You can see that \(x=a+\d x\).
We use a method called “linear approximation” to estimate the value of a (complicated) function at a given point.
Given a function, a linear approximation is a fancy phrase for something you already know:
The line tangent to the graph of a function at a point is very close to the graph of the function near that point.
This tangent line is the graph of a linear function, called the linear approximation.
Let \(f\) be a function that is differentiable on some interval I that contains the point \(a\).
The graph of a function \(f\) and the line tangent to the curve \(y=f(x)\) at the point
where \(x=a\) are given in the figure below. Find the equation of the tangent line.
![[Picture]](digInLinearApproximation.online-dbd9c8b07a853abb06db5a382a15f56e.svg)
First, find the
expression for \(m\), the slope of the tangent line to the curve \(y=f(x)\) at the point \((a,f(a))\). Select the
correct choice.
\(m= f(a)\) \(m=f'(a)\) \(m= \frac {f(a)-f(0)}{a-0}\) \(m= \frac {f(a+h)-f(a)}{h}\) We don’t have enough information to determine the
slope.
Since we know that the point \((a,f(a))\) lies on the tangent line, we can write an
equation of the tangent line.
\[ y= f(a)+f'(a)(x-a). \]
Now, we define a function, \(L\), by \(L(x)= f(a)+f'(a)(x-a)\). This function is linear and its graph is the line
tangent to the curve \(y=f(x)\) at the point where \(x=a\). This function deserves a special name. If \(f\)
is a function differentiable at \(x=a\), then a linear approximation to the function \(f\) at \(x=a\) is
given by
\[ L(x) =f(a)+ f'(a)(x-a). \]
Note that the graph of \(L\) is just the tangent line to the graph of \(f\) at \(x=a\).
A linear approximation of \(f\) is a “good” approximation as long as \(x\) is “not too far” from \(a\). If one “zooms in” on the graph of \(f\) sufficiently, then the graphs of \(f\) and \(L\) are nearly indistinguishable.
As a first example, we will see how linear approximations allow us to approximate “difficult” computations.
Let \(f\) be a function defined by
![[Picture]](digInLinearApproximation.online-f7516818cf03619df9fbaeec15077543.svg)
From the
picture we can see that
\[ f(x) =\sqrt [3]{x}.\]
Approximate \(\sqrt [3]{50}\), using \(L\), a linear approximation to the
function \(f\) at \(a=64\).
To start, write ![[Picture]](digInLinearApproximation.online-7a280543f740f0edc358b4d62dd03fd1.svg)
Now we
evaluate \(L(50)= \frac {50}{48} +\frac {8}{3}\). This was easy to compute, since \(L\) is a linear function! Using a calculator we
can see that \(L(50)= \frac {50}{48} +\frac {8}{3}\approx 3.71\) and that \(\sqrt [3]{50}\approx 3.68\). Therefore, \(L(50)\) is not only easy to evaluate, but also gives us a
close approximation of \(\sqrt [3]{50}\).
\[ \ddx f(x) = \ddx x^{1/3} = \frac {1}{3x^{\answer [given]{2/3}}}. \]
\begin{align*} L(x) &= \answer [given]{4}+ \frac {1}{3\cdot 64^{2/3}} (x-64) \\ &=4+ \frac {1}{\answer [given]{48}} (x-64) \\ &= \frac {x}{48} +\frac {8}{3}. \end{align*}
What would happen if we chose \(a=27\) instead?
Then we would use \(L_{27}\), the linear approximation to the function \(f\) at \(a=27\). In that case, \(L_{27}=f(27)+f'(27)(x-27)=3+\frac {1}{27}(x-27)\). The graph of \(L_{27}\), together with the graphs of \(f\) and \(L=L_{64}\) is given in the figure below.
\(L_{27}(50)>L_{64}(50)>f(50)\).
So, our choice, \(a=64\), was better!
With modern calculators and computing software, it may not appear necessary to use linear approximations. In fact they are quite useful. In cases requiring an explicit numerical approximation, they allow us to get a quick rough estimate which can be used as a “reality check” on a more complex calculation. In some complex calculations involving functions, the linear approximation makes an otherwise intractable calculation possible, without serious loss of accuracy.
Use a linear approximation of \(f(x) =\sin (x)\) at \(a=0\) to approximate \(\sin (0.3)\).
To start, write ![[Picture]](digInLinearApproximation.online-0e12a541627f7bbaf5e81d25252390b5.svg)
Hence, a linear
approximation for \(\sin (x)\) at \(a=0\) is \(L(x) = x\), and so \(L(0.3) = 0.3\). Comparing this to \(\sin (.3) \approx 0.295\), we see that the approximation
is quite good. For this reason, it is common to approximate \(\sin (x)\) with its linear
approximation \(L(x) = x\) when \(x\) is near zero.
\[ \ddx f(x) = \answer [given]{\cos (x)}, \]
so our linear
approximation is \begin{align*} L(x) &= 0+\answer [given]{\cos (0)}\cdot (x-0)\\ &= x. \end{align*}
1 Overestimates or Underestimates
Remember that geometrically, the actual function value corresponds to the \(y\)-coordinate of a point on the graph of the function and the approximate value corresponds to the \(y\)-coordinate of a point on the graph of the tangent line. We have terminology to describe the sitation when the graph of the function is above or below the tangent line: Concavity.
Suppose that \(L\) is the linear approximation of a differentiable function \(f\) at \(a\).
Suppose \(b\) is a different \(x\)-value near \(a\) and \(I\) is the open interval between \(a\) and \(b\).
Then:
- If \(f\) is concave up on the interval \(I\), then the approximation value \(L(b)\) is an underestimate of \(f(b)\).
- If \(f\) is concave down on the interval \(I\), then the approximation value \(L(b)\) is an overestimate of \(f(b)\).
To determine whether \(L(b)\) is an overestimate or an underestimate, it is not enough to
check the value of \(f''\) at a single point. It’s the concavity over the entire interval
between the points \(a\) and \(b\) that has to be checked.
Approximate the value of \((0.9)^3\). Is this an overestimate or an underestimate?
To make a linear approximation, we need to determine a function and the point \(a\) to
use. In this example, it would make sense to use \(f(x) = x^3\) for the function which means the
question is asking for us to approximate \(f(0.9)\).
For the point \(a\) we need a value near \(0.9\) that we can calculate \(f(a)\) and \(f'(a)\). Using \(a=1\) works for this since \(f(1) = \answer [given]{1}\) and \(f'(x) = 3x^2\) so \(f'(1) = \answer [given]{3}\).
The linear approximation function is \(L(x) = 3(x-1)+1\). That makes \(L(0.9) = \answer [given]{0.7}\).
That is our approximation. Now we need to determine whether this value was an overestimate or an underestimate. \(f''(x) = 6x\) which is positive in the interval \((0, \infty )\). That means \(f\) is concave up on the interval \((0.9, 1)\) (since this interval is inside \((0,\infty )\)).
The approximation value is \(L(0.9)=0.7\) which is an underestimate because \(f\) is concave up on the interval \((0.9,1)\).
Notice that we only indicated the interval \((0.9, 1)\) in this answer. Even though the function \(f\) is concave up on the entire \((0, \infty )\), it is only the interval \((0.9, 1)\) that is relevant to this question, so that is what was included in the response.
2 Differentials
The graph of a function \(f\) and the graph of \(L\), the linear approximation of \(f\) at \(a\), are shown in the figure below. Also, two quantities, \(\d x\) and \(\d f\), and a point \(P\) are marked in the figure. Look carefully at the figure when answering the questions below.
![[Picture]](digInLinearApproximation.online-87f0928e95f2f710bc0a262e788f4242.svg)
Select all
the correct expressions for the quantity \(\d x\).
Select all the correct expressions for the quantity \(\d f\).
You can see that \(L(x)=f(a)+\d f\).
Recall: \(L(a)=f(a)\).
Based on the figure and the expression for \(L(x)\), select all the correct expressions for \(\d f\).
Recall: \(\d f=L(x)-f(a)=f(a)+f'(a)(x-a)-f(a)=f'(a)(x-a)=f'(a)\d x\).
So, we can write \(\d f=f'(a)\d x\) and call it a differential of \(f\) at \(a\). Notice that we can define a differential at any point \(x\) of the domain of \(f\), provided that \(f'(x)\) exists.
Students frequently confuse the differential \(df\) with the derivative \(f'(a)\). Remember that the derivative tells us the rate of change at the point, while the differential \(df\) is an estimated amount of change.
We will do that in our next definition.
Let \(f\) be a differentiable function, let \(x\) be a
point in the domain of \(f\), and let \(\d x\) be some quantity, called a differential of \(x\). We
define \(\d f\), a differential of \(f\), at a point \(x\) by
![[Picture]](digInLinearApproximation.online-ab3aab72b1de35665c27379bb74be345.svg)
\[ \d f=f'(x)\cdot \d x \]
Geometrically, differentials can be interpreted via the diagram below.
Note, it is now the case (by definition!) that
\[ f'(x)=\frac {\d f}{\d x}. \]
We should not be surprised, since the slope of the tangent line in the figure is \(f'(x)\), and this slope is also given by \(\frac {\d f }{\d x}\).
Essentially, differentials allow us to solve the problems presented in the previous examples from a slightly different point of view. Recall, when \(h\) is near but not equal zero,
\[ f'(x) \approx \frac {f(x+h)-f(x)}{h}. \]
Hence,
\[ f'(x)h \approx f(x+h)-f(x). \]
We can replace a quantity \(h\) with a quantity \(\d x\) to write
\begin{align*} f'(x)\cdot \d x &\approx f(x+\d x)-f(x)\\ \d f &\approx f(x+\d x)-f(x). \end{align*}
Adding \(f(x)\) to both sides we see
\[ f(x) + \d f\approx f(x+\d x) \]
or, equivalently
\[ f(x+\d x)\approx f(x) + \d f . \]
There are contexts where the language of differentials is common. Here is the basic strategy:
![[Picture]](digInLinearApproximation.online-0c27bddf9a54813842df1c7377f674b3.svg)
We will repeat our previous examples using differentials.
Use differentials to approximate \(\sqrt [3]{50}\).
Set \(f(x) = \sqrt [3]{x}\). We want to know \(\sqrt [3]{50}\). Since \(4^3 = 64\), we set \(x=64\). Setting \(\d x=50-64=-14\),
we have
\begin{align*} \sqrt [3]{50} = f(x + \d x) &\approx f(x) + \d f\\ &\approx \sqrt [3]{64} + \d f. \end{align*}
Here we see a plot of \(y=\sqrt [3]{x}\) with the differentials above marked:
Now we must compute \(\d f\):![[Picture]](digInLinearApproximation.online-22ca46d612d58957ada7e90135f00ddf.svg)
\begin{align*} \d f &= f'(x) \cdot \d x\\ &= \answer [given]{\frac {1}{3x^{2/3}}} \cdot \d x\\ &= \frac {1}{3\cdot 64^{2/3}} \cdot (\answer [given]{-14})\\ &= \frac {1}{3\cdot 64^{2/3}} \cdot (-14)\\ &= \frac {-7}{24} \end{align*}
Hence \(f(50) \approx f(64) + \frac {-7}{24} \approx 3.71\).
Use differentials to approximate \(\sin (0.3)\).
Set \(y = \sin (x)\). We want to know \(\sin (0.3)\). Since \(\sin (0) = 0\), we will set \(x = 0\) and \(\d x=0.3\).
Write with me
\begin{align*} \sin (0.3) = \sin (x + \d x) &\approx \sin (x) + \d y\\ &\approx 0 + \d y. \end{align*}
Here we see a plot of \(y=\sin (x)\) with the differentials above marked:
Now we must compute \(\d y\):![[Picture]](digInLinearApproximation.online-54a69f6186a478e65d59e42a08b36d5b.svg)
\begin{align*} \d y &= \left (\ddx \sin (x)\right ) \cdot \d x\\ &=\answer [given]{\cos (0)} \cdot \d x\\ &= 1 \cdot (\answer [given]{0.3})\\ &= 0.3 \end{align*}
Hence \(\sin (0.3) \approx \sin (0) + 0.3 \approx 0.3\).
The upshot is that linear approximations and differentials are simply two slightly different ways of doing the exact same thing.
3 Error approximation
Differentials also help us estimate error in real life settings.
A glass containing water is shown in the figure below.
![[Picture]](digInLinearApproximation.online-c9a29b3bfe0b6de9399a22a9c6cb2ec5.svg)
Let \(y\) denote the
height of water (measured in centimeters). The volume of the water in the glass (in
milliliters), as a function of \(y\), is given by
\[ V(y) = \frac {2\pi y^{3/2}}{\sqrt {3}}. \]
There is a mark indicating when the glass is filled to \(250\) ml at \(16.8\) cm from the base of the glass. If the glass is filled within \(\pm 2\) millimeters of the mark, what are the bounds on the volume?
We want to know what a small change in the height, \(y\), does to the volume \(V\).
These small changes can be modeled by the differentials \(\d V\) and \(\d y\). Since
\[ \d V = V'(y) \d y \]
and \(V'(y) = \answer [given]{\pi \sqrt {3 y}}\) we use the
fact that \(\d y = \pm 0.2\) with \(y=\answer [given]{16.8}\) to see \[ \d V = \answer [given]{\pi \sqrt {3 \cdot 16.8}} \cdot 0.2. \]
Hence the volume will vary by roughly \(\pm \answer [given,tolerance=.01]{4.46062}\) milliliters. 4 New and old friends
You might be wondering, given a plot \(y=f(x)\),
What’s the difference between \(\Delta x\) and \(\d x\)? What about \(\Delta y\) and \(\d y\)?
Regardless, it is now a pressing question. Here’s the deal:
\[ \frac {\Delta y}{\Delta x} \]
is the average rate of change of \(y=f(x)\) with respect to \(x\). On the other hand:
\[ \frac {\d y}{\d x} \]
is the instantaneous rate of change of \(y=f(x)\) with respect to \(x\). Essentially, \(\Delta x\) and \(\d x\) are the same type of thing, they are (usually small) changes in \(x\). However, \(\Delta y\) and \(\d y\) are very different things.
- \(\Delta y=f(x+\Delta x)-f(x)\); it is the change in \(y=f(x)\) associated to \(\Delta x\).
-
\(\d y=L(x+\d x)-L(x)\), it is the change in \(y=L(x)\) associated to \(\Delta x=\d x\).
\[ \d y =f'(x)\d x \]Note: \( L(x+\d x)= f(x)+f'(x)\d x\).So, the change
\begin{align*} \d y &= L(x+\d x)-L(x)\\ &= f(x)+f'(x)\d x-L(x)\\ &= f(x)+f'(x)\d x-f(x)\\ &=f'(x)\d x \end{align*}
Suppose \(f(x) = x^2\). If we are at the point \(x=1\) and \(\Delta x =\d x = 0.1\), what is \(\Delta y\)? What is \(\d y\)?
\( \Delta y=f(1+\Delta x)-f(1)=f(1.1)-f(1)\)
\(\d y=f'(1)\cdot \d x=f'(1)\cdot 0.1\)
\[ \Delta y = \answer {0.21}\qquad \d y = \answer {0.2} \]
Differentials can be confusing at first. However, when you master them, you will have a powerful tool at your disposal.