Problem 59 - Essential Calculus

Question

The deflection \(D\) of a beam of length \(L\) is \(D=2 x^{4}-5 L x^{3}+3 L^{2} x^{2},\) where \(x\) is the distance from one end of the beam. Find the value of \(x\) that yields the maximum deflection.

Step-by-Step Solution

Answer

Solve the equation \(8x^3 - 15Lx^2 + 6L^2x = 0\) to get \(x = 0\) or the solutions of the quadratic formula \(x = [-(-15L) ± \sqrt{(-15L)^2 - 4*8*6L^2}]/(2*8)\). The value that gives \(D'' < 0\) is the \(x\) value that yields maximum deflection.

Step 1: Find the derivative of the deflection equation

The first step is to find the derivative of the equation \(D = 2x^4 - 5Lx^3 + 3L^2x^2\) with respect to \(x\). The derivative \(D'\) can be determined as: \(D' = 8x^3 - 15Lx^2 + 6L^2x\).

Step 2: Equate the derivative to zero

In order to find the value of \(x\) that yields the maximum deflection, set \(D' = 0\). So, \(8x^3 - 15Lx^2 + 6L^2x = 0\). You can factor out \(x\) from this equation to simplify. So, \(x(8x^2 - 15Lx + 6L^2) = 0\). From this equation, \(x = 0\) or the zeros of the quadratic equation \(8x^2 - 15Lx + 6L^2 = 0\).

Step 3: Solve the quadratic equation

To find the zeros of the quadratic equation \(8x^2 - 15Lx + 6L^2 = 0\), use the quadratic formula: \(x = [-(-15L) ± \sqrt{(-15L)^2 - 4*8*6L^2}]/(2*8)\). Consider only the positive root because it's not meaningful for \(x\) to be negative in the context of the problem.

Step 4: Check the second derivative

Calculate the second derivative of the equation \(D''\), and check its sign for the solved \(x\) values from Step 3. If \(D'' < 0\), then the deflection is maximum at that point. If \(D'' > 0\), it's a minimum. The second derivative \(D'' = 24x^2 - 30Lx + 6L^2\). Substitute the values of \(x\) obtained in the second derivative, and the sign will indicate whether it's a maxima or minima.

Key Concepts

Derivatives in Calculus

In calculus, the concept of derivatives is fundamental for understanding how functions change. They are the tool that measures the rate at which a quantity changes. For instance, when considering the deflection of a beam, the function describing that deflection is often a polynomial. In this case, D=2x^4-5Lx^3+3L^2x^2, represents the deflection based on the position x along the beam.

To analyze this function and determine where it has maximum or minimum values, we calculate its derivative with respect to x. The derivative, D', tells us the slope of the function D at any given point. Where this derivative equals zero, we find what are called 'critical points'. These points indicate where the function's rate of change switches direction, which can signal either a peak or a trough - in simpler terms, a maximum or minimum deflection.

To find the derivative, we apply power rules, which say that the derivative of x^n is n*x^(n−1). Applying this to each term in the deflection equation, we get D' = 8x^3 - 15Lx^2 + 6L^2x. Investigating where this derivative equals zero will lead us to potential points of maximum deflection on the beam.

Quadratic Equation

The quadratic equation is a second-order polynomial equation of the form ax^2 + bx + c = 0. It is a fundamental element of algebra and appears in various physics and engineering problems, including the study of beam deflections. In our example, after factoring out an x from the first derivative, we are left with a quadratic equation that we need to solve to find the critical points of the function.

The solutions to the quadratic equation are found using the quadratic formula: x = [-b ± sqrt(b^2 - 4ac)]/(2a). This formula ensures that we consider both the positive and negative roots of the equation. However, for physical problems like beam deflection, we often disregard negative roots since they lack practical meaning.

In our exercise, after setting the first derivative to zero, we have a quadratic in the form 8x^2 - 15Lx + 6L^2 = 0. Applying the quadratic formula allows us to find the possible values of x that could lead to maximum deflection. Remembering to take only the positive root as the beam cannot have a negative length, we can then proceed to check which of these roots indeed gives the maximum deflection.

Optimization in Calculus

Optimization is a significant area of calculus that involves finding the maximum or minimum values of functions. It is used extensively in various fields like engineering, economics, and science to determine the most efficient or cost-effective scenario. In terms of finding the maximum deflection in a beam, we apply optimization techniques to the function describing the deflection.

After locating the critical points by setting the first derivative to zero and solving the resulting quadratic equation, we need to determine whether these points represent maxima or minima. This is done using the second derivative test. It involves calculating the second derivative of our original function and then substituting the critical values x into this second derivative.

If D'' < 0 for a given critical point, it's a maximum. Conversely, if D'' > 0, it's a minimum. For the beam deflection scenario, we're interested in where the second derivative is negative since we want to find the maximum deflection. This process of optimization ensures we've located the point along the beam that will experience the greatest amount of deflection under the given conditions.