- 4.2.1 Describe the linear approximation to a function at a point.
- 4.2.2 Write the linearization of a given function.
- 4.2.3 Draw a graph that illustrates the use of differentials to approximate the change in a quantity.
- 4.2.4 Calculate the relative error and percentage error in using a differential approximation.
We have just seen how derivatives allow us to compare related quantities that are changing over time. In this section, we examine another application of derivatives: the ability to approximate functions locally by linear functions. Linear functions are the easiest functions with which to work, so they provide a useful tool for approximating function values. In addition, the ideas presented in this section are generalized later in the text when we study how to approximate functions by higher-degree polynomials Introduction to Power Series and Functions.
Linear Approximation of a Function at a Point
Consider a function that is differentiable at a point Recall that the tangent line to the graph of at is given by the equation
For example, consider the function at Since is differentiable at and we see that Therefore, the tangent line to the graph of at is given by the equation
Figure 4.7(a) shows a graph of along with the tangent line to at Note that for near 2, the graph of the tangent line is close to the graph of As a result, we can use the equation of the tangent line to approximate for near 2. For example, if the value of the corresponding point on the tangent line is
The actual value of is given by
Therefore, the tangent line gives us a fairly good approximation of (Figure 4.7(b)). However, note that for values of far from 2, the equation of the tangent line does not give us a good approximation. For example, if the -value of the corresponding point on the tangent line is
whereas the value of the function at is
In general, for a differentiable function the equation of the tangent line to at can be used to approximate for near Therefore, we can write
We call the linear function
the linear approximation, or tangent line approximation, of at This function is also known as the linearization of at
To show how useful the linear approximation can be, we look at how to find the linear approximation for at
Linear Approximation of
Find the linear approximation of at and use the approximation to estimate
Using a calculator, the value of to four decimal places is 3.0166. The value given by the linear approximation, 3.0167, is very close to the value obtained with a calculator, so it appears that using this linear approximation is a good way to estimate at least for near At the same time, it may seem odd to use a linear approximation when we can just push a few buttons on a calculator to evaluate However, how does the calculator evaluate The calculator uses an approximation! In fact, calculators and computers use approximations all the time to evaluate mathematical expressions; they just use higher-degree approximations.
Find the local linear approximation to at Use it to approximate to five decimal places.
Linear Approximation of
Find the linear approximation of at and use it to approximate
First we note that since rad is equivalent to using the linear approximation at seems reasonable. The linear approximation is given by
We see that
Therefore, the linear approximation of at is given by Figure 4.9.
To estimate using we must first convert to radians. We have radians, so the estimate for is given by
Find the linear approximation for at
Linear approximations may be used in estimating roots and powers. In the next example, we find the linear approximation for at which can be used to estimate roots and powers for real numbers near 1. The same idea can be extended to a function of the form to estimate roots and powers near a different number
Approximating Roots and Powers
Find the linear approximation of at Use this approximation to estimate
The linear approximation at is given by
the linear approximation is given by Figure 4.10(a).
We can approximate by evaluating when We conclude that
Find the linear approximation of at without using the result from the preceding example.
We have seen that linear approximations can be used to estimate function values. They can also be used to estimate the amount a function value changes as a result of a small change in the input. To discuss this more formally, we define a related concept: differentials. Differentials provide us with a way of estimating the amount a function changes as a result of a small change in input values.
When we first looked at derivatives, we used the Leibniz notation to represent the derivative of with respect to Although we used the expressions dy and dx in this notation, they did not have meaning on their own. Here we see a meaning to the expressions dy and dx. Suppose is a differentiable function. Let dx be an independent variable that can be assigned any nonzero real number, and define the dependent variable by
It is important to notice that is a function of both and The expressions dy and dx are called differentials. We can divide both sides of Equation 4.2 by which yields
For each of the following functions, find dy and evaluate when and
The key step is calculating the derivative. When we have that, we can obtain dy directly.
- Since we know and therefore
- Since This gives us
We now connect differentials to linear approximations. Differentials can be used to estimate the change in the value of a function resulting from a small change in input values. Consider a function that is differentiable at point Suppose the input changes by a small amount. We are interested in how much the output changes. If changes from to then the change in is (also denoted and the change in is given by
Instead of calculating the exact change in however, it is often easier to approximate the change in by using a linear approximation. For near can be approximated by the linear approximation
Therefore, if is small,
In other words, the actual change in the function if increases from to is approximately the difference between and where is the linear approximation of at By definition of this difference is equal to In summary,
Therefore, we can use the differential to approximate the change in if increases from to We can see this in the following graph.
We now take a look at how to use differentials to approximate the change in the value of the function that results from a small change in the value of the input. Note the calculation with differentials is much simpler than calculating actual values of functions and the result is very close to what we would obtain with the more exact calculation.
Approximating Change with Differentials
Let Compute and dy at if
The actual change in if changes from to is given by
The approximate change in is given by Since we have
For find and at if
Calculating the Amount of Error
Any type of measurement is prone to a certain amount of error. In many applications, certain quantities are calculated based on measurements. For example, the area of a circle is calculated by measuring the radius of the circle. An error in the measurement of the radius leads to an error in the computed value of the area. Here we examine this type of error and study how differentials can be used to estimate the error.
Consider a function with an input that is a measured quantity. Suppose the exact value of the measured quantity is but the measured value is We say the measurement error is dx (or As a result, an error occurs in the calculated quantity This type of error is known as a propagated error and is given by
Since all measurements are prone to some degree of error, we do not know the exact value of a measured quantity, so we cannot calculate the propagated error exactly. However, given an estimate of the accuracy of a measurement, we can use differentials to approximate the propagated error Specifically, if is a differentiable function at the propagated error is
Unfortunately, we do not know the exact value However, we can use the measured value and estimate
In the next example, we look at how differentials can be used to estimate the error in calculating the volume of a box if we assume the measurement of the side length is made with a certain amount of accuracy.
Volume of a Cube
Suppose the side length of a cube is measured to be 5 cm with an accuracy of 0.1 cm.
- Use differentials to estimate the error in the computed volume of the cube.
- Compute the volume of the cube if the side length is (i) 4.9 cm and (ii) 5.1 cm to compare the estimated error with the actual potential error.
- The measurement of the side length is accurate to within cm. Therefore,
The volume of a cube is given by which leads to
Using the measured side length of 5 cm, we can estimate that
- If the side length is actually 4.9 cm, then the volume of the cube is
If the side length is actually 5.1 cm, then the volume of the cube is
Therefore, the actual volume of the cube is between 117.649 and 132.651. Since the side length is measured to be 5 cm, the computed volume is Therefore, the error in the computed volume is
We see the estimated error is relatively close to the actual potential error in the computed volume.
Estimate the error in the computed volume of a cube if the side length is measured to be 6 cm with an accuracy of 0.2 cm.
The measurement error dx and the propagated error are absolute errors. We are typically interested in the size of an error relative to the size of the quantity being measured or calculated. Given an absolute error for a particular quantity, we define the relative error as where is the actual value of the quantity. The percentage error is the relative error expressed as a percentage. For example, if we measure the height of a ladder to be 63 in. when the actual height is 62 in., the absolute error is 1 in. but the relative error is or By comparison, if we measure the width of a piece of cardboard to be 8.25 in. when the actual width is 8 in., our absolute error is in., whereas the relative error is or Therefore, the percentage error in the measurement of the cardboard is larger, even though 0.25 in. is less than 1 in.
Relative and Percentage Error
An astronaut using a camera measures the radius of Earth as 4000 mi with an error of mi. Let’s use differentials to estimate the relative and percentage error of using this radius measurement to calculate the volume of Earth, assuming the planet is a perfect sphere.
If the measurement of the radius is accurate to within we have
Since the volume of a sphere is given by we have
Using the measured radius of 4000 mi, we can estimate
To estimate the relative error, consider Since we do not know the exact value of the volume use the measured radius to estimate We obtain Therefore the relative error satisfies
which simplifies to
The relative error is 0.06 and the percentage error is
Determine the percentage error if the radius of Earth is measured to be 3950 mi with an error of mi.
Section 4.2 Exercises
What is the linear approximation for any generic linear function
Determine the necessary conditions such that the linear approximation function is constant. Use a graph to prove your result.
Explain why the linear approximation becomes less accurate as you increase the distance between and Use a graph to prove your argument.
When is the linear approximation exact?
For the following exercises, find the linear approximation to near for the function.
For the following exercises, compute the values given within 0.01 by deciding on the appropriate and and evaluating Check your answer using a calculator.
For the following exercises, determine the appropriate and and evaluate Calculate the numerical error in the linear approximations that follow.
For the following exercises, find the differential of the function.
For the following exercises, find the differential and evaluate for the given and
For the following exercises, find the change in volume or in surface area
if the sides of a cube change from 10 to 10.1.
if the sides of a cube change from to
if the radius of a sphere changes from by
if the radius of a sphere changes from by
if a circular cylinder with changes height from 3 cm to
if a circular cylinder of height 3 changes from to
For the following exercises, use differentials to estimate the maximum and relative error when computing the surface area or volume.
A spherical golf ball is measured to have a radius of with a possible measurement error of What is the possible change in volume?
A pool has a rectangular base of 10 ft by 20 ft and a depth of 6 ft. What is the change in volume if you only fill it up to 5.5 ft?
An ice cream cone has height 4 in. and radius 1 in. If the cone is 0.1 in. thick, what is the difference between the volume of the cone, including the shell, and the volume of the ice cream you can fit inside the shell?
For the following exercises, confirm the approximations by using the linear approximation at