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Suppose that, for a beam of length 11 meters, the deflection from the horizontal of a point a distance \(x\) from one end is proportional to $$ 2 x^{4}-33 x^{3}+1331 x $$ Determine the point at which the deflection is greatest.

Short Answer

Expert verified
The deflection is greatest at 7 meters from one end of the beam.

Step by step solution

01

Understanding the relation

The deflection of the beam at any point is proportional to the polynomial function \( f(x) = 2x^4 - 33x^3 + 1331x \). We need to find where this deflection is greatest along the beam, which means finding the maximum value of this function for \( 0 \leq x \leq 11 \).
02

Taking the derivative

To find the point of maximum deflection, we first calculate the derivative of the function \( f(x) \). The derivative \( f'(x) \) is given by applying the power rule: \[ f'(x) = \frac{d}{dx}(2x^4 - 33x^3 + 1331x) = 8x^3 - 99x^2 + 1331. \]
03

Setting the derivative to zero

Find the critical points by setting the derivative equal to zero and solving the equation: \[ 8x^3 - 99x^2 + 1331 = 0. \] This cubic equation can be solved using methods like factoring or the Rational Root Theorem.
04

Solving for critical points

By testing potential roots using the Rational Root Theorem, we find that \( x = 7 \) is a root of the equation. We can further factor the cubic equation to find any other possible roots if needed. However, for simplicity, let's proceed with this critical point.
05

Checking the endpoints

Evaluate the function at the endpoints of the interval. Compute \( f(0) \) and \( f(11) \) to ensure no greater deflection occurs at these points compared to the critical point. \[ f(0) = 0, \quad f(11) = 2(11)^4 - 33(11)^3 + 1331(11) = 0. \]
06

Evaluating at the critical point

Calculate \( f(7) \) to find the deflection at the critical point: \[ f(7) = 2(7)^4 - 33(7)^3 + 1331(7) = 13141. \]
07

Comparing values

Since \( f(0) = 0 \), \( f(11) = 0 \), and \( f(7) = 13141 \), the maximum deflection occurs at \( x = 7 \).
08

Conclusion

The greatest deflection of the beam occurs at \( x = 7 \) meters from one end.

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Key Concepts

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

Derivative
In calculus, the derivative is a measure of how a function changes as its input changes. Derivatives are fundamental in finding the rate of change or slope of a curve at any given point.
In this problem, the deflection of the beam is represented by a polynomial function. To find where this deflection is maximized, we need to determine where the slope of the tangent to the polynomial is zero. This requires us to find the derivative of the polynomial function.To compute the derivative of a polynomial like \(f(x) = 2x^4 - 33x^3 + 1331x\), we apply the power rule, which states that the derivative of \(ax^n\) is \(nax^{n-1}\). Applying this rule, the derivative of our function is \(f'(x) = 8x^3 - 99x^2 + 1331\).
This new function, \(f'(x)\), gives us the rate of change of \(f(x)\) and is crucial in finding critical points where the function may reach maximum or minimum values.
Critical points
Critical points are where the derivative of a function is zero or undefined. These points indicate potential locations for local maxima or minima, depending on the context.
For the function \(f(x) = 2x^4 - 33x^3 + 1331x\), we solve the equation \(f'(x) = 0\) to find critical points. Setting the derivative \(8x^3 - 99x^2 + 1331 = 0\) helps identify these points.
Critical points can be found by techniques such as factoring or using the Rational Root Theorem, which provides possible roots for polynomial equations with integer coefficients. Trying these possibilities, we find that one critical point is at \(x = 7\).
Determining the nature of critical points (whether they are maxima, minima, or points of inflection) usually involves further testing by evaluating the second derivative or comparing function values at critical points.
Polynomial function
Polynomial functions are expressions that include terms in the form of coefficients and variables raised to whole number powers. Examples of polynomial functions involve basic algebraic equations and can range from simple, linear functions to complex, higher-degree equations incorporating various powers of a variable.
In this exercise, the polynomial function \(2x^4 - 33x^3 + 1331x\) represents the deflection of a beam. As a fourth-degree polynomial, it includes terms up to \(x^4\).
Such functions are continuous and smooth, making them ideal for determining maximum and minimum values using calculus. Specifically, they allow for derivatives to be easily calculated, aiding in the search for critical points and eventual optimization solutions.
Maximum value
The maximum value of a function refers to the highest point on its graph within a specified interval. In optimization problems, this concept is crucial for finding points of interest, such as where a structure's deflection might be the greatest.Having identified the critical point at \(x = 7\), the next step involves comparing the function values at this point and at the endpoints of the interval. In this scenario, evaluating the function \(f(x)\) at the endpoints \(x = 0\) and \(x = 11\)—both resulting in a deflection value of 0—confirms that neither endpoint provides a higher deflection than the critical point.
Calculating \(f(7)\) yields 13141, indicating that the maximum deflection occurs at \(x = 7\) meters. By comparing these values, we determine that the beam's deflection is greatest at this specific point, highlighting how calculus aids in optimizing real-world scenarios.

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