Motion of a Particle - Max Acceleration

[Jason76]

Post Edited

Given: \(\displaystyle v(t) = t^{3} - 3t^{2} + 12t + 8\)

\(\displaystyle \dfrac{dv}{dt} = a(t) = 3t^{2} - 6t + 12\)

What is the max acceleration on the interval: \(\displaystyle 0 \le t \le 3\)? - Answer = 21

Logic: When the graph has a maximum value, then it's derivative will be \(\displaystyle 0\). So \(\displaystyle y'\) of \(\displaystyle a(t) = 0\) at maximum points. You simply find which of these points lies within the given interval.

So, \(\displaystyle \dfrac{da}{dt} = b(t) = 6t - 6\) - Note \(\displaystyle b(t)\) is random made up term.

\(\displaystyle b(t) = 6t - 6 = 0\) - Solve for t. See which \(\displaystyle t\) values like within \(\displaystyle 0 \le t \le 3\)

\(\displaystyle t = 1\) and \(\displaystyle 9\) would be our acceleration value after plugging into \(\displaystyle a(t)\)

Now test for max or min using 2nd derivative test:

\(\displaystyle \dfrac{d^{2}a}{dt} = c(t) = 6\)

That is positive so our answer of \(\displaystyle t = 1\) is a minimum and the graph is concave up.

Ok, in this case, how to find the maximum?

 

[MarkFL]

For this problem, since the acceleration is a quadratic, you could just complete the square to obtain:

\(\displaystyle a(t)=3(t-1)^2+9\)

and so we know the minimal acceleration is:

\(\displaystyle a_{\min}=a(1)=9\)

Since the minimum is closer to the left end-point than the right, we know the maximum will occur at the right end-point, given the bilateral symmetry of a parabola.

The bottom line here, is when you are asked for an absolute extremum on an interval, you must check the end-points in addition to any critical values within the interval.

 

[Jason76]

For this problem, since the acceleration is a quadratic, you could just complete the square to obtain:

\(\displaystyle a(t)=3(t-1)^2+9\)

and so we know the minimal acceleration is:

\(\displaystyle a_{\min}=a(1)=9\)

Since the minimum is closer to the left end-point than the right, we know the maximum will occur at the right end-point, given the bilateral symmetry of a parabola.

The bottom line here, is when you are asked for an absolute extremum on an interval, you must check the end-points in addition to any critical values within the interval.

I'm getting \(\displaystyle t = 1\) However, when you plug that into \(\displaystyle a(t)\) you get \(\displaystyle 9\) NOT \(\displaystyle 21\) (the answer)

I'm wondering, if setting the derivative to 0 finds all critical points, then why it only located one critical point, and a minimum (not the one were looking for) at that?

Also, original post edited.

 

[MarkFL]

As I stated, the absolute maximum on the given interval is at the right endpoint, \(\displaystyle t=3\)...\(\displaystyle a(3)=?\)

 

[Bob Brown MSEE]

I'm wondering, if setting the derivative to 0 finds all critical points, then why it only located one critical point, and a minimum (not the one were looking for) at that?

Critical points only give LOCAL maximum or minimum. If a region is specified, always check the end-points of the region, in case the function is greater (or less) than all of the critical points (local max or min) at the left and right of the region.

 

[Jason76]

One way is to do the way I showed in the original post. But realizing that the answer (located between the intervals) is a minimum, you look at the two intervals and conclude one must be a maximum (as it's a multiple choice test). You could look at a test point to the left of the left interval, and one to the right of the right interval (making sure the test points are very close). But then again, factoring, as another poster showed, might be the way.

Yet Another Way

1. Take the 1st derivative of the function and set to 0 (twice).

2. Evaluate each derivative copy at each interval (in this case \(\displaystyle x= 0\) and \(\displaystyle x = 3\)) If either statement is true, and passes the 2nd derivative test, then it's the answer.

3. Otherwise, take the 1st derivative of the function and solve for \(\displaystyle x \)(to see where in the middle of the interval it lies). Also do the 2nd derivative test on it for conformation.

 

[DrPhil]

One way is to do the way I showed in the original post. But realizing that the answer (located between the intervals) is a minimum, you look at the two intervals and conclude one must be a maximum (as it's a multiple choice test). You could look at a test point to the left of the left interval, and one to the right of the right interval (making sure the test points are very close). But then again, factoring, as another poster showed, might be the way.

You ALWAYS have to check the end points of the interval, if you want absolute min or max. Since the function is continuous at the two endpoints, you can just evaluate at those points. You know that the derivative of the acceleration is zero ONLY at t=1. Evaluate
a(0), a(1), a(3)
and choose the largest to be the absolute maximum on the interval [0,3].

 

[Jason76]

You ALWAYS have to check the end points of the interval, if you want absolute min or max. Since the function is continuous at the two endpoints, you can just evaluate at those points. You know that the derivative of the acceleration is zero ONLY at t=1. Evaluate
a(0), a(1), a(3)
and choose the largest to be the absolute maximum on the interval [0,3].

The "Yet Another Way" method in my post above seems like the best way. Perhaps, that's what your saying.

 

[DrPhil]

The "Yet Another Way" method in my post above seems like the best way. Perhaps, that's what your saying.

Even simpler than "Yet Another Way." No 2nd derivative test is needed - just look at the three values of a and take the largest.

 

[Jason76]

Even simpler than "Yet Another Way." No 2nd derivative test is needed - just look at the three values of a and take the largest.

We know that 1 is the minimum value, but, really how do we know what is the maximum? The answer of \(\displaystyle 3\) does NOT have a derivative of \(\displaystyle 0\).

 

[Deleted member 4993]

We know that 1 is the minimum value, but, really how do we know what is the maximum? The answer of \(\displaystyle 3\) does NOT have a derivative of \(\displaystyle 0\).

There is a difference between the "local extremum" and "global extremum". If:

f(x) = x(x-1)(x-2) for -10≤ x ≤ 10

then

the maximum value of this function (within the given domain) occurs at x = 10

the maximum value of this function (within the given domain) occurs at x = -10

and at those points f'(x)\(\displaystyle \ne\)0