Tangent Lines and Local Linearity: Using Derivatives to Approximate

From Derivative Value to Tangent Line

At a specific input x=a, the derivative f'(a) tells you the slope of the tangent line to the graph y=f(x) at the point (a, f(a)). Once you know the slope and the point, you can write the tangent line equation using point-slope form.

Point-slope form for the tangent line at x=a

The tangent line (also called the linearization line) at x=a is:

y = f(a) + f'(a)(x-a)

This is just point-slope form y-y_1=m(x-x_1) written with (x_1,y_1)=(a,f(a)) and m=f'(a).

Step-by-step: building the tangent line equation

• Step 1: Choose the point of tangency x=a (usually near where you want an approximation).
• Step 2: Compute f(a) (the actual function value at that input).
• Step 3: Compute f'(a) (the slope at that input).
• Step 4: Substitute into y = f(a) + f'(a)(x-a).
• Step 5: Use the line to estimate f(x) for x near a.

Local Linearity and Linearization

Many smooth curves look almost like a straight line when you zoom in close enough around a point. The tangent line captures that “best local straight-line behavior.” Using the tangent line to estimate nearby values is called linearization.

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Linearization function

Define the linear approximation near a as:

L(x) = f(a) + f'(a)(x-a)

Then for x close to a, you use:

f(x) ≈ L(x)

The approximation is usually better the closer x is to a.

Graph-based visual idea (what to imagine)

• Full view: A curved graph y=f(x) with a straight tangent line touching at (a,f(a)). Away from a, the curve and line separate.
• Zoomed-in view around x=a: The curve and tangent line are nearly indistinguishable; the curve looks almost linear.

When you use linearization, you are essentially using the “zoomed-in” behavior even though you may not actually zoom a graph.

Worked Example 1: Estimating \sqrt{10}

Let f(x)=\sqrt{x}. We want \sqrt{10}=f(10). Choose a nearby value with an easy square root, such as a=9.

Step-by-step

• 1) Compute f(a): f(9)=\sqrt{9}=3.
• 2) Compute f'(x): f'(x)=\frac{1}{2\sqrt{x}}.
• 3) Evaluate the derivative at a: f'(9)=\frac{1}{2\cdot 3}=\frac{1}{6}.
• 4) Write the linearization: L(x)=3+\frac{1}{6}(x-9).
• 5) Estimate at x=10: L(10)=3+\frac{1}{6}(1)=3+\frac{1}{6}=3.1666\ldots.

So \sqrt{10} \approx 3.1667 using the tangent line at x=9.

Worked Example 2: Tangent Line Equation from a Given Derivative Value

Suppose f(2)=5 and f'(2)=-3. Find the tangent line at x=2.

Step-by-step

• Point of tangency: (2,5).
• Slope: m=f'(2)=-3.
• Point-slope form: y = 5 + (-3)(x-2).
• Simplify (optional): y = 5 - 3x + 6 = 11 - 3x.

So the tangent line is y=5-3(x-2) (or y=11-3x).

Worked Example 3: Small Change in Area (Measurement/Rounding Context)

Let a circle have radius r and area A(r)=\pi r^2. If you measure a radius as r=10 cm but the measurement could be off by a small amount, you can estimate how much the area changes using the derivative.

Derivative as “area sensitivity” to radius

A'(r)=2\pi r. At r=10, A'(10)=20\pi (square cm per cm). This means near r=10, area changes about 20\pi for each 1 cm change in radius.

Linearization for area near r=10

Use a=10. Then A(10)=100\pi and A'(10)=20\pi.

L(r)=100\pi + 20\pi(r-10)

Example: radius measured 0.1 cm too large

If the true radius is r=10.1, then:

A(10.1) \approx L(10.1)=100\pi + 20\pi(0.1)=100\pi+2\pi=102\pi

So a 0.1 cm increase in radius gives about a 2\pi square cm increase in area.

Using differentials language (optional but common in applications)

For small changes, you may see:

\Delta A \approx A'(10)\,\Delta r

Here \Delta r=0.1, so \Delta A \approx 20\pi(0.1)=2\pi.

Worked Example 4: Small Change in Volume (Practical Approximation)

For a sphere, V(r)=\frac{4}{3}\pi r^3. If r=5 cm and the radius changes slightly, estimate the volume change.

Step-by-step

• Differentiate: V'(r)=4\pi r^2.
• Evaluate at r=5: V'(5)=4\pi(25)=100\pi (cubic cm per cm).
• Small change estimate: if \Delta r=0.02 cm, then \Delta V \approx 100\pi(0.02)=2\pi cubic cm.

This is a fast way to estimate the effect of tiny measurement errors without recomputing the full volume formula.

How to Choose a Good Linearization Point a

• Pick a close to the target x so the tangent line stays close to the curve.
• Pick a where f(a) is easy to compute (like perfect squares for square roots).
• Avoid points where the function is not smooth or where the derivative is undefined (the tangent line idea may fail there).

Common Errors and How to Avoid Them

1) Confusing f(a) with f'(a)

• What goes wrong: Using the slope as the y-value or the y-value as the slope.
• Fix: Remember: f(a) is a point coordinate (height). f'(a) is a slope (steepness).
• Quick check: In y=f(a)+f'(a)(x-a), the constant term should be a y-value, and the coefficient multiplying (x-a) should be a slope.

2) Using the wrong point in (x-a)

• What goes wrong: Writing (x-f(a)) or (x-2) when the tangency point is actually a=3.
• Fix: The subtraction inside parentheses always uses the x-coordinate of the tangency point: (x-a), not (x-f(a)).

3) Mixing up x and y when identifying the tangency point

• What goes wrong: Treating (f(a),a) as the point instead of (a,f(a)).
• Fix: The graph is y=f(x), so x comes first: (a,f(a)).

4) Linearizing around the wrong input value

• What goes wrong: Wanting \sqrt{10} but linearizing at a=16 because it’s a perfect square, even though it’s far away.
• Fix: Choose a nearby convenient point (for \sqrt{10}, a=9 is close).

5) Forgetting that the approximation is local

• What goes wrong: Using L(x) far from a and expecting accuracy.
• Fix: Treat L(x) as reliable only for x near a. If you must go farther, choose a new point a closer to the new x-value.