Tangent Lines as Linear Approximations

1. Motivation. So we have some procedure to construct a line tangent to a curve at a given point. What good is it?

Well, suppose we want to compute $\sqrt{4.1}$. What is it? I don’t know!

Lets introduce the function $f(x)=\sqrt{4+x}$. Then we want to evaluate $f(0.1)$. What to do?

Why not construct $t(x)$ the line tangent to $f$ at $x=4$, then make an approximation $f(0.1)\approx t(0.1)$?

Lets see if this really works. We calculate
(1)$\displaystyle t(h)=\sqrt{4}+f'(0)\cdot(h-0)=2+\frac{h}{4}$
This gives us
(2)$\displaystyle \sqrt{4.1}\approx t(0.1)=2+\frac{1/10}{4}=\frac{81}{40}$.
Is this a good approximation? I don’t know, how can we find out?

Why not square it, and see how far off we are from the original quantity $4.1$? Let us try! We see
(3)$\displaystyle \left(\frac{81}{40}\right)^{2}=\frac{6561}{1600}=\frac{6400}{1600}+\frac{161}{1600}$.
So what? Well, basic arithmetic tells us
(3′)$\displaystyle \frac{6400}{1600}+\frac{161}{1600} = 4 + 0.1 + \frac{1}{1600}$.
Thus the difference between this and $4.1$ is $1/1600=0.000625$. So the approximation appears to be decent to a couple of digits.

Reflection. What did we just do? We observed we wanted to calculate some quantity, and expressed it as a function $f(x)$. We know what $f(0)$ is, and we wanted to find some approximate solution for small $x$.

Since $x$ is “small enough”, we just used the line tangent to a given point (for us $(0,f(0))$) as the approximation. The key moment is using a “small enough” $x$. Right now we do not have the proper tools to determine what is “small enough”, but we will develop the tools later on.

2. Approximations. If we don’t have a calculator and want to calculate stuff, we may use calculus to help us.

We describe the quantity as a point on the curve $(x_{1}, f(x_{1}))$. So we want to find some approximate value “close” to $f(x_{1})$.

We do this by finding some point on the curve $(x_{0},y_{0})$ that’s “close enough” to $(x_{1},f(x_{1}))$. (That’s why in our motivating example we chose $f(x)=\sqrt{4+x}$ instead of $g(x)=\sqrt{1+x}$: we have $f(0.1)$ be dealing with $h=0.1$ instead of something 300 times bigger when calculating $g(3.1)$!)

Then we construct the line tangent to the curve at $(x_{0},y_{0})$. We’ll describe this line $\ell$ as the points $(x,t(x))$.

Now our approximation takes $t(x_{1}-x_{0})\approx f(x_{1})$.

So what’s the algorithm?

Algorithm A.
Step 0: we want to calculate some quantity $q$.
Step 1: construct some function $f(x)$ such that at some point $x_{1}$ we have $f(x_{1})=q$.
Step 2: Find a point $x_{0}$ “close enough” to $x_{1}$.
Step 3: Construct the tangent line $t(h)=f(x_{0})+f'(x_{0})\cdot h$.
Step 4: Make the approximation $t(x_{1}-x_{0})\approx q$.

Remark. We call this a linear approximation because the tangent line is a linear function.

3. Reflection. What if we work with $g(x)=\sqrt{1+x}$ when attempting an approximation of $q=\sqrt{4.1}$? What happens? Lets try!

We see the tangent line to $(0,g(0))$ is described by
(4)$\displaystyle t(x)=1+g'(0)\cdot(x-1)=1+\frac{x}{2}$
Lets now try to approximate $t(3.1)\approx\sqrt{4.1}$. Is it any good?

Well, $t(3.1)=1+1.55=2.55$. Is this a good approximation? No! We see that $t(2.55)^{2} = 6.5025\gg q^{2}$, which tells us: $t(3.1)\approx q$ a horrible approximation!

Remark. When we have a linear approximation $t(h)$, it generally gets worse as $h$ gets bigger.