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Math 220-1 Single-variable differential calculus: Kursobjekt

Aaron Greicius

Section 19 Linearization

Definition 19.1. Linearization of a function.

Assume \(f\) is differentiable at the input \(a\text{.}\) The linearization of \(f\) centered at \(a\) is the affine function \(L(x)\) defined as
\begin{equation} L(x)=f'(a)(x-a)+f(a)\text{.}\tag{19.1} \end{equation}

Remark 19.2. Linearization.

It is important to observe the “centered at \(a\)” modifier in Definition 19.1. In other words, the function \(L(x)\) defined in (19.1) depends both on \(f\) and the specific input \(a\text{.}\)

Example 19.3. Linearization: quadratic.

Let \(f(x)=3-(x-2)^2\text{.}\) Compute the linearization \(L_1\) of \(f\) centered at \(a=1\text{,}\) and the linearization \(L_2\) of \(f\) centered at \(b=2\text{.}\)
Solution.
First compute \(f'(x)=-2(x-2)\text{.}\) Using (19.1), we have
\begin{align*} L_1(x) \amp = f'(a)(x-a)+f(a)\\ \amp = 2(x-1)+2 \amp (f'(1)=2, f(1)=2)\\ L_2(x) \amp = f'(b)(x-b)+f(b)\\ \amp = 0(x-2)+3 \amp (f'(2)=0, f(2)=3)\\ \amp = 3\text{.} \end{align*}
Thus the linearization of \(f\) centered at \(1\) is \(L_1(x)=2(x-1)+2\text{,}\) and the linearization centered at \(2\) is the constant function \(L_2(x)=3\text{.}\)

Remark 19.4. Linearization and tangent lines.

You may have noticed a resemblance to the formula for the linearization \(L\) of \(f\) centered at \(a\text{,}\) and the tangent line to the graph of \(f\) at \(x=a\text{.}\) To be precise: the graph of the linearization function \(L\) is precisely the tangent line to the graph of \(f\) at the point \(P=(a,f(a))\text{.}\)
By way of illustration, we graph the function \(f(x)=3-(x-2)^2\) in Example 19.3 along with the linearizations \(L_1\) and \(L_2\) centered at \(1\) and \(2\text{,}\) respectively.
Linearizations of a quadratic function
Figure 19.5. Linearizations of \(f(x)=3-(x-2)^2\)

Remark 19.6. Linear approximation.

Assume \(f\) is differentiable at \(a\text{,}\) and let \(L(x)=f'(a)(x-a)+f(a)\) be the linearization of \(f\) centered at \(a\text{.}\) As we can show, the differentiability of \(f\) at \(a\) ensures that values of the linearization \(L\) are close to values of \(f\) for inputs \(x\) near \(a\text{.}\) Indeed, we have
\begin{align*} f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a} \amp \implies 0=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}-\lim\limits_{x\to a}f'(a) \amp \left(\lim\limits_{x\to a}f'(a)=f'(a)\right)\\ \amp \implies \lim_{x\to a}\frac{f(x)-f(a)-f'(a)(x-a)}{x-a}=0\\ \amp \implies \lim_{x\to a}\frac{f(x)-L(x)}{x-a}=0 \amp (L(x)=f'(a)(x-a)-f(a))\text{.} \end{align*}
Consider what the limit statement
\begin{equation} \lim_{x\to a}\frac{f(x)-L(x)}{x-a}=0\tag{19.2} \end{equation}
tells us. Since \(\lim\limits_{x\to a}x-a=0\text{,}\) the numerator \(f(x)-L(x)\) must be close to zero for \(x\) close to \(a\text{.}\) Furthermore, we must have \(f(x)-L(x)\) approaching zero faster than \(x-a\) approaches zero. More precisely, invoking the epsilon-delta definition of the limit, we can show that for any \(\epsilon > 0\text{,}\) we have
\begin{equation*} \abs{f(x)-L(x)}< \epsilon \abs{x-a} \end{equation*}
for all \(x\) sufficiently close to \(a\text{.}\) After we introduce the mean value theorem, we will be able to give an even better quantitative description of just how good an approximation \(L(x)\) is to \(f(x)\text{.}\)

Example 19.8. Linear approximation: quadratic.

Let \(f(x)=3-(x-2)^2\text{.}\)
  1. Use the linearization \(L_1(x)\) of \(f\) centered at \(1\) to estimate \(f(0.5)\text{.}\)
  2. Compare your estimate of \(f(0.5)\) with the actual value. How close are the two?
Solution.
  1. Recall that we computed \(L_1(x)=2(x-1)+2\text{.}\) Thus we have
    \begin{align*} f(1/2) \amp \approx L_1(1/2)\\ \amp = 2(-1/2)+2\\ \amp = 1\text{.} \end{align*}
  2. The exact value is \(f(1/2)=3-(1/2-2)^2=3-9/4=3/4\text{.}\) Thus our estimate is off by \(1/4\text{:}\) i.e., \(\abs{L_1(1/2)-f(1/2)}=1/4\text{.}\)

Example 19.9. Linear approximation: cube-root.

Use linear approximation to estimate \(\sqrt[3]{8.05}\text{.}\)
Solution.
To use linear approximation, we need to compute the linearization of a function \(f\) centered at some input \(a\text{.}\) We take \(f(x)\sqrt[3]{x}\text{,}\) and \(a=8\text{.}\) Why \(a=8\text{?}\) I can compute \(f(8)\) and \(f'(8)\) easily by hand, and \(8.05\) is reasonably close to \(8\text{.}\) We first compute
\begin{align*} f'(x) \amp =(x^{1/3})'\\ \amp =\frac{1}{3}x^{-2/3}\text{.} \end{align*}
The linearization of \(f\) centered at \(a=8\) is then
\begin{align*} L(x) \amp = f'(8)(x-8)+f(8)\\ \amp =\frac{1}{12}(x-8)+2 \amp (8^{1/3}=2, 8^{-2/3}=1/(\sqrt[3]{8})^2=1/4) \end{align*}
Lastly, we estimate
\begin{align*} f(8.05) \amp \approx L(8.05)\\ \amp = \frac{1}{12}\cdot 0.05+2\\ \amp \frac{1}{240}+2 \amp (0.05=1/20)\\ \amp = \frac{481}{240}\text{.} \end{align*}
Using technology we see that
\begin{align*} f(8.05) \amp = 2.0041580\dots \\ \frac{481}{240} \amp = 2.0041666\dots\text{.} \end{align*}
Our estimate ended up being pretty close!

Example 19.10. Linear approximation: trig.

Let \(f(x)=\tan(\pi x)\text{.}\) Use linear approximation to estimate \(f(0.251)\text{.}\)
Solution.
We compute the linearization \(L\) of \(f\) centered at \(a=\frac{1}{4}=0.25\text{.}\) First we compute the derivative of \(f\text{:}\)
\begin{align*} f'(x) \amp = \sec^2(\pi x) (\pi x)' \amp \text{(chain)}\\ \amp = \pi \sec^2(\pi x)\text{.} \end{align*}
It follows that the linearization centered at \(\frac{1}{4}\) is
\begin{align*} L(x) \amp = f'(1/4)(x-1/4)+f(1/4)\\ \amp = 2\pi(x-1/4)+1 \amp (\tan(\pi/4)=1, \sec^2(\pi/4)=1/(\sqrt{2}/2)^2=2) \text{.} \end{align*}
Lastly we estimate
\begin{align*} f(0.251) \amp \approx L(0.251)\\ \amp = 2\pi (0.251-0.25)+1 \amp (1/4=0.25)\\ \amp = 2\pi(0.001)+1\\ \amp = \frac{\pi}{500}+1\text{.} \end{align*}

Example 19.11. Linear approximation: marshmallow.

Dudley places a cylindrical marshmallow in the microwave, causing it to expand such a manner that the ratio of its height and radius is preserved. Initially the height and radius of the marshmallow are both equal to 2 centimeters; when Dudley removes the marshmallow the height and radius are both equal to 2.1 centimeters.
  1. Use linear approximation to estimate the change in volume of the marshmallow.
  2. Compare your estimate with the actual change in volume.
Solution.
Since the radius \(r\) and height \(h\) of the marshmallow are initially equal, and since the marshmallow expands in such a way that the ratio of radius to height is preserved, we see that we always have \(r=h\text{,}\) and thus
\begin{equation*} V=\pi r^2 h=\pi x^3\text{,} \end{equation*}
where \(x=r=h\text{.}\) We thus treat volume \(V=f(x)=\pi x^3\) as a function of the common dimension \(x\) specifying both radius and height.
  1. To estimate the change in volume, we compute the linearization \(L\) of \(V=f(x)\) centered at \(2\text{.}\) Since \(f'(x)=3\pi x^2\text{,}\) we have
    \begin{equation*} L(x)=f'(2)(x-2)+f(2)=12\pi (x-2)+8\pi\text{.} \end{equation*}
    It follows that the change of volume \(\Delta V=f(2.1)-f(2)\) can be estimated as
    \begin{align*} \Delta V \amp = f(2.1)-f(2)\\ \amp \approx L(2.1)-f(2) \amp (L(2.1)\approx f(2.1)\\ \amp = 12\pi(0.1)=\frac{12\pi }{10}=1.2\, \pi \end{align*}
  2. The exact change of volume is
    \begin{align*} f(2.1)-f(2) \amp = \pi (2.1)^3-8\pi \\ \amp = (1.261) \pi \text{.} \end{align*}
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