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Chapter 11: Problem 52

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

Expert verified
Answer: The marble will remain at the stable equilibrium point of \(x=2\).

Step by step solution

01

Finding the first derivative of \(h(x)\)

To find the critical points, we first need to find the derivative of the height function with respect to \(x\). $$h'(x) = \frac{d}{dx}(x^{3}+3 x^{2}-24 x+16)$$ Now differentiate \(h(x)\) term by term: $$h'(x) = 3x^{2}+6 x-24$$
02

Setting the first derivative equal to zero

Now we will set this first derivative equal to zero and solve for \(x\) to find the potential critical points: $$3x^{2}+6 x-24 = 0$$
03

Solving for the critical points

Now let's solve the quadratic equation for \(x\): $$3x^{2}+6 x-24 = 0$$ We can first divide the equation by 3 to simplify: $$x^{2}+2 x-8=0$$ Now we can factor the quadratic equation: $$(x + 4)(x - 2) = 0$$ The critical points are: $$x = -4, 2$$
04

Finding the second derivative of \(h(x)\)

Now, to classify these critical points as stable, unstable, or semi-stable equilibrium points, we need to find the second derivative of the height function \(h(x)\): $$h''(x) = \frac{d^{2}}{dx^{2}}(x^{3}+3 x^{2}-24 x+16)$$ Differentiate the first derivative, \(h'(x)\): $$h''(x) = 6x+6$$
05

Classifying the critical points using the second derivative test

Now, we will use the second derivative test to classify the critical points. The second derivative test states that: - If \(h''(x) > 0\) at a critical point, the equilibrium is stable. - If \(h''(x) < 0\) at a critical point, the equilibrium is unstable. - If \(h''(x) = 0\) at a critical point, the test is inconclusive. Evaluating the second derivative, \(h''(x)\) at the critical points: - At \(x=-4\), \(h''(-4) = 6(-4) + 6 = -18 < 0 \Rightarrow\) unstable equilibrium - At \(x=2\), \(h''(2) = 6(2) + 6 = 18 > 0 \Rightarrow\) stable equilibrium So, the marble will remain where it is placed at the stable equilibrium point of \(x=2\).

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