Difficulty: Intermediate to Advanced | Prerequisites: All differentiation rules (Ch. 2 Sec. 1–6), implicit differentiation.
Section 7 applies the derivative to motion problems: velocity, acceleration, direction, and total distance. Section 8 introduces related rates, where two or more quantities change simultaneously and you use implicit differentiation with respect to time to connect their rates. Section 9 uses the tangent line as a local linear approximation of a function.
Velocity is the derivative of position; acceleration is the derivative of velocity. Whether an object speeds up or slows down depends on whether velocity and acceleration share the same sign. Related rates problems connect the rates of change of two or more quantities through an equation you differentiate with respect to time. Linear approximation uses the tangent line to estimate a function's value near a known point.
Velocity, v(t)
The derivative of the position function: v(t) = s'(t). Positive velocity means moving forward (in the positive direction); negative means moving backward.
Acceleration, a(t)
The derivative of velocity (second derivative of position): a(t) = v'(t) = s''(t). It describes how the velocity itself is changing.
Speed
The absolute value of velocity: |v(t)|. Speed is always non-negative. It tells you how fast the object is moving regardless of direction.
Speeding up / slowing down
The object is speeding up when velocity and acceleration have the same sign (both positive or both negative). It is slowing down when they have opposite signs. If acceleration is zero, it is neither speeding up nor slowing down at that instant.
Think of it as: acceleration in the same direction as motion = speeding up. Acceleration opposing motion = slowing down.
Total distance traveled
The sum of the absolute values of displacement on each sub-interval between turning points (where v(t) = 0). Unlike net displacement (which can be zero if the object returns to its start), total distance counts every metre covered.
Related rates
A class of problems where two or more quantities change with time, and you need to find one rate of change given information about the others. You set up a geometric or algebraic equation, differentiate both sides with respect to time t (using implicit differentiation and the chain rule), then substitute known values.
Linear approximation (linearisation)
Using the tangent line at x = a to approximate f(x) for x near a: f(x) ≈ f(a) + f'(a)(x - a). The tangent line is the best straight-line estimate of the function near the point of tangency.
Differential
If y = f(x), the differential dy = f'(x) dx. It gives the approximate change in y resulting from a small change dx in x.
Position, velocity, acceleration:
s(t) = position. v(t) = s'(t) = velocity. a(t) = v'(t) = s''(t) = acceleration.
Units of v(t) = units of s / units of t (e.g. metres per second). Units of a(t) = units of v / units of t (e.g. m/s²).
Interpreting the sign of velocity:
v(t) > 0: moving in the positive direction (forward).
v(t) < 0: moving in the negative direction (backward).
v(t) = 0: the particle is (momentarily) at rest. These are the turning points.
Maximum height: For a projectile thrown upward, maximum height occurs when v(t) = 0. Example: h(t) = 15 + 10t - 5t². v(t) = 10 - 10t. Set to zero: t = 1 second. Maximum height: h(1) = 20 m.
Speeding up vs. slowing down:
If v(t) and a(t) have the same sign, the object is speeding up (acceleration reinforces motion).
If v(t) and a(t) have opposite signs, the object is slowing down (acceleration opposes motion).
If a(t) = 0, the object is neither speeding up nor slowing down at that instant.
To solve: find where v(t) = 0 and a(t) = 0, build a sign chart for each, compare signs on each interval.
Total distance traveled on [a, b]:
Find all times where v(t) = 0 in the interval (these are the turning points).
Break [a, b] into sub-intervals at these turning points.
Compute the displacement |s(end) - s(start)| on each sub-interval.
Add them all up.
Example: s(t) = t³/3 - 5t²/2 + 5t on [0, 3]. v(t) = t² - 5t + 5. Turning points at t = 1 and t = 5 (only t = 1 is in [0, 3]). Total distance = |s(1) - s(0)| + |s(3) - s(1)|.
General technique:
Draw a picture and label all quantities that change with time as functions of t.
Write an equation relating those quantities (geometric formula, Pythagoras, trig, etc.).
Differentiate both sides with respect to t using implicit differentiation and the chain rule.
Substitute all known values (at the specific instant in question) and solve for the unknown rate.
Key formulas to know: circle (A = πr², C = 2πr), sphere (V = (4/3)πr³, SA = 4πr²), cone (V = (1/3)πr²h), Pythagoras (a² + b² = c²), triangle area (A = (1/2)bh), similar triangles, basic trig.
Worked examples:
Expanding sphere: V = (4/3)πr³, dV/dt = 10 in³/s, r(t₁) = 2 in. Differentiate: dV/dt = 4πr² · (dr/dt). Plug in: 10 = 4π(4)(dr/dt), so dr/dt = 10/(16π) = 5/(8π) in/s.
Sliding ladder: A 13-ft ladder with the top sliding down at 3 ft/s. x² + y² = 169. Differentiate: 2x(dx/dt) + 2y(dy/dt) = 0. When x = 5: y = 12, dy/dt = -3. Solve: 10(dx/dt) + 24(-3) = 0, so dx/dt = 72/10 = 36/5 ft/s.
Filling cone: V = (πh³)/12 (after relating r and h by similar triangles: r = h/2 when the cone is 10 ft tall with base radius 5 ft). dV/dt = 9 ft³/min. Differentiate: dV/dt = (πh²/4)(dh/dt). At h = 6: 9 = (36π/4)(dh/dt) = 9π(dh/dt), so dh/dt = 1/π ft/min.
Two cars: One south at 20 mph, one west at 30 mph. After 2 hours: a = 40, b = 60, d = √(1600 + 3600) = √5200. Differentiate d² = a² + b²: 2d(dd/dt) = 2a(da/dt) + 2b(db/dt). Plug in: dd/dt = [40(20) + 60(30)] / √5200.
Linear approximation: The tangent line at x = a gives the best linear estimate of f(x) near a:
f(x) ≈ L(x) = f(a) + f'(a)(x - a).
Differentials: For y = f(x), the differential is dy = f'(x) dx. This approximates the actual change Δy = f(x + dx) - f(x) when dx is small. The approximation improves as dx gets smaller.
v(t) = s'(t) \qquad a(t) = v'(t) = s''(t) \qquad \text{speed} = |v(t)|f(x) \approx L(x) = f(a) + f'(a)(x - a) \qquad dy = f'(x)\,dxRelated rates: differentiate your geometric equation with respect to t, attaching d(·)/dt to every time-dependent variable via the chain rule.
Students often confuse "slowing down" with "negative velocity." An object moving backward (v < 0) with negative acceleration (a < 0) is speeding up, not slowing down, because v and a have the same sign.
Confusing displacement with total distance traveled. If a particle goes 5 m forward then 3 m back, displacement is 2 m but total distance is 8 m.
In related rates, using specific values too early. You must differentiate the general equation first, then substitute the numbers. Plugging in before differentiating kills the variable you need.
Forgetting that related-rates quantities are functions of time even if they are not written that way. When differentiating A = πr², both A and r depend on t, so dA/dt = 2πr · (dr/dt).
⚠️ "When is the particle speeding up / slowing down" is a classic exam question. Build sign charts for v(t) and a(t) and compare signs on each interval.
⚠️ Total distance traveled requires finding turning points (v = 0) and summing absolute displacements. Students who compute only s(b) - s(a) get net displacement, not total distance.
⚠️ Related rates problems follow a rigid four-step structure. Exam grading often awards partial credit for each step, so show your work even if you get stuck partway.
⚠️ Common related-rates setups: sliding ladder, filling/draining tank (cone or cylinder), expanding circle/sphere, two objects moving apart, angle of elevation. Know the geometry for each.
⚠️ Linear approximation is usually a shorter exam question. Memorise L(x) = f(a) + f'(a)(x - a) and practise identifying a good base point a.
True or False: If v(t) < 0 and a(t) < 0, the object is slowing down.
Fill in the blank: To find total distance on [0, 5], first find all t in (0, 5) where v(t) = ___.
True or False: In a related-rates problem, you should substitute known values before differentiating.
Fill in the blank: The linear approximation of f near x = a is L(x) = f(a) + ___(x - a).
True or False: Speed can be negative.
Answers: 1. False (same sign means speeding up). 2. 0 (turning points). 3. False (differentiate first, substitute second). 4. f'(a). 5. False (speed = |v(t)| ≥ 0).
Q: A projectile has height h(t) = 15 + 10t - 5t². When does it reach maximum height, and what is that height?
A: v(t) = 10 - 10t. Set to zero: t = 1 s. h(1) = 15 + 10 - 5 = 20 m.
Q: For s(t) = t² - 6t, when is the particle speeding up on [0, 4]?
A: v(t) = 2t - 6, a(t) = 2 (constant, always positive). v(t) = 0 at t = 3. On (0, 3): v < 0, a > 0 (opposite signs, slowing down). On (3, 4): v > 0, a > 0 (same sign, speeding up). Speeding up on (3, 4).
Q: A 13-ft ladder slides with the top moving down at 3 ft/s. How fast is the bottom moving when it is 5 ft from the wall?
A: x² + y² = 169. When x = 5, y = 12. Differentiate: 2x(dx/dt) + 2y(dy/dt) = 0. 10(dx/dt) + 24(-3) = 0. dx/dt = 72/10 = 36/5 ft/s.
Q: A sphere's volume increases at 10 in³/s. Find dr/dt when r = 2.
A: V = (4/3)πr³. dV/dt = 4πr²(dr/dt). 10 = 4π(4)(dr/dt). dr/dt = 10/(16π) = 5/(8π) in/s.
Q: Use linear approximation to estimate √(9.1).
A: f(x) = √x, a = 9, f(9) = 3, f'(x) = 1/(2√x), f'(9) = 1/6. L(9.1) = 3 + (1/6)(0.1) = 3 + 1/60 ≈ 3.0167.
Motion analysis connects to optimisation problems in the next chapter (finding maxima and minima). Related rates reuse implicit differentiation (Section 2.6) and foreshadow integration-based rate problems in Calculus II. Linear approximation is the first-order case of Taylor polynomials, which are developed fully in Calculus II.
velocity, acceleration, speed, speeding up, slowing down, total distance traveled, displacement, turning points, sign chart, related rates, implicit differentiation with time, ladder problem, cone filling, expanding sphere, similar triangles, linear approximation, linearisation, tangent line approximation, differential, dy dx, MTH 132, Calculus I, Chapter 2, Sections 2.7, 2.8, 2.9, Stewart Calculus