Linearization – AP Calculus

Sometimes we need a quick estimate of a function's value near a point where we know the exact value. The tangent line at that point provides a linear approximation (linearization) of the function. This is especially useful for complicated functions like $\sqrt{x}$, $\sin x$, or $e^x$ near a known point.

🎯 In this section you will learn

The formula for linearization: $L(x) = f(a) + f'(a)(x-a)$
How to approximate function values using tangent lines
How to estimate changes using differentials $dy = f'(x) dx$
When linear approximations are accurate (and when they're not)

📌 The Linearization Formula

The linearization of $f$ at $x = a$ is the tangent line approximation:

$ L(x) = f(a) + f'(a)(x - a) $$ For $x$ close to $a$, $f(x) \approx L(x)

💡 Intuition

The tangent line is the best linear approximation to the function near $x = a$. The slope of the tangent line matches the function's slope at that point.

Example 1Linearization of $\sqrt{x}$

Find the linearization of $f(x) = \sqrt{x}$ at $x = 4$. Use it to approximate $\sqrt{4.1}$.

①

Find $f(a)$ and $f'(a)$

$f(4) = 2$, $f'(x) = \frac{1}{2\sqrt{x}}$, $f'(4) = \frac{1}{4}$

②

Write $L(x)$

$L(x) = f(4) + f'(4)(x-4) = 2 + \frac{1}{4}(x-4)$

③

Approximate $\sqrt{4.1}$

$L(4.1) = 2 + \frac{1}{4}(0.1) = 2 + 0.025 = 2.025$

④

Compare with actual

Actual $\sqrt{4.1} \approx 2.02485$ — error is only $0.00015$!

Example 2Linearization of $\sin x$

Find the linearization of $f(x) = \sin x$ at $x = 0$. Estimate $\sin(0.1)$.

①

Find $f(0)$ and $f'(0)$

$f(0) = 0$, $f'(x) = \cos x$, $f'(0) = 1$

②

Write $L(x)$

$L(x) = 0 + 1(x-0) = x$

③

Approximate $\sin(0.1)$

$L(0.1) = 0.1$

④

Actual

$\sin(0.1) \approx 0.099833$ — very close!

Example 3Linearization of $e^x$

Find the linearization of $f(x) = e^x$ at $x = 0$. Estimate $e^{0.2}$.

$f(0) = 1$, $f'(0) = 1$
$L(x) = 1 + 1(x-0) = 1 + x$
$L(0.2) = 1.2$, actual $e^{0.2} \approx 1.2214$

📌 Differentials: Estimating Change

Another way to think about linear approximation is using differentials. If $y = f(x)$, then:

dy = f'(x) \, dx \quad \text{and} \quad \Delta y \approx dy \quad \text{when } dx \text{ is small}

Here $dx$ is a small change in $x$, and $dy$ estimates the corresponding change in $y$.

Example 4Using Differentials

The radius of a circle increases from 10 cm to 10.1 cm. Estimate the increase in area.

①

Area formula

$A = \pi r^2$

②

Find $dA$

$dA = 2\pi r \, dr$

③

Plug in $r=10$, $dr=0.1$

$dA = 2\pi (10)(0.1) = 2\pi \approx 6.283$ cm²

④

Actual change

$\Delta A = \pi(10.1^2 - 10^2) = \pi(102.01 - 100) = 2.01\pi \approx 6.315$ cm²

Example 5Error Estimation

The edge of a cube is measured as 5 cm with a possible error of 0.02 cm. Estimate the maximum possible error in the volume.

$V = s^3$, $dV = 3s^2\, ds$
$s = 5$, $ds = \pm 0.02$
$dV = 3(25)(\pm 0.02) = \pm 1.5$ cm³
The volume error is approximately $\pm 1.5$ cm³.

📊 Visualizing Linear Approximation

Interactive graph below. The orange line is the tangent line at the selected point. Notice how close it stays to the blue curve near the point of tangency:

⚠️ Important Notes

The linear approximation is most accurate when $x$ is very close to $a$.
The farther $x$ is from $a$, the larger the error.
Differentials $dy$ estimate the change in $y$, not the new value.
Error estimation using differentials assumes $dx$ is small and $f'$ doesn't change much.

🔍 Key Takeaways

Linearization: $L(x) = f(a) + f'(a)(x-a)$ is the tangent line approximation.
Used to estimate function values near known points.
Differentials: $dy = f'(x)dx$ estimates the change in $y$ for a small change $dx$ in $x$.
Common applications: error estimation, propagation of uncertainty.
The closer $x$ is to $a$, the better the approximation.

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