Problem 55 - Calculus Single Variable

Question

A function \(f,\) a point \(c,\) an increment \(\Delta x,\) and a positive integer \(n\) are given. Use the method of increments to estimate \(f(c+\Delta x)\). Then let \(h=\Delta x\) / \(N\). Use the method of increments to obtain an estimate \(y_{1}\) of \(f(c+h) .\) Now, with \(c+h\) as the base point and \(y_{1}\) as the value of \(f(c+h),\) use the method of increments to obtain an estimate \(y_{2}\) of \(f(c+2 h)\). Continue this process until you obtain an estimate \(y_{N}\) of \(f(c+N \cdot h)=f(c+\Delta x) .\) We say that we have taken \(N\) steps to obtain the approximation. The number \(h\) is said to be the step size. Use a calculator or computer to evaluate \(f(c+\Delta x)\) directly. Compare the accuracy of the one step and \(N\) -step approximations. $$ f(x)=x^{1 / 3}, c=27, \Delta x=0.9, N=3 $$

Step-by-Step Solution

Answer

The direct calculation gives 3.026, while the 3-step increment method gives 3.0308, closer than the one-step approximation 3.0111.

Step 1: Calculate Step Size

To begin, calculate the step size \( h \) by using the formula \( h = \frac{\Delta x}{N} \). Given that \( \Delta x = 0.9 \) and \( N = 3 \), we find:\[ h = \frac{0.9}{3} = 0.3 \]

Step 2: Estimate f(c + Δx) directly

Calculate \( f(c + \Delta x) = f(27 + 0.9) = f(27.9) \) using the function \( f(x) = x^{1/3} \):\[ f(27.9) = 27.9^{1/3} \approx 3.026 \]

Step 3: Calculate f(c)

Find the value of \( f(c) \), which is the starting point of our increment process, where \( c = 27 \):\[ f(27) = 27^{1/3} = 3 \]

Step 4: Estimate y1 using one step

Using the method of increments, estimate \( y_1 = f(c + h) \) where \( c = 27 \) and \( h = 0.3 \):Calculate the derivative \( f'(x) = \frac{1}{3} x^{-2/3} \). Evaluate \( f'(27) \) using the initial value:\[ f'(27) = \frac{1}{3} (27)^{-2/3} = \frac{1}{3} \times \frac{1}{9} = \frac{1}{27} \] Then estimate \( f(27 + 0.3) \) using the formula:\[ y_1 = f(27) + f'(27) \times 0.3 = 3 + \frac{1}{27} \times 0.3 = 3 + 0.0111 = 3.0111 \]

Step 5: Estimate y2 using increments

Now estimate \( y_2 = f(c + 2h) \) using \( y_1 \) as the base point:Recalculate \( f'(27.3) \): \[ f'(27.3) = \frac{1}{3} (27.3)^{-2/3} \approx \frac{0.099}{\underline{\phantom{xx}}} \] Then calculate:\[ y_2 = y_1 + f'(27.3) \times 0.3 \approx 3.0111 + 0.0099 \approx 3.021 \]

Step 6: Estimate y3 using increments

Finally, calculate \( y_3 = f(c + 3h) \), using \( y_2 \) as the base point:Recalculate \( f'(27.6) \):\[ f'(27.6) = \frac{1}{3} (27.6)^{-2/3} \approx \frac{0.0983}{\underline{\phantom{xx}}} \] Then calculate:\[ y_3 = y_2 + f'(27.6) \times 0.3 \approx 3.021 + 0.0098 \approx 3.0308 \]

Step 7: Compare Results

We have two values to compare. The direct computation yields \( f(27.9) \approx 3.026 \), while our increment method over \( N = 3 \) steps gives \( y_3 = 3.0308 \). The increment method over one step gives \( y_1 = 3.0111 \).

Key Concepts

Method of Increments

The method of increments is a powerful tool used in calculus for approximating the value of a function at a given point by taking small steps from a known base point. Imagine you are on a path and you want to know what the landscape looks like after a short walk. By taking small steps and considering tiny changes in the function's value, you can estimate what the function's output will be at a new point.

• Start at a known point, such as where the function is already calculated.
• Use increments, small steps of size \( h \), to move towards the point you want to evaluate.
• This technique uses function values and derivatives at the known point.

By making these incremental changes, the method provides a step-by-step way to estimate values, making it especially useful in understanding complex functions.

Derivatives

In calculus, derivatives play a crucial role in analyzing how functions change. Think of a derivative as a measure of change or a tool that tells us the tendency of a function to increase or decrease at a certain point. Simply put, it is like knowing the speed on a speedometer while driving.

• The derivative of a function at a point gives the rate of change at that point.
• For the function \( f(x) = x^{1/3} \), the derivative \( f'(x) = \frac{1}{3}x^{-2/3} \) shows how rapidly the cube root function changes as \( x \) varies.
• By evaluating the derivative at specific points, you can predict how the function behaves near those points.

Using derivatives, we can modify our understanding of the function's shape and make accurate approximations in small steps.

Function Approximation

Function approximation is a technique in calculus that allows us to estimate the value of a function when exact computation is difficult or impossible. It involves using simpler functions, like linear or polynomial representations, to mimic the behavior of more complicated ones.

• The approximation allows us to "predict" or "guess" the function's value at new points.
• By using the method of increments, we linearize the function, which helps us estimate the output efficiently without needing to compute complex operations.
• This technique is particularly beneficial when direct evaluation is computationally intensive.

The ability to approximate functions accurately is essential in fields like engineering and physics, where exact solutions might be hard to come by.

Numerical Methods

Numerical methods encompass a broad area of techniques in mathematics, designed to provide approximate solutions to problems involving continuous variables. These methods come into play when dealing with problems that are challenging to solve analytically.

• These methods allow for calculations that are otherwise too complex for simple arithmetic.
• Numerical methods use stepwise iteration, breaking down problems into manageable parts.
• For example, in our exercise, the method of increments is a type of numerical method that helps estimate the function by considering incremental steps.

Numerical methods are essential, providing solutions where traditional algebraic methods fall short, and allowing computers to solve complex calculus problems swiftly and efficiently.