/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 Find the rate of change of \(y\)... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the rate of change of \(y\) with respect to \(x\) at the given value of \(x\). \(y=2 x^{3}+2 ; \quad x=2\)

Short Answer

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The rate of change of \(y\) with respect to \(x\) at \(x=2\) is 24.

Step by step solution

01

Differentiate the function with respect to \(x\)

The given function is \(y = 2x^3 + 2\). To find the derivative with respect to \(x\), we will apply the power rule of differentiation to each term of \(y\): \(\frac{dy}{dx} = \frac{d}{dx} (2x^3 + 2)\) The power rule states that \(\frac{d}{dx} (x^n) = nx^{n-1}\), where \(n\) is a constant. Applying this rule to each term, we get: \(\frac{dy}{dx} = 2\cdot \frac{d}{dx} (x^3) + \frac{d}{dx} (2)\) Now, differentiating each term with respect to \(x\): \(\frac{dy}{dx} = 2(3x^2) + 0\) Simplifying, we have: \(\frac{dy}{dx} = 6x^2\)
02

Find the rate of change at the given value of \(x\)

The given value of \(x\) is 2. To find the rate of change of \(y\) with respect to \(x\) at this point, we plug in \(x=2\) into the derivative \(\frac{dy}{dx}\): \(\left.\frac{dy}{dx}\right|_{x=2} = 6(2^2)\) Calculating the value: \(\left.\frac{dy}{dx}\right|_{x=2} = 6(4) = 24\) So, the rate of change of \(y\) with respect to \(x\) at \(x=2\) is 24.

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Key Concepts

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

Understanding the Rate of Change
The rate of change is a concept that describes how one quantity changes in relation to another. When you find the rate of change of a function, you are essentially discovering how quickly the dependent variable responds to changes in the independent variable. For example, in the exercise, we are looking at the function \(y = 2x^3 + 2\) and are asked to find how \(y\) changes as \(x\) changes when \(x = 2\).

In mathematics, the rate of change is often represented by the derivative. The derivative of a function at a given point provides the slope of the tangent line at that point, symbolizing the instantaneous rate of change. In this case, when \(x = 2\), the rate at which \(y\) changes with respect to \(x\) is the derivative of \(y\) evaluated at that point, which tells us how \(y\) is increasing or decreasing as \(x\) increments by a tiny amount.

Therefore, the rate of change at a specific point offers a precise snapshot of the behavior of the function at that instant, making it a powerful tool for understanding trends and making predictions based on functional relationships.
Applying the Power Rule
One of the most efficient techniques in differentiation is the power rule. It is particularly useful when dealing with polynomial functions. The power rule states that if you have a term in the form of \(x^n\), its derivative will be \(nx^{n-1}\), where \(n\) is a constant exponent. This method simplifies the differentiation process by offering a straightforward, consistent formula to apply.

In the given problem, the function \(y = 2x^3 + 2\) can be differentiated using the power rule. By treating \(2x^3\) as a separate term, you apply the power rule: multiplying the coefficient 2 by the derivative of \(x^3\), which is \(3x^2\), resulting in \(6x^2\).
  • Differentiate \(x^3\) to get \(3x^2\)
  • Multiply by the coefficient 2 to obtain \(6x^2\)
The constant \(2\) differentiates to zero, as constants don't change regardless of \(x\). Thus, the derivative \(\frac{dy}{dx} = 6x^2\) provides the rate of change of the function \(y\) concerning \(x\).

Utilizing the power rule simplifies the calculation, turning what could be a cumbersome task into a quick and efficient process. This makes it a fundamental tool in single-variable calculus.
Introduction to Single Variable Calculus
Single variable calculus is a branch of mathematics focusing on functions of a single independent variable. This stream is often seen as the foundation for more advanced calculus concepts. Its primary focus is on understanding concepts of limits, derivatives, and integrals with respect to one variable.

In the context of the given exercise, how a simple polynomial function such as \(y = 2x^3 + 2\) behaves as \(x\) changes demonstrates prime applications of single variable calculus. Here, the derivative of the function concerning \(x\) epitomizes how calculus aids in determining the rate of change.
  • **Limits**: Establish how functions behave near certain points.
  • **Derivatives**: Measure how functions change, central to finding rates of change.
  • **Integrals**: Determine the area under curves, useful for cumulative measurements.
Each of these components is crucial for analyzing real-world scenarios where relationships between quantities need to be accurately modeled and understood. The study of single variable calculus thus opens doors to a thorough comprehension of such dynamics within and beyond mathematics.

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Most popular questions from this chapter

Current in a Circuit A series \(R L\) circuit including a resistor \(R\) and inductance \(L\) is shown in the schematic. Suppose that the electromotive force \(E(t)\) is \(V\) volts, the resistance is \(R\) ohms, and the inductance is \(L\) henries, where \(V, R\), and \(L\) are positive constants. Then the current at time \(t\) is given by $$ I(t)=\frac{V}{R}\left(1-e^{-R t / L}\right) $$ amperes. Using l'Hôpital's Rule, evaluate \(\lim _{R \rightarrow 0^{+}} I\) to find an expression for the current in a circuit in which the resistance is 0 ohms.

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Two families of curves are orthogonal trajectories of each other if every curve of one family is orthogonal to every curve in the other family. In Exercises \(93-96\), (a) show that the given families of curves are orthogonal to each other, and (b) sketch a few members of each family on the same set of axes. $$ 2 x^{2}+y^{2}=c, \quad y^{2}=k x, \quad c, k \text { constants } $$

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