function:
v(t) = s'(t)
To find s'(t), we differentiate each term of the position function separately:
s(t) = t^3/3 - 12t^2/2 + 36t
Differentiating the first term:
d/dt(t^3/3) = (3t^2)/3 = t^2
Differentiating the second term:
d/dt(-12t^2/2) = -24t
Differentiating the third term:
d/dt(36t) = 36
Combining the derivatives, we have:
v(t) = t^2 - 24t + 36
To find the intervals where v(t) is positive (particle moves to the right), we need to solve the inequality v(t) > 0.
t^2 - 24t + 36 > 0
To solve this inequality, we can use the factored form of the quadratic equation. The solutions are the points where the quadratic function equals zero:
(t - 2)(t - 18) > 0
Setting each factor equal to zero:
t - 2 = 0 --> t = 2
t - 18 = 0 --> t = 18
Now we create a sign chart to determine the intervals where v(t) is positive:
(-∞) |----(-)-----0----(+)-----| (∞)
|---------+---------+----------|
| (-) | (+) | (-) |
From the sign chart, we can see that the velocity is positive (moving to the right) when t < 2 and 18 < t.
Thus, the interval in seconds where the particle is moving to the right is t ∈ (2, 18).
To find the interval in seconds where the particle is moving to the right, we need to determine where the particle's velocity is positive. In other words, we need to find the values of t where the derivative of the position function, s'(t), is greater than zero.
First, let's find the derivative of the position function, s(t), with respect to t to obtain the velocity function, v(t).
s(t) = (t^3)/3 - 12(t^2)/2 + 36t
Taking the derivative of each term:
s'(t) = (1/3) * 3t^2 - 12 * (2/2)t + 36
Simplifying:
s'(t) = t^2 - 6t + 36
Our goal is to find the values of t where s'(t) > 0.
To solve this quadratic inequality, we can find the t-intercepts by setting s'(t) equal to zero and determining the values of t that make it equal to zero:
t^2 - 6t + 36 = 0
Using the quadratic formula, we can solve for t:
t = (-b ± √(b^2 - 4ac)) / 2a
Plugging in the values from our quadratic equation:
t = (-(-6) ± √((-6)^2 - 4*1*36)) / (2*1)
Simplifying:
t = (6 ± √(36 - 144)) / 2
t = (6 ± √(-108)) / 2
Since the square root of a negative number is not defined in the real number system, it means that there are no real solutions to the quadratic equation t^2 - 6t + 36 = 0. Hence, there are no t-intercepts, which means the velocity function s'(t) > 0 for all values of t.
Therefore, the particle is always moving to the right between t = 0 and t = 15, given the position function.
s(t) = t^3/3 - 12t^2/2 + 36t
Strange, saying 12t^2/2 rather than 6t^2, but hey ...
Assuming "moving to the right" means s(t) is increasing, then we just have to find when ds/dt > 0. That is,
t^2 - 12t + 36 > 0
Hmmm. Something tells me you had a typo, and it was really
s(t) = t^3/3 - 13t^2/2 + 36t
Then
ds/dt = t^2 - 13t + 36 > 0
(t-9)(t-4) > 0
Now I suspect another typo, because ds/dt > 0 when t<4 or t>9.
Fix your mistakes and follow the logic to a correct solution.