Concept understanding — Meaning of Derivative and Rate of Change
The derivative f′(x) of a function carries two equivalent readings, and this chapter leans on both throughout.
As a slope. For the curve y=f(x), the slope of the chord joining (x,f(x)) and (x+h,f(x+h)) is the Newton quotient
hf(x+h)−f(x).
Taking h→0 gives the slope of the curve at (x,f(x)) itself:
f′(x)=limh→0hf(x+h)−f(x).
If θ is the angle the tangent makes with the positive x-axis (measured anticlockwise), then f′(x)=tanθ.
As a rate of change.f′(x)=dxdy is also the instantaneous rate of change of y with respect to x; over an interval [a,b] the average rate of change is the ordinary difference quotient b−af(b)−f(a) (a chord slope), while the derivative at a single point is the instantaneous rate.
Motion along a line. If s=f(t) is the position of an object at time t (measured from a fixed origin, positive direction = forward):
v(t)=dtds,a(t)=dtdv=dt2d2s.
Speed=∣v(t)∣=dtds — always non-negative, regardless of direction.
v(t)=0: the particle is momentarily at rest.
v(t)>0: moving forward; v(t)<0: moving backward.
The particle changes direction exactly where v(t) changes sign (not merely where it is zero — the sign must flip on either side).
If the particle reverses direction at time tc∈(t1,t2), the total distance travelled from t1 to t2 is ∣s(tc)−s(t1)∣+∣s(t2)−s(tc)∣ — NOT simply ∣s(t2)−s(t1)∣, since backtracking would otherwise cancel out.
Near Earth's surface a freely falling body has constant acceleration g≈9.8m/s2 (32ft/s2), giving a=−g,v=−gt+v0,s=−21gt2+v0t+s0.
Related rates. A related-rates problem links two or more time-varying quantities through a single equation (areas, volumes, distances, …); differentiating that equation with respect to time t (using the chain rule on every quantity that depends on t) produces an equation relating the rates dtd(⋅), which is then solved for the unknown rate at the instant specified. The recurring workflow is: (1) write the governing geometric/physical relation, (2) differentiate both sides with respect to t, (3) substitute the known values (and any values found from the constraint at that instant), (4) solve for the required rate.
Tip
In a related-rates problem, only differentiate AFTER writing the general relation — substituting the specific numeric values too early (before differentiating) silently drops the terms whose rates you actually need.
v(t)=6t2−18t+12=6(t−1)(t−2) changes sign at t=1,2; total distance sums the absolute leg-distances; acceleration is a(t)=12t−18.
✓Final answer
Direction changes at t=1 and t=2.
Total distance in [0,4] is 34 m.
a(1)=−6, a(2)=6.
v(t)=s′(t) tells us when and where the particle changes direction (sign changes of v); the total distance adds the absolute value of each directional "leg"; the acceleration a(t)=v′(t) is evaluated at each time v=0.
Step 1. Differentiate for velocity and acceleration.
s(t)=2t3−9t2+12t−4⇒v(t)=6t2−18t+12=6(t2−3t+2)=6(t−1)(t−2), and a(t)=12t−18.
Step 2. Find when the particle changes direction.
v(t)=0 at t=1,2. For t<1: both factors negative ⇒v>0. For 1<t<2: (t−1)>0,(t−2)<0⇒v<0. For t>2: both positive ⇒v>0. Since v genuinely changes sign at both roots, the particle changes direction at t=1andt=2.
Step 3. Total distance in [0,4].
Compute s at 0,1,2,4: s(0)=−4, s(1)=2−9+12−4=1, s(2)=16−36+24−4=0, s(4)=128−144+48−4=28.
Total distance=∣s(1)−s(0)∣+∣s(2)−s(1)∣+∣s(4)−s(2)∣=∣1−(−4)∣+∣0−1∣+∣28−0∣=5+1+28=34 m.
Step 4. Acceleration at each time the velocity is zero.
a(1)=12(1)−18=−6; a(2)=12(2)−18=6.
✓Final answer
The particle changes direction at t=1 s and t=2 s.
Total distance travelled in the first 4 seconds is 34 m.
Acceleration is −6 m/s² at t=1 and 6 m/s² at t=2.
Sign-change of v(t) locates direction changes; sum |leg-distances| for total distance
Computing total distance as ∣s(4)−s(0)∣ (net displacement) instead of summing each leg
Reporting only one of the two direction-change times