Problem 21 - Calculus

Question

Let \(R=f(S)\) and \(f^{\prime}(10)=3.\) (a) For small \(\Delta S,\) write an approximate equation relating \(\Delta R\) and \(\Delta S\) near \(S=10.\) (b) Estimate the change in \(R\) if \(S\) changes from \(S=10\) to \(S=10.2.\) (c) Let \(f(10)=13 .\) Estimate \(f(10.2).\)

Step-by-Step Solution

Answer

(a) \(\Delta R \approx 3 \Delta S\); (b) 0.6; (c) 13.6.

Step 1: Understand the concept

We are given a function \(R = f(S)\) with its derivative at a point: \(f'(10) = 3\). The problem asks to relate \(\Delta R\) (the change in \(R\)) with \(\Delta S\) (the change in \(S\)) for small values close to \(S=10\). According to calculus, for small changes, \(\Delta R \approx f'(S)\Delta S\).

Step 2: Develop the equation for small changes

For small changes \(\Delta S\) near \(S=10\), the change in \(R\) can be approximated by the equation \(\Delta R \approx f'(10)\Delta S\). Here, \(f'(10)=3\), so the relationship is given by \(\Delta R \approx 3 \Delta S\).

Step 3: Compute \(\Delta R\) for the given \(\Delta S\)

Given \(S\) changes from \(S=10\) to \(S=10.2\), \(\Delta S = 10.2 - 10 = 0.2\). Using the equation from Step 2, \(\Delta R \approx 3 \times 0.2 = 0.6\).

Step 4: Estimate \(f(10.2)\)

We know that \(f(10) = 13\). The estimate of \(f(10.2)\) is given by \(f(10.2) \approx f(10) + \Delta R\). Substituting the known values, we get \(f(10.2) \approx 13 + 0.6 = 13.6\).

Key Concepts

Function Approximation

Function approximation is a core idea in calculus, enabling us to estimate the behaviors of functions using simpler representations. Imagine a complex curve that describes the function's behavior. This curve can be difficult to work with directly. Function approximation helps navigate this by providing a simpler version – usually a line or a polynomial – that behaves similarly to the original function over a small interval.

When you have a function \( R = f(S) \), you can approximate this function near a point by using its linear properties. This is especially useful when calculating small changes, where instead of the whole curve, a linear path "approximates" the real function over a small range.

• Use known values and derivatives of the function.
• Focus on small intervals where the approximation holds well.
• Accurate approximations simplify complicated calculations.

Function approximations are widely used to solve real-world problems where a precise model isn’t available or necessary for quick estimates.

Rate of Change

The rate of change is a concept describing how one quantity changes in response to another quantity. In the context of the exercise, when we talk about the rate of change of \( R \) with respect to \( S \), it's how much \( R \) changes when \( S \) changes by a small amount. This is mathematically expressed as the derivative \( f'(S) \).

In simpler terms, if you think of \( R \) and \( S \) like a distance and time respectively, the rate of change is akin to speed, detailing how fast \( R \) is changing as \( S \) progresses. In our example, \( f'(10) = 3 \) implies that for every small increase in \( S \) around 10, \( R \) jumps by approximately 3 times that amount.

• A positive rate of change indicates an increasing trend.
• A negative rate implies a decreasing trend.
• When the rate is zero, the function is not changing at that point.

Understanding the rate of change allows us to predict how outputs respond to shifts in inputs in various functions.

Linearization

Linearization is a strategy used to simplify complex functions into a linear form, which is much easier to analyze and use for calculations. When you linearize a function at a certain point, you are essentially creating a linear approximation of that function around that point.

Think of linearization as drawing a tangent to the curve of the function at a given point. This tangent line represents the function in a linear form near that point. It's especially useful for estimating function values when the changes in the variable are small, as demonstrated in the exercise with the approximate equation \( \Delta R \approx 3 \Delta S \).

• Linearization provides an equation that is straightforward to handle.
• It is particularly effective for small intervals.
• It simplifies the computation of approximate values at nearby points.

Linearization is an essential concept in differential calculus because it bridges the gap between a complex function and our ability to analyze and compute its changes efficiently.