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.

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About Alex Nelson

I like math. I like programming. Most of all, I love puzzles.
This entry was posted in Approximation, Calculus, Differential, Tangent Line and tagged . Bookmark the permalink.

4 Responses to Tangent Lines as Linear Approximations

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