4. Applications of Derivatives

Differentials

4. Applications of Derivatives

Differentials: Videos & Practice Problems

Topic summary

Understanding differentials involves using derivatives to approximate function values. The differential $d y$ can be calculated as $d y = f' (x) d x$ , where $d x$ represents a small change in $x$ . To assess accuracy, absolute error is the difference between the exact and approximate values, while relative error is the ratio of absolute error to the exact value, often expressed as a percentage. This process is crucial for estimating values in real-world applications.

concept

Finding Differentials

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Finding Differentials Video Summary

In calculus, understanding how to find derivatives is essential, and this knowledge extends to the concept of differentials. A differential, denoted as $ dy $ and $ dx $, represents the individual changes in the function and its variable, respectively. To find the differential $ dy $ for a function $ f(x) $, we can use the relationship $ dy = f'(x) \cdot dx $, where $ f'(x) $ is the derivative of the function.

For example, consider the function $ f(x) = x^3 + x $. To find $ dy $ when $ x = 2 $ and $ dx = 0.1 $, we first calculate the derivative $ f'(x) $. Using the power rule, we find:

\[ f'(x) = 3x^2 + 1 \]

Substituting $ x = 2 $ into the derivative gives:

\[ f'(2) = 3(2^2) + 1 = 12 + 1 = 13 \]

Now, we can find $ dy $ by multiplying the derivative by $ dx $:

\[ dy = f'(2) \cdot dx = 13 \cdot 0.1 = 1.3 \]

This means that the differential $ dy $ is 1.3 when $ dx = 0.1 $. Understanding this relationship allows us to approximate function values without direct calculation. For instance, to estimate $ f(2.1) $, we can use the formula:

\[ f(x + dx) \approx f(x) + dy \]

Substituting $ x = 2 $ and $ dx = 0.1 $, we have:

\[ f(2.1) \approx f(2) + dy \]

Calculating $ f(2) $:

\[ f(2) = 2^3 + 2 = 8 + 2 = 10 \]

Thus, we can estimate:

\[ f(2.1) \approx 10 + 1.3 = 11.3 \]

This approximation is useful in real-world applications, especially when dealing with complex functions where direct computation may be cumbersome. By using differentials, we can simplify calculations and obtain estimates that are close to the actual values.

Problem

If $f\left(x\right)=x^3-8x+6$ , find the differential $dy$ when x = 2 and $dx = 0.2$ .

0.8

0.4

3.2

-1.3

Problem

If $f (x) = \sqrt{x + 2}$ , find the differential $d y$ when $x equals 2$ and $d x equals 0.01$ .

0.02

0.0025

2.00

0.025

example

Finding Differentials Example 1

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Finding Differentials Example 1 Video Summary

To find the differential $ dy $ for the function $ f(x) = x^3 - 2x $ when $ x = 2 $ and $ dx = 0.2 $, we start by recalling that the differential $ dy $ is given by the formula:

$ dy = f'(x) \, dx $

First, we need to compute the derivative $ f'(x) $. Using the power rule, we differentiate $ f(x) $:

$ f'(x) = \frac{d}{dx}(x^3) - \frac{d}{dx}(2x) = 3x^2 - 2 $

Now, substituting $ x = 2 $ into the derivative:

$ f'(2) = 3(2^2) - 2 = 3(4) - 2 = 12 - 2 = 10 $

Next, we can calculate $ dy $ by substituting $ f'(2) $ and $ dx $ into the differential formula:

$ dy = f'(2) \cdot dx = 10 \cdot 0.2 = 2 $

Now that we have $ dy = 2 $, we can estimate $ f(2.2) $. The approximation can be expressed as:

$ f(x + dx) \approx f(x) + dy $

In this case, we want to estimate $ f(2.2) $, which can be rewritten as:

$ f(2.2) \approx f(2) + dy $

First, we calculate $ f(2) $:

$ f(2) = 2^3 - 2(2) = 8 - 4 = 4 $

Now, substituting $ f(2) $ and $ dy $ into the approximation:

$ f(2.2) \approx 4 + 2 = 6 $

Thus, the estimated value of $ f(2.2) $ is 6. This method of using differentials allows us to approximate function values effectively.

concept

Determining Error and Relative Error

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Determining Error and Relative Error Video Summary

In mathematics, particularly in fields like engineering and science, approximating values is a common practice, often achieved through the use of differentials. When we approximate a function, it is crucial to assess how close our estimated value is to the actual or exact value. This evaluation is essential for ensuring the accuracy of our measurements.

One method to determine the accuracy of an approximation is by calculating the absolute error, which is defined as the difference between the exact value and the approximate value. The formula for absolute error can be expressed as:

$$ \text{Absolute Error} = | \text{Exact Value} - \text{Approximate Value} | $$

For instance, if we approximate a function and obtain a value of 11.3, we can find the exact value by substituting the relevant input into the function. If the exact value is calculated to be 11.361, we can then compute the absolute error:

$$ \text{Absolute Error} = | 11.361 - 11.3 | = 0.061 $$

A smaller absolute error indicates a more accurate approximation, while a larger value suggests a significant discrepancy between the estimated and actual values.

Another important concept is relative error, which provides a ratio of the absolute error to the exact value. This ratio helps to contextualize the error in relation to the size of the exact value, and it can be calculated using the formula:

$$ \text{Relative Error} = \frac{\text{Absolute Error}}{\text{Exact Value}} $$

Using our previous example, the relative error would be:

$$ \text{Relative Error} = \frac{0.061}{11.361} \approx 0.005369 $$

This small relative error indicates that our approximation was quite accurate. To express this error as a percentage, we can multiply the relative error by 100%:

$$ \text{Percentage Error} = \text{Relative Error} \times 100\% \approx 0.5369\% $$

A low percentage error signifies that the approximation is close to the exact value, reinforcing the reliability of our measurement method. Understanding these concepts—absolute error, relative error, and percentage error—enables mathematicians, engineers, and scientists to evaluate the precision of their approximations effectively.

example

Determining Error and Relative Error Example 1

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Determining Error and Relative Error Example 1 Video Summary

To estimate the value of the function $ f(x) = 8x - x^2 $ at $ x = 3.1 $ using differentials, we start by recognizing that $ f(3.1) $ can be approximated as $ f(3) + dy $, where $ dy $ is the differential of $ f $. The differential $ dy $ is calculated using the formula:

$ dy = f'(x) \cdot dx $

First, we need to find the derivative $ f'(x) $. Using the power rule, we differentiate $ f(x) $:

$ f'(x) = 8 - 2x $

Next, we substitute $ x = 3 $ into the derivative:

$ f'(3) = 8 - 2(3) = 8 - 6 = 2 $

Given that $ dx = 0.1 $, we can now calculate $ dy $:

$ dy = f'(3) \cdot dx = 2 \cdot 0.1 = 0.2 $

Now, we can estimate $ f(3.1) $:

$ f(3.1) \approx f(3) + dy $

To find $ f(3) $, we substitute $ x = 3 $ into the original function:

$ f(3) = 8(3) - 3^2 = 24 - 9 = 15 $

Thus, we have:

$ f(3.1) \approx 15 + 0.2 = 15.2 $

Next, we calculate the absolute error by finding the difference between the actual value $ f(3.1) $ and our approximation. We first compute the actual value:

$ f(3.1) = 8(3.1) - (3.1)^2 = 24.8 - 9.61 = 15.19 $

The absolute error is given by:

$ \text{Absolute Error} = | \text{Actual Value} - \text{Approximation} | = | 15.19 - 15.2 | = 0.01 $

To find the relative error, we use the formula:

$ \text{Relative Error} = \frac{\text{Absolute Error}}{\text{Actual Value}} = \frac{0.01}{15.19} \approx 0.0006583 $

This small relative error indicates that our approximation was quite accurate. In summary, using differentials, we estimated $ f(3.1) $ to be approximately $ 15.2 $, with an absolute error of $ 0.01 $ and a relative error of approximately $ 0.0006583 $.

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What is a differential in calculus?

A differential in calculus refers to the infinitesimally small change in a function's output (dy) corresponding to a small change in the input (dx). It is calculated using the derivative of the function. If y = f(x), then the differential dy is given by dy = f'(x) dx. This concept is used to approximate changes in the function's value and is crucial for applications in various fields such as physics, engineering, and economics.

How do you find the differential dy for a given function?

To find the differential dy for a given function y = f(x), follow these steps: First, compute the derivative of the function, f'(x). Then, multiply the derivative by the small change in x, denoted as dx. The formula is dy = f'(x) dx. For example, if f(x) = x³ + x, then f'(x) = 3x² + 1. If dx = 0.1 and x = 2, then dy = (3(2)² + 1) * 0.1 = 1.3.

What is the difference between absolute error and relative error?

Absolute error is the difference between the exact value and the approximate value, calculated as |exact value - approximate value|. Relative error, on the other hand, is the ratio of the absolute error to the exact value, expressed as a fraction or percentage. It is calculated as (absolute error / exact value). Relative error provides a measure of the accuracy of the approximation in relation to the size of the exact value.

How can differentials be used to approximate function values?

Differentials can be used to approximate function values by using the formula f(x + dx) ≈ f(x) + dy, where dy = f'(x) dx. This method is useful when calculating the exact value is difficult. For example, to approximate f(2.1) for f(x) = x³ + x, use f(2) = 10 and dy = 1.3 (from previous calculations). Thus, f(2.1) ≈ 10 + 1.3 = 11.3.

Why are differentials important in real-world applications?

Differentials are important in real-world applications because they allow for the approximation of values that are difficult to calculate exactly. This is particularly useful in fields such as physics, engineering, and economics, where precise measurements are often challenging. By using differentials, one can estimate changes and make informed decisions based on these approximations, ensuring efficiency and accuracy in practical scenarios.