Linear Approximations for MATH 139
Exam Relevance for MATH 139
Use L(x) = f(a) + f prime a times (x-a) to approximate function values near x=a. Tangent line approximation.
This skill appears on:
What is Linear Approximation?
Let's break down the name:
- Linear = a line (straight line)
- Approximation = an estimate (close enough, not exact)
So linear approximation literally means: estimating with a line.
The Idea: Some functions are hard to calculate by hand. What's $\sqrt{16.1}$? That's annoying. But $\sqrt{16} = 4$ is easy! Linear approximation says: "Start at a nice easy point, then use a straight line to estimate the hard value."
Quick Example: You know $\sqrt{16} = 4$. You want $\sqrt{16.1}$. Since 16.1 is just a tiny bit more than 16, the answer should be just a tiny bit more than 4. Linear approximation tells you exactly how much more — by using the tangent line (a straight line that touches the curve at that point).
That's it! We're approximating (estimating) complicated values using a linear (straight line) approach.
Key Variables
- $f(x)$ = the function we want to approximate
- $a$ = the "nice" point where we know the exact value (our anchor point)
- $f(a)$ = the exact value of the function at $a$
- $f'(a)$ = the slope of the function at $a$ (this tells us how fast $f$ is changing)
- $x$ = the point where we want to estimate the value
- $L(x)$ = the linear approximation (our estimate)
The Formula
The linear approximation of $f(x)$ near $x = a$ is:
$$L(x) = f(a) + f'(a)(x - a)$$
Breaking it down:
- $f(a)$: Start at the known value
- $f'(a)$: The rate of change (slope of the tangent line)
- $(x - a)$: How far we're moving from our anchor point
- $f'(a)(x - a)$: The adjustment based on how the function is changing
This is just the equation of the tangent line at $x = a$! We're using the tangent line as our approximation.
When Does This Work Well?
Linear approximation works best when:
- $x$ is close to $a$: The closer you are to your anchor point, the better the estimate
- The function isn't curving too much: If $f$ has a small second derivative near $a$, the tangent line stays close to the actual curve
Warning: As you move farther from $a$, your estimate gets worse. The tangent line eventually diverges from the curve.
Step-by-Step Process
- Identify the function $f(x)$ you want to approximate
- Choose a nice anchor point $a$ close to $x$ where you can easily calculate $f(a)$ and $f'(a)$
- Calculate $f(a)$ — the exact value at your anchor point
- Find $f'(x)$ and then calculate $f'(a)$ — the slope at your anchor point
- Plug into the formula: $L(x) = f(a) + f'(a)(x - a)$
- Simplify to get your estimate
Problem: Use linear approximation to estimate $\sqrt{16.1}$.
Step 1: Identify the function
$f(x) = \sqrt{x} = x^{1/2}$
We want to find $f(16.1)$.
Step 2: Choose a nice anchor point
We need a point close to 16.1 where we know the exact square root. $a = 16$ is perfect because $\sqrt{16} = 4$.
Step 3: Calculate $f(a)$
$f(16) = \sqrt{16} = 4$
Step 4: Find the derivative and evaluate at $a$
$f'(x) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$
$f'(16) = \frac{1}{2\sqrt{16}} = \frac{1}{2(4)} = \frac{1}{8}$
Step 5: Apply the formula
$$L(x) = f(a) + f'(a)(x - a)$$
$$L(16.1) = 4 + \frac{1}{8}(16.1 - 16)$$
$$L(16.1) = 4 + \frac{1}{8}(0.1)$$
$$L(16.1) = 4 + 0.0125 = 4.0125$$
Answer: $\sqrt{16.1} \approx 4.0125$
How accurate is this? The actual value is $\sqrt{16.1} = 4.01248...$ so our estimate is extremely close!
Problem: Use linear approximation to estimate $\ln(1.05)$.
Step 1: Identify the function
$f(x) = \ln(x)$
We want to find $f(1.05)$.
Step 2: Choose a nice anchor point
$a = 1$ is perfect because $\ln(1) = 0$.
Step 3: Calculate $f(a)$
$f(1) = \ln(1) = 0$
Step 4: Find the derivative and evaluate at $a$
$f'(x) = \frac{1}{x}$
$f'(1) = \frac{1}{1} = 1$
Step 5: Apply the formula
$$L(x) = f(a) + f'(a)(x - a)$$
$$L(1.05) = 0 + 1(1.05 - 1)$$
$$L(1.05) = 0.05$$
Answer: $\ln(1.05) \approx 0.05$
How accurate is this? The actual value is $\ln(1.05) = 0.04879...$ so our estimate is off by about 2.5%. Still pretty good for mental math!
Problem: Use linear approximation to estimate $\sqrt[3]{8.1}$.
Step 1: Identify the function
$f(x) = \sqrt[3]{x} = x^{1/3}$
We want to find $f(8.1)$.
Step 2: Choose a nice anchor point
$a = 8$ is perfect because $\sqrt[3]{8} = 2$.
Step 3: Calculate $f(a)$
$f(8) = \sqrt[3]{8} = 2$
Step 4: Find the derivative and evaluate at $a$
$f'(x) = \frac{1}{3}x^{-2/3} = \frac{1}{3x^{2/3}} = \frac{1}{3\sqrt[3]{x^2}}$
$f'(8) = \frac{1}{3\sqrt[3]{64}} = \frac{1}{3(4)} = \frac{1}{12}$
Step 5: Apply the formula
$$L(x) = f(a) + f'(a)(x - a)$$
$$L(8.1) = 2 + \frac{1}{12}(8.1 - 8)$$
$$L(8.1) = 2 + \frac{1}{12}(0.1)$$
$$L(8.1) = 2 + \frac{0.1}{12} = 2 + 0.008\overline{3} \approx 2.0083$$
Answer: $\sqrt[3]{8.1} \approx 2.0083$
How accurate is this? The actual value is $\sqrt[3]{8.1} = 2.00829...$ — nearly perfect!
Problem: Use linear approximation to estimate $\sin(0.1)$ (where 0.1 is in radians).
Step 1: Identify the function
$f(x) = \sin(x)$
We want to find $f(0.1)$.
Step 2: Choose a nice anchor point
$a = 0$ is perfect because $\sin(0) = 0$ and $\cos(0) = 1$.
Step 3: Calculate $f(a)$
$f(0) = \sin(0) = 0$
Step 4: Find the derivative and evaluate at $a$
$f'(x) = \cos(x)$
$f'(0) = \cos(0) = 1$
Step 5: Apply the formula
$$L(x) = f(a) + f'(a)(x - a)$$
$$L(0.1) = 0 + 1(0.1 - 0)$$
$$L(0.1) = 0.1$$
Answer: $\sin(0.1) \approx 0.1$
Key Insight: For small angles (in radians), $\sin(x) \approx x$. This is a famous result that comes directly from linear approximation!
Common Mistakes to Avoid
- Choosing a bad anchor point: Pick $a$ where you can easily compute both $f(a)$ and $f'(a)$
- Forgetting the derivative: The slope $f'(a)$ is essential — don't skip this step
- Going too far from $a$: Linear approximation only works well near the anchor point
- Sign errors in $(x - a)$: Be careful whether $x > a$ or $x < a$
Quick Reference: Common Functions
| Function | Derivative | Good anchor points |
|---|---|---|
| $\sqrt{x}$ | $\frac{1}{2\sqrt{x}}$ | Perfect squares: 1, 4, 9, 16, 25, ... |
| $\sqrt[3]{x}$ | $\frac{1}{3\sqrt[3]{x^2}}$ | Perfect cubes: 1, 8, 27, 64, ... |
| $\ln(x)$ | $\frac{1}{x}$ | $x = 1$ (since $\ln(1) = 0$), or $x = e$ |
| $e^x$ | $e^x$ | $x = 0$ (since $e^0 = 1$) |
| $\sin(x)$ | $\cos(x)$ | $x = 0, \frac{\pi}{2}, \pi, ...$ |
| $\cos(x)$ | $-\sin(x)$ | $x = 0, \frac{\pi}{2}, \pi, ...$ |
Linear Approximation Formula
Estimates the value of f(x) using the tangent line at a nearby point a. This is also called the tangent line approximation or linearization.
Variables:
- $L(x)$:
- the linear approximation (estimated value)
- $f(a)$:
- the exact value of the function at the anchor point
- $f'(a)$:
- the derivative (slope) of the function at the anchor point
- $x$:
- the point where you want to estimate the value
- $a$:
- the anchor point (a 'nice' nearby value where f is easy to compute)
- $(x - a)$:
- the distance from the anchor point to x
Tangent Line Equation (Point-Slope Form)
The equation of the tangent line to f(x) at x = a. Rearranging this gives the linear approximation formula. The tangent line IS the linear approximation!
Variables:
- $y$:
- the y-coordinate on the tangent line
- $f(a)$:
- the y-coordinate of the point of tangency
- $f'(a)$:
- the slope of the tangent line
- $x$:
- the x-coordinate (input value)
- $a$:
- the x-coordinate of the point of tangency
Differential Approximation
The change in y (Δy) can be approximated by the differential dy. This is closely related to linear approximation: if dx = (x - a), then dy gives the adjustment from f(a).
Variables:
- $\Delta y$:
- the actual change in y (exact but hard to compute)
- $dy$:
- the differential (approximation of Δy)
- $f'(x)$:
- the derivative at the starting point
- $dx$:
- the change in x (same as x - a)
This skill is taught in the following courses. Create an account to access practice exercises and full course materials.