Local linearity and linearization
If a function is differentiable at x = a, its graph looks almost straight when you zoom in near a. That straight line is the tangent line.
Linearization: L(x) = f(a) + f′(a)(x − a).
Use it for x close to a, where f(a) and f′(a) are easy. Example: √4.1 ≈ 2 + (1/4)(0.1) = 2.025.
Over or under?
If the curve bends down (f″ < 0, concave down) the tangent lies above the curve, so the estimate is too big. If f″ > 0 (concave up) the estimate is too small. For √x, f″ < 0, so 2.025 is a little too big.
The further x is from a, the bigger the error.
Behaviour of implicit relations
An implicit relation mixes x and y, like x² + y² = 25 or x³ + y³ = 6xy. We do not solve for y. We differentiate both sides with respect to x and treat y as a function of x (chain rule: d/dx of y² is 2y·dy/dx).
For the circle: 2x + 2y·dy/dx = 0, so dy/dx = −x/y.
Horizontal and vertical tangents
- Horizontal tangent: dy/dx = 0, so the top of the fraction is 0 (here x = 0, points (0, ±5)).
- Vertical tangent: the bottom of the fraction is 0 (here y = 0, points (±5, 0)).
You can also find d²y/dx² by differentiating again and substituting dy/dx back in. This tells you where the curve is concave up or down.
Vector-valued functions and their derivatives
A vector-valued function gives a position for every time t: r(t) = ⟨x(t), y(t)⟩. Think of a drone's position.
Differentiate each part on its own:
- Velocity: v(t) = r′(t) = ⟨x′(t), y′(t)⟩. It points along the path (tangent).
- Speed: |v(t)| = √(x′(t)² + y′(t)²).
- Acceleration: a(t) = r″(t) = ⟨x″(t), y″(t)⟩.
- Slope of the path: dy/dx = y′(t) / x′(t), when x′(t) ≠ 0.
For r(t) = ⟨3cos t, 2sin t⟩: v(t) = ⟨−3sin t, 2cos t⟩. At t = 0, v = ⟨0, 2⟩: moving straight up at speed 2.
The nth term test for divergence
A series a₁ + a₂ + a₃ + … adds infinitely many terms. If the terms themselves do not shrink to 0, the sum keeps growing (or swinging), so it cannot settle.
Test: if lim aₙ ≠ 0 (or the limit does not exist), the series diverges.
Careful: if lim aₙ = 0, the test says nothing. The harmonic series 1 + 1/2 + 1/3 + … has terms going to 0 but still diverges. Example: Σ n/(2n + 1): terms go to 1/2 ≠ 0, so it diverges.
Try it: estimate a root with a ruler
Draw y = √x carefully on graph paper from x = 0 to 9. Place a ruler so it just touches the curve at (4, 2). Read the ruler's height at x = 5. Compare with 2 + 1/4 = 2.25 and with a calculator's √5 = 2.236. Now try x = 4.2: the ruler gets much closer.
Key formulas and definitions
- L(x) = f(a) + f′(a)(x − a)
- Δy ≈ f′(a)·Δx
- Implicit: differentiate both sides, collect dy/dx terms
- r′(t) = ⟨x′(t), y′(t)⟩
- Speed = √(x′(t)² + y′(t)²)
- dy/dx = y′(t)/x′(t)
- If lim aₙ ≠ 0 then Σaₙ diverges
Worked examples
1. Use the tangent line at x = 4 to estimate √4.1.
f(x) = √x, f(4) = 2, f′(x) = 1/(2√x) so f′(4) = 1/4. L(x) = 2 + (1/4)(x − 4). L(4.1) = 2 + 0.025 = 2.025.
2. Is the estimate in Example 1 too big or too small?
f″(x) = −1/(4x^(3/2)) < 0, so the curve is concave down and lies below its tangent. The estimate 2.025 is slightly too big (true value 2.0248).
3. Estimate (1.02)^5 using linearization at x = 1.
f(x) = x⁵, f(1) = 1, f′(x) = 5x⁴, f′(1) = 5. L(1.02) = 1 + 5(0.02) = 1.10. (True value ≈ 1.104.)
4. Find dy/dx for x² + xy + y² = 7 and the slope at (1, 2).
2x + (y + x·dy/dx) + 2y·dy/dx = 0. So dy/dx(x + 2y) = −(2x + y), dy/dx = −(2x + y)/(x + 2y). At (1, 2): −(2 + 2)/(1 + 4) = −4/5.
5. On x² + y² = 25, where are the tangents vertical?
dy/dx = −x/y. Vertical when the denominator y = 0 (and x ≠ 0): x² = 25, so at (5, 0) and (−5, 0).
6. r(t) = ⟨t², t³ − 3t⟩. Find velocity, speed and the slope dy/dx at t = 2.
v(t) = ⟨2t, 3t² − 3⟩. At t = 2: v = ⟨4, 9⟩. Speed = √(16 + 81) = √97 ≈ 9.85. dy/dx = 9/4 = 2.25.
7. Does Σ (3n² + 1)/(n² + 5) converge?
aₙ → 3 as n → ∞. Since the limit is not 0, by the nth term test the series diverges.
Common mistakes
- Using linearization far from x = a. The tangent is only good close to the point.
- Forgetting dy/dx when differentiating y terms: d/dx(y²) is 2y·dy/dx, not 2y.
- Thinking that lim aₙ = 0 proves a series converges. The nth term test can only prove divergence.
- Mixing up speed and velocity: velocity is a vector ⟨x′, y′⟩; speed is its length, a number.