Problem 27 - Calculus

Question

Find \(c\) in \(G-I \Leftrightarrow c(\Delta x)^{4}\left[y^{\prime \prime \prime}(1)-y^{\prime \prime \prime}(-1)\right]\) by taking \(y=x^{4}, \Delta x=2,\) and \(G=(-1 / \sqrt{3})^{4}+(1 / \sqrt{3})^{4}\)

Step-by-Step Solution

Answer

\(c = \frac{1}{3456}\) assuming \(I = 0\).

Step 1: Substitute Given Values

We are given the functions and values: \(y = x^4\), \(\Delta x = 2\), and \(G = (-1 / \sqrt{3})^4 + (1 / \sqrt{3})^4\). First, calculate \(G\).

Step 2: Calculate \(G\)

Substituting values into the equation for \(G\):\[G = \left(-\frac{1}{\sqrt{3}}\right)^4 + \left(\frac{1}{\sqrt{3}}\right)^4\]Since both terms are the same, we calculate one and multiply by 2:\[\left(\frac{1}{\sqrt{3}}\right)^4 = \frac{1}{(\sqrt{3})^4} = \frac{1}{9}\]Thus,\[ G = 2 \times \frac{1}{9} = \frac{2}{9}\].

Step 3: Differentiate \(y = x^4\)

Differentiate \(y = x^4\) to find the third derivative:\[y^{\prime}(x) = 4x^3\]\[y^{\prime\prime}(x) = 12x^2\]\[y^{\prime\prime\prime}(x) = 24x\].

Step 4: Calculate \(y^{\prime\prime\prime}(1)\) and \(y^{\prime\prime\prime}(-1)\)

Substitute \(1\) and \(-1\) into \(y^{\prime\prime\prime}(x)\):\[y^{\prime\prime\prime}(1) = 24 \times 1 = 24\]\[y^{\prime\prime\prime}(-1) = 24 \times (-1) = -24\].

Step 5: Compute Expression Involving Derivatives

Calculate \(y^{\prime\prime\prime}(1) - y^{\prime\prime\prime}(-1)\):\[24 - (-24) = 24 + 24 = 48\].

Step 6: Substitute Everything Back into Expression for \(c\)

The expression given is:\[c(\Delta x)^4[y^{\prime\prime\prime}(1) - y^{\prime\prime\prime}(-1)] = G - I\]We know \(G = \frac{2}{9}\) and \(\Delta x = 2\), then:\[c(2)^4(48) = \frac{2}{9} - I\].

Step 7: Solve for \(c\)

Since the exercise doesn't provide the value of \(I\), assume \(I = 0\) to isolate \(c\):\[c \times 16 \times 48 = \frac{2}{9}\]So, solve for \(c\):\[c = \frac{2}{9 \times 16 \times 48}\]\[c = \frac{2}{6912}\]\[c = \frac{1}{3456}\].

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus. This mathematical operation is used to calculate the derivative of a function. When finding a derivative, you are looking into how a function changes as the input changes. In simpler terms, it allows you to find the rate at which one quantity changes with respect to another.
The first derivative of a function gives you the slope of the tangent line at any point on the curve. This tells you the rate of change of the function. Higher-order derivatives, like the second and third, give more information about the function's curvature and behavior.
In the given problem, we started by differentiating the function \( y = x^4 \). To find the third derivative, we followed a specific order of differentiation: first derive once to determine the slope, then again to uncover more about how the slope changes, and a third time to see how these changes evolve further. Each derivative builds upon the previous, helping dissect the function's full dynamic behavior.

Third Derivative

The third derivative of a function provides insights into the rate of change of the rate of change of the rate of change. Let’s break it down: if the first derivative gives the function's slope, and the second derivative gives the curvature (how the slope changes), then the third derivative tells you how the curvature itself is changing.
For the function\( y = x^4 \), the process followed was:

• First derivative: \( y' = 4x^3 \)
• Second derivative: \( y'' = 12x^2 \)
• Third derivative: \( y''' = 24x \)

The third derivative here, \( 24x \), reveals how quickly the parabola itself is steepening or flattening as \( x \) changes. This derivative is essential in exercises where detailed information about the function's behavior is necessary, such as finding \( y'''(1) \) and \( y'''(-1) \) in our problem.

Function Substitution

Function substitution is a technique often used in calculus to simplify expressions or solve equations. It involves replacing variables with given values or expressions to make the calculations manageable.
In the context of our exercise, we substituted specific values into different parts of the expression to find \( G \) and relate it to other components. Substituting \( y = x^4 \) directly impacts the derivatives calculated and how these are used later on.

• For \( G = \left(-\frac{1}{\sqrt{3}}\right)^4 + \left(\frac{1}{\sqrt{3}}\right)^4 \), substituting helped us compute that \( G = \frac{2}{9} \).
• Similarly, substituting \( 1 \) and \(-1\) into the third derivative \( y''' = 24x \) provided the values \( y'''(1) = 24 \) and \( y'''(-1) = -24 \).

Substitution here allowed us to break a complex problem into simpler, easily computable parts, gradually building toward the final solution.

Step-by-Step Solution

The step-by-step solution method is an accessible approach to solving mathematical problems. It helps in visualizing each step clearly, making complex tasks simple and more understandable.
In the original exercise, we started by substituting given values and calculating individual components separately. Step-by-step approaches included:

• Computing \( G \) by substituting and simplifying given numbers.
• Differentiating \( y = x^4 \) to get the third derivative.
• Substituting into the derivatives to find specific values \( y'''(1) \) and \( y'''(-1) \).
• Substituting everything back to solve for the unknown \( c \).

This systematic strategy ensures no detail is overlooked, and students can follow the logic used at each stage. Such clarity is beneficial when tackling similar problems in calculus.