Linear approximation
Straight up close
Without a calculator, what is \sqrt{4.1}? You know \sqrt4=2 exactly, and 4.1 is close to 4. How could knowing the slope at 4 get you the rest of the way?
Zoom in on any smooth curve and it starts to look like a straight line, its tangent line. Move the ochre point away from the base point and watch the gap between the curve and the tangent grow.
Drag the ochre point along the curve. The ochre segment is the error between the curve and the tangent line.
| Δx | f(a + Δx) | L(a + Δx) | Error | Error ÷ (Δx)² |
|---|
The linearization
The tangent line to f at a is
Near a, f(x)\approx L(x). For \sqrt{4.1}, take f(x)=\sqrt x and a=4. Then f(4)=2 and f'(4)=\tfrac{1}{2\sqrt4}=\tfrac14, so
The true value is 2.02485 to five places. Look at the last column of the table above. Each time \Delta x shrinks by 10, the error shrinks by about 100. What does that suggest about the size of the error?
The approximation is good only near a. Choose a where you know f and f' exactly and the point you want is close by.
Solve it step by step
Pick a function and a base point, find the tangent line, and use it. Give approximations to at least four decimal places. The score counts steps you answer correctly on the first try.
Common mistakes
- Choosing a base point you cannot evaluate
For \sqrt{24}, use a=25, not a=24. The point of the method is that f(a) is easy.
- Using the slope at the wrong point
The slope is f'(a), at the base point, not at the point you are estimating.
- Trusting it far from a
Drag the point far away in the graph. The error grows roughly with the square of the distance.