Q.A police jeep, approaching an orthogonal intersection from the northern direction, is chasing a speeding car that has turned and moving straight east. When the jeep is 0.6 km north of the intersection and the car is 0.8 km to the east, the police determine with a radar that the distance between them and the car is increasing at 20 km/hr. If the jeep is moving at 60 km/hr at the instant of measurement, what is the speed of the car?

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

Taking $h\to0$ gives the slope of the curve at $(x,f(x))$ itself:

If $\theta$ is the angle the tangent makes with the positive $x$-axis (measured anticlockwise), then $f'(x)=\tan\theta$.

As a rate of change. $f'(x)=\dfrac{dy}{dx}$ 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 $\dfrac{f(b)-f(a)}{b-a}$ (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):

• Speed $=|v(t)|=\left|\dfrac{ds}{dt}\right|$ — 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 $t_c\in(t_1,t_2)$, the total distance travelled from $t_1$ to $t_2$ is $|s(t_c)-s(t_1)|+|s(t_2)-s(t_c)|$ — NOT simply $|s(t_2)-s(t_1)|$, since backtracking would otherwise cancel out.
• Near Earth's surface a freely falling body has constant acceleration $g\approx9.8\ \text{m/s}^2$ ($32\ \text{ft/s}^2$), giving $a=-g,\ v=-gt+v_0,\ s=-\tfrac12gt^2+v_0t+s_0$. …