Linear Approximations and Differentials

Linear Approximations #

The motivation for this section is that we will want to be approximating a function near a point by their tangent lines at that point.

Sample (very) basic example: #

We’re going to approximate $f(x)=x^2$ around $x=2$.

1. Find the tangent line at $x=2$

   The derivative is $y=2x$. The tangent line then is going to be: $\begin{array}{rlr}y-4& =4(x-2)y& =4x-4\end{array}$

2. Try to approximate (1.9)^2 using the tangent line:

   Since 1.9 is “close” to 2, we can use the tangent line at 2:

   $\begin{array}{rlrl}y& =4(1.9)-4& =7.6-4& =3.6\end{array}$

   Since the actual value $(1.9{)}^2=3.61$, that’s pretty close. Not bad!

3. What about approximating $(1{)}^2$?

   …. 1 is not “close” to 2.

   $\begin{array}{rlr}y& =4(1)-4& =0\end{array}$

   and obviously $1^2\ne 0$, so this is really crappy approximation.

Linear Approximations / Linearization #

In general, the tangent line to a function at the point $(a,f(a))$ is: $$y-f(a)=f^{\prime }(a)(x-a)$$

We’ll solve for $y$, label it $L(x)$, and give it the name linearization: $$L(x)=f(a)+f^{\prime }(a)(x-a)$$

Note This is called linearization, a linear approximation, or the first-order Taylor Polynomial¹ for $f$ at $x=a$.

Example #

Approximate $y=\sin (x)$ for $x$ values near 0.

Since $$f(x)\approx f(a)+f^{\prime }(a)(x-a)$$

Find $a,f(a),f^{\prime }(x),$ and $f^{\prime }(a)$ to find the linear approximation.

Give it a try before revealing the solution here:

Now let’s use the approximation to approximate $\sin (1^{\circ })$.

Example #

1. Find the linearization of $\sqrt[5]{32-4x}$ around $x=0$.
2. Use that linearization to approximate $\sqrt{27.99}$

Example: #

Evaluating $\sqrt{3.99}$ in two ways, once, centering on $x=0$, and second, centering on $x=4$:

To approximate around $x=0$, we need the function that you’re using to approximate to be sqrt(4-x):

Here’s using the given function: #

and in order to get $\sqrt{3.99}$, you need to plug in $x=0.01$.

On the other hand, here’s using the function $\sqrt{x}$ around x=4. Notice that you’ll still get the same answers, but the $x$ values you’re using are very different

here to approximate $\sqrt{3.99}$ you need to plug in $x=3.99$.

Differentials #

Recall that $f^{\prime }(x)={\displaystyle \frac{\text{ change in y’s }}{\text{change in x’s}}}$

so $f^{\prime }(x)\approx {\displaystyle \frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}}$ and let $\mathrm{\Delta }x\to 0$, we get Leibniz notation:

$$f^{\prime }(x)={\displaystyle \frac{dy}{dx}}$$ and solving for $dy$ we get the differentials $$dy=f^{\prime }(x)dx$$

This defines the differential $dy$ in terms of the variables $x$ and $dx$.

Example #

If $f(x)=\sqrt{x+1}$, find $dy$.

Solution: (try it before revealing the spoiler)

Connection of differentials to linear approximations #

$$f(a+dx)\approx f(a)+dy$$ a tiny, tiny change in $dx$ changes the function by the tiny, tiny change $dy$.

For concrete (less tiny) changes, we’ll write $\mathrm{\Delta }x$ and $\mathrm{\Delta }y$ to have:

$$f(a+\mathrm{\Delta }x)\approx f(a)+\mathrm{\Delta }y$$

Example #

For example, using $f(x)=\sqrt{x+1}$, let $a=3$ and $\mathrm{\Delta }x=0.1$, find the approximation of $\sqrt{4.1}$

Solution: set up $$\begin{array}{rlrl}f(3+\mathrm{\Delta }x)& =f(3+0.1)& =f(3.1)& =\sqrt{3.1+1}\end{array}$$ so $$\begin{array}{rlrlrl}f(3+\mathrm{\Delta }x)=\sqrt{4.1}& \approx f(3)+\mathrm{\Delta }y& \approx f(3)+{\displaystyle \frac{\mathrm{\Delta }x}{2\sqrt{3+1}}}& =\sqrt{3+1}+{\displaystyle \frac{0.1}{2\sqrt{3+1}}}& =2+{\displaystyle \frac{0.1}{4}}& =2.025\end{array}$$

… note that the calculator says $\sqrt{4.1}\approx 2.02484$, so we’re pretty close!