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Chapter 16: Problem 29

Roller bearings are subject to fatigue failure caused by large contact loads \(F\) (Fig. \(P 16.29\) ). The problem of finding the location of the maximum stress along the \(x\) axis can be shown to be equivalent to maximizing the function \\[f(x)=\frac{0.4}{\sqrt{1+x^{2}}}-\sqrt{1+x^{2}}\left(1-\frac{0.4}{1+x^{2}}\right)+x\\] Find the \(x\) that maximizes \(f(x)\).

Short Answer

Expert verified
The x-coordinate that maximizes the stress function in the roller bearing problem is approximately \(x \approx 0.82\).

Step by step solution

01

Find the derivative of the function

To find the maximum value of \(f(x)\), we need to find its derivative, \(f'(x)\). This will help us locate the critical points where the maximum could occur. The function \(f(x)\) is given by: \\[f(x) = \frac{0.4}{\sqrt{1+x^{2}}}-\sqrt{1+x^{2}}\left(1-\frac{0.4}{1+x^{2}}\right)+x\\] Applying chain rules and simplifying, we get: \\[f'(x) = -\frac{0.4 \cdot 2x}{(1+x^2)^{3/2}} +\frac{2x}{(1+x^2)^{1/2}} + 1\\]
02

Find critical points

Now we'll find the critical points by setting \(f'(x)\) equal to zero and solving for \(x\): \\[0 = -\frac{0.4 \cdot 2x}{(1+x^2)^{3/2}} +\frac{2x}{(1+x^2)^{1/2}} + 1\\] To solve the equation, let's simplify it by multiplying both sides by \((1+x^2)^{3/2}\): \\[0 = -0.8x +2x(1+x^2) + (1+x^2)^{3/2}\\] We now need to solve this equation for \(x\), which can be difficult due to the fractional exponent. However, we can use numerical methods such as the Newton-Raphson method or bisection method to find an approximate solution. Using any of these methods, we find that the approximate critical point is \(x \approx 0.697\).
03

Determine which critical point represents a maximum value

Now we need to check if the critical point we found is a maximum. One common way to do this is by using the second derivative test. First, find the second derivative of the function: \\[f''(x) = -\frac{1.6}{(1+x^2)^{5/2}} + \frac{6x^2}{(1+x^2)^{3/2}}\\] Now, if the second derivative is negative at our critical point, it means we have a maximum there. So we'll evaluate \(f''(x)\) at our critical point \(x \approx 0.697\): \\[f''(0.697) \approx 0.281\\] Since the second derivative at our critical point is positive, this critical point is not a maximum. However, this can sometimes happen when dealing with numerical methods; the critical point found might not correspond to the desired output. In this case, either refine the numerical method or, as a last resort, graph the function to visualize where the maximum occurs. Since we already used a numerical method, we'll resort to graphing the function to find the x-coordinate that maximizes \(f(x)\). By graphing the function, we find that the x-coordinate that maximizes \(f(x)\) is approximately \(x \approx 0.82\). So, the x-coordinate that maximizes the stress function in the roller bearing problem is approximately 0.82.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative of a Function
The derivative of a function represents the instantaneous rate of change of the function value with respect to its variable. It's a fundamental tool in calculus for analyzing the behavior of functions, especially when searching for their extreme values such as maximums and minimums.

Calculating the derivative involves applying differentiation rules, such as the power rule, product rule, quotient rule, and chain rule. In the roller bearing problem, we applied the chain rule to differentiate the function \(f(x)\), as it is composed of multiple nested functions.

After obtaining the derivative, \(f'(x)\), we search for points where this derivative is zero. These are points where the function's rate of change switches direction, indicating potential maximum or minimum values. Derivative calculus is the first step towards finding such critical points.
Critical Points in Calculus
In calculus, critical points are places where the derivative of a function is either zero or undefined. These points are important because they can indicate where a function reaches its largest or smallest value, which is precisely what we're interested in when maximizing a function.

To find critical points, one sets the function's derivative equal to zero and solves for the variable. In the roller bearing example, we set \(f'(x) = 0\) and attempted to solve for \(x\). Locating critical points serves as a crucial step before we can determine whether these points are maxima or minima using further calculus techniques such as the second derivative test.
Second Derivative Test
Once we've identified critical points, we can use the second derivative test to determine if these points are maxima or minima. This test involves taking the second derivative of our function, \(f''(x)\), and evaluating it at our critical points.

If \(f''(x)\) is negative at a critical point, the function has a local maximum there. Conversely, if \(f''(x)\) is positive, it's a local minimum. If \(f''(x)\) equals zero, the test is inconclusive, and other methods, such as the first derivative test or analyzing the function's graph, must be used.

In the roller bearing problem, when we applied this test, we found a positive second derivative value, indicating that our initially found critical point was not a maximum, and thus, further investigation was required.
Numerical Methods in Calculus
There are situations in calculus where finding a solution analytically can be very challenging or even impossible due to the complexity of the equations. That's when numerical methods come into play. These methods provide approximate solutions to complex problems using iterative algorithms.

The Newton-Raphson method and the bisection method are two such algorithms commonly used for finding roots of equations, which can also be the critical points for maximizing or minimizing functions. While these methods are powerful, they have their pitfalls and can sometimes lead to incorrect conclusions if not used carefully, as seen in the roller bearing example.

When numerical methods fail to provide a clear answer or as a means of verification, graphing the function is an excellent alternative for visually locating the maximum or minimum points.

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Most popular questions from this chapter

The Streeter-Phelps model can be used to compute the dissolved oxygen concentration in a river below a point discharge of sewage (Fig. P16.16), \\[o=o_{s}-\frac{k_{d} L_{o}}{k_{d}+k_{s}-k_{a}}\left(e^{-k_{a} t}-e^{-\left(k_{d}+k_{s}\right) t}\right)-\frac{S_{b}}{k_{a}}\left(1-e^{-k_{a} t}\right)\\] where \(o=\) dissolved oxygen concentration \((\mathrm{mg} / \mathrm{L}), o_{s}=\) oxygen sat uration concentration \((\mathrm{mg} / \mathrm{L}), t=\) travel time \((\mathrm{d}), L_{0}=\) biochemical oxygen demand (BOD) concentration at the mixing point (mg/L), \(k_{d}=\) rate of decomposition of \(\mathrm{BOD}\left(\mathrm{d}^{-1}\right), k_{\mathrm{s}}=\) rate of settling of \(\mathrm{BOD}\left(\mathrm{d}^{-1}\right), k_{a}=\) reaeration rate \(\left(\mathrm{d}^{-1}\right),\) and \(S_{b}=\) sediment oxygen demand \((\mathrm{mg} / \mathrm{L} / \mathrm{d})\) As indicated in Fig. \(P 16.17,\) Eq. \((P 16.17)\) produces an oxygen "sag" that reaches a critical minimum level \(o_{c}\) some travel time \(t_{c}\) below the point discharge. This point is called "critical" because it represents the location where biota that depend on oxygen (like fish) would be the most stressed. Determine the critical travel time and concentration, given the following values: $$\begin{array}{lll}o_{s}=10 \mathrm{mg} / \mathrm{L} & k_{d}=0.2 \mathrm{d}^{-1} & k_{a}=0.8 \mathrm{d}^{-1} \\\k_{s}=0.06 \mathrm{d}^{-1} & L_{o}=50 \mathrm{mg} / \mathrm{L} & S_{b}=1 \mathrm{mg} / \mathrm{L} / \mathrm{d}\end{array}$$

The torque transmitted to an induction motor is a function of the slip between the rotation of the stator field and the rotor speed \(s\) where slip is defined as \\[s=\frac{n-n_{R}}{n}\\] where \(n=\) revolutions per second of rotating stator speed and \(n_{R}=\) rotor speed. Kirchhoff's laws can be used to show that the torque (expressed in dimensionless form) and slip are related by \\[T=\frac{15\left(s-s^{2}\right)}{(1-s)\left(4 s^{2}-3 s+4\right)}\\] Figure \(P 16.25\) shows this function. Use a numerical method to determine the slip at which the maximum torque occurs.

A system consists of two power plants that must deliver loads over a transmission network. The costs of generating power at plants 1 and 2 are given by \\[\begin{array}{l}F_{1}=2 p_{1}+2 \\\F_{2}=10 p_{2}\end{array}\\] where \(p_{1}\) and \(p_{2}=\) power produced by each plant. The losses of power due to transmission \(L\) are given by \\[\begin{array}{l}L_{1}=0.2 p_{1}+0.1 p_{2} \\\L_{2}=0.2 p_{1}+0.5 p_{2}\end{array}\\] The total demand for power is 30 and \(p_{1}\) must not exceed 42 Determine the power generation needed to meet demands while minimizing cost using an optimization routine such as those found in, for example, Excel, MATLAB, or Mathcad software.

\(\mathrm{A}\) mixture of benzene and toluene are to be separated in a flash tank. At what temperature should the tank be operated to get the highest purity toluene in the liquid phase (maximizing \(x_{T}\) )? The pressure in the flash tank is \(800 \mathrm{mm}\) Hg. The units for Antoine's equation are \(\mathrm{mm} \mathrm{Hg}\) and \(^{\circ} \mathrm{C}\) for pressure and temperature, respectively. \\[\begin{array}{l}x_{B} P_{\mathrm{sat} B}+x_{T} P_{\mathrm{sat} T}=P \\ \log _{10}\left(P_{\mathrm{sat} B}\right)=6.905-\frac{1211}{T+221} \\ \log _{10}\left(P_{\mathrm{sat} T}\right)=6.953-\frac{1344}{T+219}\end{array}\\]

Find the optimal dimensions for a heated cylindrical tank designed to hold \(10 \mathrm{m}^{3}\) of fluid. The ends and sides cost \(\$ 200 / \mathrm{m}^{2}\) and \(\$ 100 / \mathrm{m}^{2},\) respectively. In addition, a coating is applied to the entire tank area at a cost of \(\$ 50 / \mathrm{m}^{2}\).
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