Question

A track has a height that is a function of horizontal position \(x\), given by \(h(x)=x^{3}+3 x^{2}-24 x+16\). Find all the positions on the track where a marble will remain where it is placed. What kind of equilibrium exists at each of these positions?

Short answer

Answer: The marble will remain in equilibrium at \(x = -4\) (unstable) and \(x = 2\) (stable).

Step-by-step solution

Find the derivative of the height function

To find the critical points, we first need to calculate the derivative of the height function, which will give us the slope of the tangent at any point. We are given \(h(x) = x^3 + 3x^2 - 24x + 16\). Find the derivative with respect to \(x\): \(h'(x) = \frac{d}{dx}(x^3 + 3x^2 - 24x + 16) = 3x^2 + 6x - 24\)

Find the critical points

Now, we need to find the critical points by setting the derivative equal to zero and solving for \(x\). \(3x^2 + 6x - 24 = 0\) We notice that all terms have a common factor of 3, which can be factored out: \(x^2 + 2x - 8 = 0\) This equation can be factored further: \((x+4)(x-2) = 0\) So, the critical points are at \(x = -4\) and \(x = 2\).

Find the second derivative of the height function

In order to classify the equilibrium at each critical point, we'll use the second derivative test. First, we need to find the second derivative of the height function: \(h''(x) = \frac{d^2}{dx^2}(x^3 + 3x^2 - 24x + 16) = \frac{d}{dx}(3x^2 + 6x - 24) = 6x + 6\)

Classify the critical points using the second derivative test

Now, we'll analyze the sign of the second derivative at each critical point to determine whether each point corresponds to a stable or unstable equilibrium: \(h''(-4) = 6(-4) + 6 = -24 + 6 = -18\) Since the second derivative is negative at \(x = -4\), we have a local maximum, which means the point corresponds to an unstable equilibrium. \(h''(2) = 6(2) + 6 = 12 + 6 = 18\) Since the second derivative is positive at \(x = 2\), we have a local minimum, which means the point corresponds to a stable equilibrium.

Conclusion

The marble will remain in place at two positions on the track: at \(x = -4\), we have an unstable equilibrium, and at \(x = 2\), we have a stable equilibrium.

Key concepts

Critical Points Calculation

Understanding critical points is essential for analyzing functions in physics and mathematics. In our exercise, we are given a height function of a track, represented by a cubic polynomial:

To find the critical points where a marble will remain where it's placed, we first calculate the derivative of the height with respect to the horizontal position, x. This derivative provides the slope of the tangent at any point along the track. Critical points occur where this slope is zero - indicating either a peak, trough, or an inflection point on the track.

By setting the first derivative equal to zero and solving for x, we locate these critical points. Through factoring, it's revealed that the marble can potentially come to rest at two positions, x = -4 and x = 2. It is important to provide these detailed steps to ensure students can follow the logic and understand the techniques used to find critical points.

Second Derivative Test

Once we've determined the critical points, the next concept is the second derivative test, a method to examine the concavity of the function at these points. Evaluating the concavity helps in determining whether a critical point is a local minimum, maximum, or neither. The sign of the second derivative at the critical point tells us the kind of equilibrium that exists: a positive second derivative (h''(x) > 0) indicates a local minimum with stable equilibrium, whereas a negative second derivative (h''(x) < 0) implies a local maximum with unstable equilibrium.

Describing the second derivative test step by step, as seen in the exercise's solution, brings clarity to how it applies to each critical point found beforehand. In the exercise's context, this test shows us that the marble is in stable equilibrium at one point, and in unstable equilibrium at the other.

Stability of Equilibrium

In the final analysis of our exercise, understanding the stability of equilibrium is imperative. Physically, it encapsulates the behaviors we'd expect to see when a marble is placed on different parts of the track. If a marble is at a stable equilibrium point, it will return to this position after experiencing a small displacement. This is consistent with a local minimum and is linked to a positive second derivative, as found at the point x = 2 in the exercise.

On the other hand, an unstable equilibrium means that upon any small disturbance, the marble will move away from the equilibrium position. This is related to a local maximum and a negative second derivative, experienced at x = -4. By walking students through the implications of the second derivative test, they gain a deeper insight into the dynamic aspects of equilibrium in physical situations.