Linear Approximation and Differentials

Linear Approximation

Since this section uses tangent lines frequently, it is worthwhile to recall how we find the equation of the line tangent to $f$ at a point $x=a$ . The line tangent to $f$ at $x=a$ goes through the point $(a, f(a))$ and has slope $f^{\prime}(a)$ so, using the point-slope form $y-y_{0}=m\left(x-x_{0}\right)$ for linear equations, we have

$y-f(a)=f^{\prime}(a) \cdot(x-a) \text { and } y=f(a)+f^{\prime}(a) \cdot(x-a)$

This final result is the equation of the line tangent to $\mathrm{f}$ at $\mathrm{x}=\mathrm{a}$ .

If $\quad \mathrm{f}$ is differentiable at $\mathrm{x}=\mathrm{a}$ ,

then the equation of the line $L$ tangent to $f$ at $x=a$ is

$\mathbf{L}(\mathbf{x})=\mathbf{f}(a)+\mathbf{f}^{\prime}(\mathbf{a}) \cdot(\mathbf{x}-\mathbf{a})$

Example 1: Find the equation of the line $\mathrm{L}(\mathrm{x})$ which is tangent to the graph of $\mathrm{f}(\mathrm{x})=\sqrt{\mathrm{x}}$ at the point $(9,3)$ . Evaluate $\mathrm{L}(9.1)$ and $\mathrm{L}(8.88)$ to approximate $\sqrt{9.1}$ and $\sqrt{8.88}$ .

Solution: $f(x)=\sqrt{x}=x^{1 / 2}$ and $f^{\prime}(x)=\frac{1}{2} x^{-1 / 2}=\frac{1}{2 \sqrt{x}}$ so $f(9)=3$ and $f^{\prime}(9)=\frac{1}{2 \sqrt{9}}=\frac{1}{6}$ . Then

$\mathrm{L}(\mathrm{x})=\mathrm{f}(9)+\mathrm{f}^{\prime}(9) \cdot(\mathrm{x}-9)=3+\frac{1}{6}(\mathrm{x}-9)$ . If $\mathrm{x}$ is close to $\mathrm{9}$ , then the value of $\mathrm{L}(\mathrm{x})$ is a good approximation of the value of $\sqrt{x}$ (Fig. 2). The number $9.1$ is close to 9 so $\sqrt{9.1}=\mathrm{f}(9.1) \approx \mathrm{L}(9.1)=3+\frac{1}{6}(9.1-9)=3.016666$

Similarly, $\sqrt{8.88}=\mathrm{f}(8.88) \approx \mathrm{L}(8.88)=3+\frac{1}{6}(8.88-9)=2.98.$ In fact, $\sqrt{9.1} \approx 3.016621$ , so our estimate, using $\mathrm{L(9.1)}$ , is within $0.000045$ of the exact answer. $\sqrt{8.88} \approx 2.979933$ (accurate to 6 decimal places) and our estimate is within $0.00007$ of the exact answer.

In each example, we got a good estimate of a square root with very little work. The graph indicates the tangent line $\mathrm{L}(\mathrm{x})$ is slightly above $\mathrm{f}(\mathrm{x})$ , and each estimate is slightly larger than the exact value.

Practice 1: Find the equation of the line $\mathrm{L}(\mathrm{x})$ tangent to the graph of $\mathrm{f}(\mathrm{x})=\sqrt{\mathrm{x}}$ at the point $(16,4)$ (Fig. 3). Evaluate $\mathrm{L}(16.1)$ and $\mathrm{L(15.92)}$ to approximate $\sqrt{16.1}$ and $\sqrt{15.92}$ . Are your approximations using $\mathrm{L}$ larger or smaller than the exact values of the square roots?

Practice 2: Find the equation of the line $\mathrm{L}(\mathrm{x})$ tangent to the graph of $\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}$ at the point $(1,1)$ and use $\mathrm{L}(\mathrm{x})$ to approximate $(1.02)^{3}$ and $(0.97)^{3}$ . Do you think your approximations using $\mathrm{L}$ are larger or smaller than the exact values?

The process we have used to approximate square roots and cubics can be used to approximate any differentiable function, and the main result about the linear approximation follows from the two statements in the boxes. Putting these two statements together, we have the process for Linear Approximation.

Linear Approximation Process: (Fig. 4)
If

$\mathrm{f}$ is differentiable at

$\mathrm{a}$ and $\mathbf{x}$ is close to $\mathrm{a}$ ,

then (geometrically) the graph of the tangent line $\mathrm{L}(\mathrm{x})$ is close to the graph of $\mathrm{f}(\mathrm{x})$ , and

(algebraically) the values of the tangent line function

$\mathrm{L}(\mathrm{x})=\mathrm{f}(\mathrm{a})+\mathrm{f}^{\prime}(\mathrm{a}) \cdot(\mathrm{x}-\mathrm{a})$ approximate the values of $\mathrm{f}(\mathrm{x})$

$f(x) \approx L(x)=f(a)+f^{\prime}(a) \cdot(x-a)$

Sometimes we replace "

$\mathrm{x}-\mathrm{a}$ " with "

$\Delta \mathrm{x}$ " in the last equation, and the statement becomes

$\mathrm{f}(\mathrm{x}) \approx \mathrm{f}(\mathrm{a})+\mathrm{f}^{\prime}(\mathrm{a}) \cdot \Delta \mathrm{x}$ .

Example 2: Use the linear approximation process to approximate the value of $\mathrm{e}^{0.1}$ .

Solution: $\quad f(x)=e^{x}$ so $f^{\prime}(x)=e^{x}.$ We need to pick a value a near $\mathrm{x}=0.1$ for which we know the exact value of $\mathrm{f}(\mathbf{a})$ and $\mathrm{f}^{\prime}(\mathbf{a})$ , and $\mathbf{a}=0$ is the obvious choice. Then

$\begin{aligned} \mathrm{e}^{0.1}=\mathrm{f}(0.1) \approx \mathrm{L}(0.1) &=\mathrm{f}(0)+\mathrm{f}^{\prime}(0) \cdot(0.1-0) \\ &=\mathrm{e}^{0}+\mathrm{e}^{0} \cdot(0.1)=1+1 \cdot(0.1)=1.1 \end{aligned}$

This approximation is within $0.0052$ of the exact value of $e^{0.1}$

Practice 3: Approximate the value of $(1.06)^{4}$ , the amount $\$ 1$ becomes after 4 years in a bank which pays $6 \%$ interest compounded annually. (Take $f(x)=x^{4}$ and $a=1$ ).

Practice 4: Use the linear approximation process and the values in the table to estimate the value of $\mathrm{f}$ when $\mathrm{x}=1.1,1.23$ and $1.38$ .

$\begin{array}{l|c|c} x & f(x) & f^{\prime}(x) \\ \hline 1 & 0.7854 & 0.5 \\ 1.208761 & 0.4098 & \\ 1.4 & 0.9505 & 0.3378 \end{array}$

We can also approximate functions as well as numbers.

Example 3: Find a linear approximation formula $L(x)$ for $\sqrt{1+x}$ when $x$ is small. Use your result to approximate $\sqrt{1.1}$ and $\sqrt{.96}$ .

Solution: $\mathrm{f}(\mathrm{x})=\sqrt{1+\mathrm{x}}=(1+\mathrm{x})^{1 / 2}$ so $\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2}(1+\mathrm{x})^{-1 / 2}=\frac{1}{2 \sqrt{1+\mathrm{x}}}$ . Since "x is small", we know that $\mathrm{x}$ is close to $\mathrm{0}$ , and we can pick $a=0.$ Then $f(a)=f(0)=1$ and $f^{\prime}(a)=f^{\prime}(0)=\frac{1}{2}$ so

$\sqrt{1+\mathrm{x}} \approx \mathrm{L}(\mathrm{x})=\mathrm{f}(0)+\mathrm{f}^{\prime}(0) \cdot(\mathrm{x}-0)=1+\frac{1}{2} \mathrm{x}=1+\frac{\mathrm{x}}{2}$

If $x$ is small, then $\sqrt{1+\mathbf{x}} \approx 1+\frac{\mathbf{x}}{2}. \sqrt{1.1}=\sqrt{1+0.1} \approx 1+\frac{0.1}{2}=1.05$ and $\sqrt{0.96}=\sqrt{1+(-.04)} \approx 1+\frac{-.04}{2}=0.98$ . Use your calculator to determine by how much each estimate differs from the true value.