Problem 20 - Calculus

Question

$$\text { Let }\left.\frac{d V}{d r}\right|_{r=2}=16$$. (a) For small \(\Delta r,\) write an approximate equation relating \(\Delta V\) and \(\Delta r\) near \(r=2\) (b) Estimate \(\Delta V\) if \(\Delta r=0.1\) (c) Let \(V=32\) when \(r=2 .\) Estimate \(V\) when \(r=2.1\)

Step-by-Step Solution

Answer

\(\Delta V \approx 1.6\) when \(\Delta r=0.1\); estimated \(V\) at \(r=2.1\) is 33.6.

Step 1: Understand the Given Derivative

We are given that \( \left.\frac{d V}{d r}\right|_{r=2}=16 \). This means that the rate of change of \( V \) with respect to \( r \) at \( r=2 \) is 16. This is important for setting up the equation relating \( \Delta V \) and \( \Delta r \).

Step 2: Write the Approximate Relationship for \( \Delta V \) and \( \Delta r \)

For small changes in \( r \), the change in \( V \) can be approximated by the derivative: \( \Delta V \approx \frac{d V}{d r} \cdot \Delta r \). At \( r=2 \), this becomes \( \Delta V \approx 16 \cdot \Delta r \).

Step 3: Estimate \( \Delta V \) for \( \Delta r = 0.1 \)

Using the equation from Step 2, substitute \( \Delta r = 0.1 \) into \( \Delta V \approx 16 \cdot \Delta r \). Thus, \( \Delta V \approx 16 \cdot 0.1 = 1.6 \).

Step 4: Calculate New Volume when \( r=2.1 \)

Since we are given \( V=32 \) when \( r=2 \), the estimated new volume when \( r=2.1 \) is \( V + \Delta V = 32 + 1.6 = 33.6 \).

Key Concepts

Derivative

The derivative is a fundamental concept in differential calculus and expresses the rate at which one variable changes with respect to another. It is commonly denoted as \( \frac{dV}{dr} \) in calculus, symbolizing the rate of change of \( V \) with respect to \( r \). This can be imagined like the speed at which something is changing. For example, if the derivative is high at a certain point, it means that the quantity \( V \) is changing quickly as \( r \) changes.
In our exercise, the derivative \( \left.\frac{dV}{dr}\right|_{r=2} = 16 \) tells us that around \( r=2 \), the volume \( V \) is increasing at a rate of 16 units for every one unit increase in \( r \). Thus, it provides an understanding of how the volume is behaving without needing to know the exact formula for \( V \). Understanding derivatives is crucial, as they help us predict changes and make calculations simpler when dealing with small differences.

Rate of Change

Rate of change describes how one quantity changes in relation to another. In our problem, it is shown through the derivative \( \frac{dV}{dr} \), and it explains how volume \( V \) changes when the radius \( r \) changes. The rate of change helps in identifying whether a quantity is increasing or decreasing and how rapidly that change is taking place.
For instance, a rate of change of 16 at \( r=2 \) indicates that the volume increases by 16 cubic units as the radius increases by 1 unit. This tells us that even small changes in \( r \) can have a significant impact on the volume \( V \). By understanding this rate, we can make informed predictions about how adjustments in radius will affect the volume, which is vital in scientific and engineering applications.

Approximation

Approximation in calculus allows us to estimate values for functions without needing an exact solution. For small changes in the independent variable, the derivative provides a good approximation of the change in the dependent variable.The formula \( \Delta V \approx \frac{dV}{dr} \cdot \Delta r \) is an example of using this concept for approximation.
In our step-by-step solution, this is used to approximate the change in volume \( \Delta V \) when the radius changes by a small amount \( \Delta r \). We calculated \( \Delta V \approx 16 \cdot 0.1 = 1.6 \). This approximation is useful when an exact calculation is difficult or unnecessary, allowing for quicker and often sufficiently accurate solutions to real-world problems.

Small Changes

Small changes are vital in understanding calculus, as they are the foundation of approximations. When changes in the independent variable \( r \) are small, the changes in \( V \) can predictably be approximated with ease using derivatives. This is because the derivative gives a linear estimate which is most accurate over small intervals.
For example, with \( \Delta r = 0.1 \), the small change results in a relatively small \( \Delta V \approx 1.6 \). Such approximations assume linear behavior close to the point of calculation and are quite precise enough for practical use cases. Small changes provide a straightforward way to estimate large complexities by focusing on tiny, understandable parts.