Problem 34 - Basic Technical Mathematics with Calculus

Question

The rate of change of the vertical deflection \(y\) with respect to the horizontal distance \(x\) from one end of a beam is a function of \(x\) For a particular beam, the function is \(k\left(x^{5}+1350 x^{3}-7000 x^{2}\right)\) where \(k\) is a constant. Find \(y\) as a function of \(x\) if \(y=0\) when \(x=0\)

Step-by-Step Solution

Answer

The deflection is given by \( y = k \left( \frac{x^6}{6} + 337.5x^4 - \frac{7000x^3}{3} \right) \).

Step 1: Understand the Given Problem

We are given that the rate of change of the vertical deflection \( y \) with respect to the horizontal distance \( x \) is a function \( k(x^5 + 1350x^3 - 7000x^2) \). We need to find \( y \) as a function of \( x \), i.e., integrate the given derivative, and use the condition \( y(0) = 0 \) to find the constant of integration.

Step 2: Set Up the Integral

The rate of change of \( y \) is given as the derivative \( \frac{dy}{dx} = k(x^5 + 1350x^3 - 7000x^2) \). To find \( y \), integrate this with respect to \( x \).

Step 3: Integrate the Function

Perform the integration:\[y = \int k(x^5 + 1350x^3 - 7000x^2) \, dx\]This can be broken into three parts:\[y = k \int x^5 \, dx + k \int 1350x^3 \, dx - k \int 7000x^2 \, dx\]

Step 4: Calculate Each Integral

Evaluate each integral separately:1. \( \int x^5 \, dx = \frac{x^6}{6} \)2. \( \int 1350x^3 \, dx = 1350 \cdot \frac{x^4}{4} = 337.5x^4 \)3. \( \int 7000x^2 \, dx = 7000 \cdot \frac{x^3}{3} = \frac{7000x^3}{3} \)

Step 5: Combine Results and Simplify

Substitute the results of integration into the equation for \( y \):\[y = k \left( \frac{x^6}{6} + 337.5x^4 - \frac{7000x^3}{3} \right) + C\]

Step 6: Apply the Initial Condition

Use the condition \( y(0) = 0 \) to find the constant \( C \):\( y(0) = k \left( \frac{0^6}{6} + 337.5 \times 0^4 - \frac{7000 \times 0^3}{3} \right) + C = 0 \)Thus, \( C = 0 \).

Step 7: Express the Final Function

The final function, using \( C = 0 \), is:\[y = k \left( \frac{x^6}{6} + 337.5x^4 - \frac{7000x^3}{3} \right)\]This is \( y \) as a function of \( x \).

Key Concepts

Integration

Integration is a fundamental concept in calculus that involves finding a function given its derivative. This process is the reverse operation of differentiation and helps us understand how a rate of change accumulates over time or distance.
For example, in the given problem, the vertical deflection of a beam needs to be found from its rate of change provided by the function. This means we need to integrate this function to obtain a formula that describes the vertical deflection. Integration often involves solving definite or indefinite integrals, in our case an indefinite one, meaning we solve it generally without limits at first.

• To integrate, you sum the areas under the curve of the derivative function.
• When integrating, each term of the polynomial in the function must be handled separately.
• Add the constant of integration at the end since the antiderivative of a function is not unique without it.

Understanding the basics of integration is key to solving many calculus problems effectively.

Quadratic function

A quadratic function is one that can be characterized by the highest degree of 2, usually written as \( ax^2 + bx + c \). In calculus, we often deal with polynomials that include quadratic terms like these, but also many other higher degree terms.
In this exercise, even though the given function is of higher degree polynomial, it includes quadratic-like terms that we work with during integration. This means we perform integral steps on terms such as \( x^3 \) or \( x^5 \), treating each separately like you would in a quadratic polynomial.

• Recognize quadratic and polynomial terms in functions.
• Understand that integration will increase the degree of each term by 1.
• When working through integration, coefficients like constants multiply directly.

This makes operations on quadratic functions and similar polynomial terms relatively straightforward once you have practiced.

Constant of integration

The constant of integration, often denoted by \( C \), is crucial in the process of indefinite integration. When you integrate a function, you are essentially reversing the operation of differentiation, which loses constant information.
Upon integrating, we introduce \( C \) because there can be infinitely many antiderivatives; they only differ by a constant. Here, solving for \( C \) involves initial conditions such as \( y(0) = 0 \) to find the specific antiderivative that fits the context of the problem.

• Use the given condition to solve for \( C \) — plugging values into the integrated function permits finding \( C \).
• The condition simplifies locating the specific solution among infinite possibilities.
• Without finding \( C \), your problem solution would remain generally valid but not specifically applicable.

By addressing the constant of integration, you ensure the solution fits the particular scenario described by the problem conditions.