Local Linearization

Let's say you're on a long car trip and there's a mountain in the distance. Looks steep, right? But when you get there, you feel, oh, this isn't that steep!

Similarly, if you take a curve, if you keep zooming into it, it will look more and more like a line. We can use this to approximate the value of a function at annoying points.

Main article: secant and tangent lines

The above graph represents a function $f(x).$ We want to figure an approximation for $f(x)$ at $x$ because $x$ is a difficult number to deal with. But it is easy to compute $f(a)$, where $a$ is very close to $x.$ So, what we can do is draw a line tangent to $(a, f(a))$. Remember, as we zoom into the curve, it will look more and more like the tangent line, whose slope is $f'(a)$.

In order to approximate $f(x)$ at $(x, f(x))$, we're going to find the point on the tangent line at the $x$-value $x$ because as long as $x$ is really close to $a,$ the difference should be minuscule. We can draw a right triangle, with one leg (we'll call it the differential $dx$ ) having a length $x-a$ and another leg (we'll call it the differential $dy$ with the $y$-value ranging from the point $(x, f(a))$ to the point on the tangent line at $x$. Then, the slope of the hypotenuse is $\dfrac{dy}{dx}$, or $f'(a)$. Over here, what's important to us is $dy$, because that gets us closer to the approximate value we're looking for. We can first write

$$\dfrac{dy}{dx} = f'(a) \Rightarrow dy = f'(a) dx.$$

But that isn't the whole thing yet. We're missing some of the height. Since the "easy" value to deal with was at the point $(a, f(a))$, and our triangle started at the height of $f(a)$, we need to add $f(a)$ to our approximation equation. As a result, we can write

$$f(x) \approx f(a) + f'(a) dx .$$

But what was $dx$? Since $dx$ was $x -a$, we have

$$f(x) \approx f(a)+f'(a)(x-a).$$

Approximate $\sqrt{4.01}$.

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We have $f(x) = \sqrt{x}$. Since we have $x= 4.01$, a hairy number to deal with, what's a number close to 4.01 that we know the square root of? Let's try 4.

Our formula says $f(x) \approx f(a) + f'(a) (x-a )$, so

$$\begin{aligned} f(4.01) &\approx \sqrt{4} + f'(4) (4.01 - 4 ) \\ f'(4) &= \dfrac{1}{2\sqrt{4}} = \dfrac{1}{4} \\ \Rightarrow f(4.01) &\approx 2.0025. \ _\square \end{aligned}$$

Note: If you try $\sqrt{4.01}$ on a calculator, you'll get 2.002498439, which is pretty close to the approximate value.

Approximate $\sin 44^{\circ}$.

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When is the approximate value less than the actual? When is the approximate value greater than the actual?

If we look at the concavity of functions, we'll see that a tangent line is always "under" a curve that is concave up, and a tangent line is always "above" a curve that is concave down, except at the points of tangency. We can thus conclude that the approximation will be less than the actual value if the function is concave up at the point of tangency, and that the approximation will be greater than the actual value if the function is concave down at the point.

I. 2.0009 $\quad$ II. 2.0008 $\quad$ III. 2.0007

Which of the three numbers above could be a local linear approximation of $8.01^{\frac{1}{3}}$?

A. I only $\quad$ B. II only $\quad$ C. I & II only $\quad$ D. II & III only $\quad$ E. I, II & III $\quad$ F. None of them

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As $f(x) = x^{1/3}$ is concave down, you can say that any local linear approximation you make must be greater than the actual value. The actual value is 2.000832986, so the only possible approximation is option I, or choice A. $_\square$

In multivariable calculus, we extend local linear approximation to derive many important formulas, such as those for multivariable approximation and multivariable chain rule.

Given a function $z= f(x, y)$, we can say

$$\Delta z \approx \dfrac{\partial{z}}{\partial{x}} \Delta x + \dfrac{\partial{z}}{\partial{y}} \Delta y.$$

That is the multivariable approximation formula. Basically, we are adding the following quantities:

• the change in $z$ when $x$ is changed a little, with $y$ held constant
• the change in $z$ when $y$ is changed a little, with $x$ held constant.

By the way, an important thing to keep in mind: $$\Delta z \neq dz.$$ We will use $\Delta z$ to refer to an actual number, and $dz$ to refer to a differential. As we make $\Delta x$ and $\Delta y$ both approach zero, then we can make them into the differentials $dx$ and $dy$, and $\Delta z$ also will become $dz$.

We can use multivariable approximation to get the equation of a plane tangent to a three dimensional surface given by $z = f(x, y)$. In order to get an approximate tangent plane, we first need two lines tangent to the same point on that plane. These two lines could be

$$\left\{\begin{array}{c}x z = z_{0} + a \Delta x \\ y =y_{0} \end{array}\right.$$

$$\left\{\begin{array}{c}x z = z_{0} + b \Delta y \\ x = x_{0} , \end{array}\right.$$

where $a = \dfrac{\partial{f}}{\partial{x}}$ and $b = \dfrac{\partial{f}}{\partial{y}}$.

Then the equation of the tangent plane becomes

$$z- z_{0} = a \left(x - x_{0} \right) + b \left( y - y_{0} \right) .$$

Given a function of two variables, we could make any changes in $x$ and $y$ approach zero, and rewrite the approximation equation as

$$dz = \dfrac{\partial{z}}{\partial{x}} dx + \dfrac{\partial{z}}{\partial{y}} dy .$$

Then, we might want to make the whole function dependent on $t$ in order to get the values of $x, y$ using the parameter $t$. We can divide that whole equation by $dt$ and get the multivariable chain rule:

$$\dfrac{dz}{dt} = \dfrac{\partial{z}}{\partial{x}} \dfrac{dx}{dt} + \dfrac{\partial{z}}{\partial{y}} \dfrac{dy}{dt} .$$