Difficulty: Intermediate | Prerequisites: All differentiation rules from Sections 1 through 5 (power, product, quotient, chain, trig). Section 7 uses these in physics contexts; Section 9 uses the tangent line concept from Section 1.
Section 7 applies derivatives to motion: figuring out when an object is moving forward, backward, speeding up, slowing down, and how far it travels in total. Section 9 reverses the usual question: instead of finding exact slopes, you use the tangent line to approximate function values near a known point. Both sections are about interpreting the derivative, not just computing it.
Velocity is the derivative of position; acceleration is the derivative of velocity. An object speeds up when velocity and acceleration share the same sign, and slows down when they have opposite signs. Total distance requires breaking the interval at turning points and summing absolute displacements. Linear approximation uses the tangent line at a known point to estimate the function at a nearby point. The differential dy = f'(x)·dx approximates how much the output changes for a small change in input.
Velocity
v(t) = s'(t), the derivative of the position function. Positive velocity means moving forward (in the positive direction); negative velocity means moving backward.
Acceleration
a(t) = v'(t) = s''(t), the derivative of velocity (or the second derivative of position). It measures how the velocity itself is changing.
Speed
The absolute value of velocity: |v(t)|. Speed is always non-negative. Think of it as how fast, with no regard for direction.
Speeding up
The speed is increasing. This happens when velocity and acceleration have the same sign (both positive or both negative).
Slowing down
The speed is decreasing. This happens when velocity and acceleration have opposite signs.
Total distance travelled
The sum of the absolute displacements over each sub-interval between turning points. Unlike net displacement (which can be zero if you return to the start), total distance counts every metre covered.
Linearisation
L(x) = f(a) + f'(a)(x − a). The tangent line at x = a used as an approximation of f(x) for x near a. Also called linear approximation or tangent line approximation.
Differential
dy = f'(x)·dx, where dx = Δx is a small change in x. The differential dy approximates the actual change Δy = f(x + dx) − f(x). Think of it as the tangent-line estimate of how much the output moves.
Given s(t) (position), differentiate to get v(t) = s'(t) (velocity) and a(t) = v'(t) = s''(t) (acceleration).
Moving forward (positive direction): v(t) > 0
Moving backward (negative direction): v(t) < 0
At rest (standing still): v(t) = 0
For a projectile thrown upward, maximum height occurs when v(t) = 0. Find that time, then evaluate s(t) at that time.
Example: h = 15 + 10t − 5t². Velocity: v(t) = 10 − 10t. Set v = 0: t = 1 s. Maximum height: h(1) = 15 + 10 − 5 = 20 m.
The clean method: compare the signs of v(t) and a(t).
Same sign (both + or both −): speeding up
Opposite signs: slowing down
Either is zero: neither speeding up nor slowing down at that instant
This avoids the messy approach of differentiating |v(t)| directly.
Example: s(t) = t³/3 − 3t² + 5t. Then v(t) = t² − 6t + 5 = (t−1)(t−5) and a(t) = 2t − 6. Analyse sign charts for v and a to determine intervals of speeding up and slowing down.
Find when v(t) = 0 on the interval [a, b]. These are the turning points.
Break [a, b] into sub-intervals at the turning points.
Compute |s(end) − s(start)| on each sub-interval.
Add them all up.
Displacement is s(b) − s(a), which can be zero even if the object moved (it came back). Total distance counts the entire path.
Example: s(t) = t³/3 − 3t² + 5t on [0, 3]. Velocity v(t) = (t−1)(t−5) = 0 at t = 1 (t = 5 is outside [0,3]). Compute s(0) = 0, s(1) = 1/3 − 3 + 5 = 7/3, s(3) = 9 − 27 + 15 = −3. Total distance = |7/3 − 0| + |−3 − 7/3| = 7/3 + 16/3 = 23/3.
Near a known point x = a, the tangent line sits close to the curve. So L(x) = f(a) + f'(a)(x − a) gives a good estimate of f(x) when x is close to a.
Approximate √4.5. Use f(x) = √x with a = 4. Then f(4) = 2, f'(x) = 1/(2√x), f'(4) = 1/4. L(x) = 2 + (1/4)(x − 4). L(4.5) = 2 + (1/4)(0.5) = 2.125. The actual value is about 2.1213, so the approximation is quite close.
At a = π/4: f(π/4) = √2/2, f'(π/4) = cos(π/4) = √2/2. L(x) = √2/2 + (√2/2)(x − π/4). Use this to approximate sin(11π/40) by plugging in x = 11π/40.
The differential is the tangent-line version of "how much does y change?"
dx = Δx = x − a (the small change in input)
dy = f'(x) · dx (the estimated change in output)
Δy = f(x + dx) − f(x) (the actual change in output)
The approximation: dy ≈ Δy, and it gets better as dx gets smaller.
Example: y = cos(πx). dy = −π sin(πx) · dx. At x = 1/3, dx = −0.02: dy = −π sin(π/3)(−0.02) = −π(√3/2)(−0.02) = 0.01π√3 ≈ 0.0544.
If a measurement has a small error dr in the input, the resulting error in the output is approximately dy = f'(r) · dr.
Example: a sphere's radius is measured as 5 in with possible error 0.01 in. V = (4/3)πr³, so dV = 4πr² · dr = 4π(25)(0.01) = π in³ ≈ 3.14 in³. The relative error dV/V = 3(dr/r) = 3(0.01/5) = 0.006, about 0.6%.
Students confuse displacement with total distance. Displacement can be zero; total distance cannot (unless the object never moved). Always check for turning points.
"Speeding up" does not mean positive acceleration. A car moving backward (v < 0) with negative acceleration (a < 0) is speeding up, because the speed |v| is increasing.
When computing total distance, some students forget to take absolute values on each sub-interval. A negative displacement on a sub-interval becomes a positive distance.
In linear approximation, students sometimes pick an anchor point a that is far from the target x. The approximation degrades quickly with distance; choose the nearest value where you know f(a) and f'(a) exactly.
⚠️ "When is the particle speeding up / slowing down?" is a very common exam question. Use the sign-chart method (compare signs of v and a), not the derivative of |v|.
⚠️ Total distance problems appear on nearly every exam. Practise the full workflow: find turning points, evaluate position, sum absolute displacements.
⚠️ Linear approximation and differentials are often combined with trig or root functions. Be ready to approximate values like √4.5, sin(33°), or ³√1001.
True or false: A particle with v(t) < 0 and a(t) < 0 is slowing down.
Fill in the blank: Total distance travelled = sum of ______ on each sub-interval.
True or false: The linearisation of f at a is the equation of the tangent line at a.
Fill in the blank: dy = ______ · dx
If s(2) = 10, s(5) = 3, and the object turned around once at t = 4 where s(4) = 15, what is the total distance from t = 2 to t = 5?
Q: A particle moves with s(t) = cos(πt/4) for 0 ≤ t ≤ 10. When is it moving in the positive direction?
A: v(t) = −(π/4)sin(πt/4). Positive direction means v(t) > 0, so sin(πt/4) < 0. This occurs when πt/4 is in (π, 2π), (3π, 4π), etc., giving t in (4, 8).
Q: A projectile has h = 15 + 10t − 5t². When does it hit the ground, and with what velocity?
A: h = 0 when −5t² + 10t + 15 = 0, so t² − 2t − 3 = 0, giving (t−3)(t+1) = 0, so t = 3 s. v(3) = 10 − 10(3) = −20 m/s (downward).
Q: Use linear approximation to estimate ³√1001.
A: f(x) = x^(1/3), a = 1000. f(1000) = 10, f'(x) = (1/3)x^(−2/3), f'(1000) = 1/300. L(1001) = 10 + (1/300)(1) = 10.00333.
Q: A sphere's radius is measured at 5 in ± 0.01 in. Estimate the maximum error in the volume.
A: dV = 4πr² dr = 4π(25)(0.01) = π ≈ 3.14 in³.
Q: s(t) = t² − 6t. Find the total distance from t = 0 to t = 4.
A: v(t) = 2t − 6 = 0 at t = 3. s(0) = 0, s(3) = 9 − 18 = −9, s(4) = 16 − 24 = −8. Total distance = |−9 − 0| + |−8 − (−9)| = 9 + 1 = 10.
Motion problems draw on every differentiation rule from Sections 3 through 5. The speeding-up/slowing-down analysis foreshadows concavity in Chapter 3 (where f'' determines the shape of the graph). Linear approximation connects forward to Taylor polynomials (studied in later courses), which extend the idea from a line to higher-degree polynomials.
velocity, acceleration, speed, speeding up, slowing down, at rest, total distance travelled, displacement vs distance, turning points, linear approximation, tangent line approximation, linearisation, differential, dy, dx, error estimation, relative error, ³√1001, sin 33 degrees, MTH 132, Calculus I, Michigan State, Chapter 2