We can use derivatives to estimate how a function behaves around a certain point. With linear approximation, we can analyse these small variations and predict values. These methods are useful in fields like physics and engineering, where precise measurements and predictions are crucial, but often challenging to obtain directly.
Sometimes a small change in one variable can cause a big change in another value. For example, a small increase (or error) in the radius of a sphere means a lot more volume is added. If you estimate the small error in one variable, you can calculate the significant change in a larger variable by using derivatives. This involves linear approximation using tangents.
Approximating functions using the tangent
The curve of any function that we can differentiate will appear linear if we look at it over a small enough interval.
If \(a\) is a point within that interval, we can use the tangent line through \((a,f(a))\) to get an approximation of values of \(f(x)\), as long as \(x\) remains near point \(a\). This forms the basis of linear approximation.
The tangent line is defined by the equation:
\[y=f(a)+f'(a)(x-a)\]
When we use this to approximate a function, we get the linear approximation formula:
\[ L(x)=f(a)+f'(a)(x-a) \]
You might be wondering: why bother using the tangent to approximate a function if you could just substitute the value into the original function and get the exact answer? In some cases, the original function might be too complex or intensive to evaluate. It can be much easier and quicker to use a linear tangent. Linear approximation is also handy when you don't know the original function or it is difficult to work with.
Remember, the tangent method is used for approximation.
Example 1 – approximating functions
Consider the graph of the function \(f(x)=e^{2-x}\) near \(x=2\). Approximate the value of the function at \(x=1.9\) and \(x=2.05\).
First, find \(f'(x)\).
\[f'(x)=-e^{2-x}\]
To find the equation for the tangent:
\[\begin{align*} y & = f(a)+f'(a)(x-a)\\
& = f(2)+f'(2)(x-2)\\
& = e^{2-2}+(-e^{2-2})(x-2)\\
& = 1+(-1)(x-2)\\
& = 1-x+2\\
& = 3-x
\end{align*}\]
Thus, the equation of the tangent at \(x=2\) is \(y=3-x\).
We can then estimate \(L(1.9)\) and \(L(2.05)\) by substituting them into the equation for the tangent.
\[\begin{align*} L(1.9) & = 3-1.9\\
& = 1.1
\end{align*}\]
Compare this to \(f(1.9)=e^{2-1.9}\approx1.105\). It's a pretty good approximation.
Compare this to \(f(2.05)=e^{2-2.05}\approx0.951\).
If \(f(x)=\sqrt[3]{x}\), approximate the value of \(f(7.98)\).
Often, you will need to make your own judgement about where to draw your tangent. Here, we want to approximate the value at \(x=7.98\). A close-enough value is \(x=8\), so let's do that.
First, find the \(f'(x)\). We will need to convert \(\sqrt[3]{x}\) to index form: \(x^{\frac{1}{3}}\).
\[f'(x)=\frac{x^{-\frac{2}{3}}}{3}\]
Now, we can find the equation for the tangent and substitute \(x=7.98\) directly into it.
\[\begin{align*} L(x) & = f(a)+f'(a)(x-a)\\
& = \sqrt[3]{8}+\frac{8^{-\frac{2}{3}}}{3}(7.98-8)\\
& = 2+\frac{1}{12}(7.98-8)\\
& = 1.998
\end{align*}\]
Differentials
Differentials can also be used to estimate the amount a function will change as a result of a small change in \(x\) or a small change in \(y\). These are denoted by \(\triangle x\) and \(\triangle y\), respectively.
You would use differentials instead of the linear approximation formula when you are looking at change, rather than to estimate specific values at a certain point.
Consider this graph of function \(y=f(x)\). Point \(P\) is at \(x=1\), and another point \(Q\) is shown.
The tangent at \(P\) can be used to estimate the change in \(x\) and \(y\) values, based on the basic gradient formula: \(y=mx\).
The gradient \(m\) is the same as rise over run or the derivative of \(y\) with respect to \(x\). In other words, \(\dfrac{dy}{dx}\).
Generally, as \(x\) approaches \(0\):
\[\frac{\triangle y}{\triangle x}\rightarrow \frac{dy}{dx}\]
For small values of \(x\), we can approximate \(y\):
\[\triangle y\approx\frac{dy}{dx}\triangle x\]
Example 1 – approximating small changes
The side of a square has length \(5\textrm{ cm}\). How much will the area of the square increase when the side length is increased by \(0.01\textrm{ cm}\)?
Let the area of the square be \(A\) and the length of a side be \(x\textrm{ cm}\). This means that \(A=x^{2}\) and \(\dfrac{dA}{dx}=2x\).
\[\begin{align*} \triangle A & \approx\frac{dA}{dx}\triangle x\\
& = 2x\triangle x\\
& = 2\times5\times0.01\\
& = 0.1
\end{align*}\]
The area of the square will increase by approximately \(0.1\textrm{ cm}^{2}\).
A \(2\%\) error is made in measuring the radius of a sphere. Find the percentage error in the volume.
If we let the radius be \(r\) and the volume be \(V\), then \(\Delta r=0.02r\). The volume of a sphere is given by \(V=\dfrac{4}{3}\pi r^{3}\), so \(\dfrac{dV}{dr}=4\pi r^{2}\).
The percentage error in the volume is:
\[\begin{align*} \textrm{Percentage error in }V & \approx\frac{\Delta V}{V}\times100\%\\
& = \frac{0.08\pi r^{3}}{\frac{4}{3}\pi r^{3}}\times100\%\\
& = \frac{0.08}{\frac{4}{3}}\times100\%\\
& = 6\%
\end{align*}\]
The percentage error in the volume is approximately \(6\%\).
Exercise – making approximations
The following formulas may be helpful for completing this exercise:
Surface area of a sphere: \(A=4\pi r^{2}\)
Volume of a cylinder: \(V=\pi r^{2}h\)
Volume of a sphere: \(V=\dfrac{4}{3}\pi r^{3}\)
Use linear approximation of the function \(f(x)=\sqrt{x}\) to find the approximate value of \(\sqrt{16.1}\).
Use linear approximation to find an approximate value of \(\cos(59^{\circ})\). Hint: Convert degrees to radians using \(1^{\circ}=\dfrac{\pi}{180}\textrm{ radians}\).
If the radius of a sphere is increased from \(10\textrm{ cm}\) to \(10.1\textrm{ cm}\), what is the approximate increase in surface area?
The height of a cylinder is \(10\textrm{ cm}\) and its radius is \(4\textrm{ cm}\). Find the approximate increase in volume when the radius increases to \(4.02\textrm{ cm}\).
An error of \(3\%\) is made in measuring the radius of a sphere. Find the percentage error in volume.
