Source: MTH 132 Calculus I, Michigan State University, Chapter 2 – Derivatives (Sections 7–9)
Difficulty: Intermediate to Advanced | Prerequisites: Parts 1 and 2 study notes (all differentiation rules including chain rule and implicit differentiation)
With all the differentiation tools in place, this part turns to applications. Section 7 translates derivative language into physics: position, velocity, acceleration, total distance. Section 8 tackles related rates, which combine implicit differentiation with real-world geometry to find how fast one quantity changes given information about another. Section 9 introduces linear approximation and differentials, using the tangent line to estimate function values near a known point. These three topics account for a large share of exam word problems.
Velocity is the derivative of position, and acceleration is the derivative of velocity. An object speeds up when velocity and acceleration share the same sign. Related rates problems use implicit differentiation with respect to time to connect rates of change of different quantities. Linear approximation replaces a curve with its tangent line near a point; differentials measure the approximate change in y.
Velocity
v(t) = s'(t), the derivative of the position function. It tells you how fast and in which direction the object is moving. Units: distance per time.
Acceleration
a(t) = v'(t) = s''(t), the derivative of velocity (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; it tells you how fast without regard to direction.
Speeding up
The speed (|v(t)|) 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 where the object does not change direction. Find when v(t) = 0, break the time interval at those points, compute |s(t_end) − s(t_start)| on each piece, and add them up.
In simple terms, displacement can cancel out if the object reverses direction, but total distance counts every metre covered.
Related rates
Problems where two or more quantities change with time and are connected by an equation. You differentiate the equation with respect to time (using implicit differentiation and the chain rule) and solve for the unknown rate.
Linearisation (linear approximation)
The tangent-line approximation of f near x = a: L(x) = f(a) + f'(a)(x − a). For x close to a, f(x) ≈ L(x).
Think of it as replacing a curve with the straight line that best fits it at one point.
Differential
If y = f(x), the differential dy = f'(x)·dx, where dx = Δx is a small change in x. The differential dy approximates the actual change Δy = f(x + Δx) − f(x).
s(t) = position at time t.
v(t) = s'(t) = velocity. Positive means moving forward (or upward); negative means moving backward (or downward); zero means the object is momentarily at rest.
a(t) = v'(t) = s''(t) = acceleration.
Forward: v(t) > 0.
Backward: v(t) < 0.
Standing still: v(t) = 0.
An object thrown upward reaches its maximum height when v(t) = 0 (the moment it stops rising before falling back down).
Speeding up: v(t) and a(t) have the same sign.
Slowing down: v(t) and a(t) have opposite signs.
At a moment when v(t) = 0, the object is neither speeding up nor slowing down.
The sign-comparison approach is cleaner than trying to differentiate |v(t)| directly, which can be algebraically unpleasant.
The technique:
Find when v(t) = 0. These are the turning points.
Break the time interval into sub-intervals at those points.
Compute the displacement |s(t_end) − s(t_start)| on each sub-interval.
Add the absolute values together.
Displacement (net change in position) can be zero even when total distance is large, if the object reverses.
Draw a diagram and label all quantities that change with time.
Write an equation relating the quantities (geometric, trigonometric, or physical).
Differentiate both sides with respect to time t (using implicit differentiation and the chain rule).
Substitute the known values and rates, then solve for the unknown rate.
Circle: C = πd, d = 2r, A = πr².
Triangle: A = (1/2)bh, Pythagorean theorem c² = a² + b².
Rectangle: P = 2l + 2w, A = lw.
Sphere: V = (4/3)πr³, SA = 4πr².
Cone: V = (h/3)(area of base) = (1/3)πr²h.
Cylinder: V = πr²h.
Expanding/shrinking sphere. Given dV/dt, find dr/dt or dSA/dt. Differentiate V = (4/3)πr³ to get dV/dt = 4πr²·(dr/dt), then solve.
Sliding ladder. A ladder of fixed length leans against a wall. The relationship is x² + y² = L². Differentiate: 2x·(dx/dt) + 2y·(dy/dt) = 0. Given one rate and the position, find the other.
Filling a cone. Given dV/dt, find dh/dt. The key trick is using similar triangles to express r in terms of h (or vice versa) so you have a single variable before differentiating.
Rising balloon / angle of elevation. A balloon rises; a person watches from a fixed distance. Use tan(θ) = height/distance, differentiate, and substitute.
Two cars diverging. Use the Pythagorean theorem (or the law of cosines for non-right angles) to relate the distance between the cars to their individual positions. Differentiate with respect to t.
The linearisation (or linear approximation, or tangent-line approximation) of f at a is:
L(x) = f(a) + f'(a)(x − a).
For x close to a, f(x) ≈ L(x). The approximation works well when x − a is small and f does not curve sharply near a.
Example: approximate √4.5 using f(x) = √x at a = 4. f(4) = 2, f'(x) = 1/(2√x), f'(4) = 1/4. So L(4.5) = 2 + (1/4)(0.5) = 2.125. The true value is about 2.1213, so the estimate is quite close.
Differentials are an equivalent way to frame the same idea. If y = f(x):
dx = Δx = x − a (the small change in x).
Δy = f(x + Δx) − f(x) (the actual change in y).
dy = f'(x)·dx (the approximate change in y, using the tangent line).
The key relationship: Δy ≈ dy when dx is small.
Example: approximate how much y = √x changes as x goes from 4 to 4.5. dy = (1/(2√4))·(0.5) = 0.125.
Differentials are used for error estimation. If a measurement x has a small error dx, the resulting error in a computed quantity y = f(x) is approximately dy = f'(x)·dx.
Example: a sphere's radius is measured as 5 in with a possible error of 0.01 in. The error in the volume is approximately dV = 4πr²·dr = 4π(25)(0.01) = π in³ ≈ 3.14 in³. Whether this is a "big" error depends on context; comparing it to the total volume V = (4/3)π(125) ≈ 523.6 in³ gives a relative error of about 0.6%.
Formula | Context |
|---|---|
v(t) = s'(t) | Velocity from position |
a(t) = v'(t) = s''(t) | Acceleration from velocity |
speed = | v(t) |
L(x) = f(a) + f'(a)(x − a) | Linearisation at x = a |
dy = f'(x)·dx | Differential |
C = πd, A = πr² | Circle |
c² = a² + b² | Pythagorean theorem |
V = (4/3)πr³, SA = 4πr² | Sphere |
V = (1/3)πr²h | Cone |
V = πr²h | Cylinder |
A = lw, P = 2l + 2w | Rectangle |
Velocity and acceleration are directly from physics: every kinematics problem in a first-year physics course relies on the same derivative relationships. Related rates are how engineers and scientists connect changing quantities in real time, for example tracking how fast a water level rises in an oddly shaped tank given a fixed inflow rate. Linear approximation underpins how engineers estimate changes in a quantity without recomputing the full function, and it is the conceptual basis for Newton's method and Taylor polynomials later in the course.
"If velocity is negative, the object is slowing down." Negative velocity means moving backward, not slowing down. Slowing down is about speed decreasing, which requires velocity and acceleration to have opposite signs.
"Total distance and displacement are the same thing." Displacement is net change in position (can be zero if the object returns). Total distance adds up every bit of ground covered.
"In a related rates problem, I can substitute numbers before differentiating." Substituting specific values before differentiating kills the variables you need. Always differentiate first, then substitute.
"The linearisation gives the exact value." It gives an approximation. The further x is from a, the worse the approximation becomes.
⚠️ Motion problems (position, velocity, acceleration, speeding up/slowing down) are a guaranteed exam topic. Know how to read a velocity graph and determine when the object is speeding up, slowing down, or at rest.
⚠️ Total distance problems require finding when v(t) = 0 to break the interval. A common error is computing s(b) − s(a) and calling it total distance (that gives displacement instead).
⚠️ Related rates: always draw a diagram. The most common error is forgetting that r and h (or x and y) both change with time. If two quantities are linked by similar triangles, eliminate one variable before differentiating.
⚠️ Linear approximation: know how to choose the right base point a (a nearby value where f and f' are easy to compute). For √4.5, use a = 4; for sin(33°), use a = 30° = π/6.
⚠️ Differentials: remember dy ≈ Δy, not dy = Δy. The question may ask you to compute dy (the approximation) or Δy (the exact change) or both.
True or false: If v(t) < 0 and a(t) < 0, the object is slowing down.
Fill in the blank: The linearisation of f at a is L(x) = ___.
True or false: In a related rates problem, you should substitute known values before differentiating.
Fill in the blank: The differential dy equals ___.
True or false: Total distance travelled equals |s(b) − s(a)|.
Answers: 1. False (same sign means speeding up). 2. f(a) + f'(a)(x − a). 3. False (differentiate first, then substitute). 4. f'(x)·dx. 5. False (that is displacement; total distance accounts for direction changes).
Q: A particle moves with position s(t) = t³/3 − 3t² + 5t. Find the velocity and determine when the particle is moving forward.
A: v(t) = s'(t) = t² − 6t + 5 = (t − 1)(t − 5). The particle moves forward when v(t) > 0, which is t < 1 or t > 5.
Q: A 13-foot ladder leans against a wall. The top slides down at 3 ft/s. How fast is the bottom moving away from the wall when the bottom is 5 ft from the wall?
A: x² + y² = 169. When x = 5, y = 12. Differentiate: 2x(dx/dt) + 2y(dy/dt) = 0. Substitute: 10(dx/dt) + 24(−3) = 0, so dx/dt = 72/10 = 7.2 ft/s.
Q: Use linearisation to approximate √(4.5).
A: f(x) = √x, a = 4. f(4) = 2, f'(4) = 1/4. L(4.5) = 2 + (1/4)(0.5) = 2.125.
Q: A sphere's radius increases at 2 cm/s. How fast is the volume increasing when r = 10?
A: V = (4/3)πr³. dV/dt = 4πr²·(dr/dt) = 4π(100)(2) = 800π cm³/s.
Q: Use a differential to approximate the change in y = x² + 8 as x goes from 1 to 1.02.
A: Wait, the source actually has y = √(x² + 8). dy = [x/√(x² + 8)]·dx. At x = 1, dx = 0.02: dy = [1/3]·(0.02) = 0.02/3 ≈ 0.00667.
Velocity and acceleration problems here use only first and second derivatives; they extend naturally to the optimisation problems in Chapter 3 (finding when velocity is maximised, for instance). Related rates problems are implicit differentiation applied to time-dependent quantities, so they draw directly on Section 6. Linear approximation is the one-term version of a Taylor polynomial, which appears in Calculus II. Differentials also underpin the notation in integration: the dx in ∫f(x)dx has its roots here.
velocity, acceleration, speed, speeding up, slowing down, position function, total distance travelled, displacement, related rates, implicit differentiation with time, similar triangles in related rates, ladder problem, cone filling problem, balloon problem, linear approximation, linearisation, tangent line approximation, differential, dy, dx, delta y, error estimation, relative error, MTH 132, Calculus I, MSU, Chapter 2