Linear Approximations for MATH 122
Exam Relevance for MATH 122
Linear approximations appear in MATH 122 for estimating changes in cost or revenue.
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)
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