3.2 Derivatives as a Rate of Change
Position, velocity, speed and acceleration — and what the signs of $v$ and $a$ together tell you about the motion.
In this section we look at applications of the derivative by focusing on its interpretation as a rate of change. These applications include acceleration and velocity in physics, population growth rates in biology, and marginal functions in economics.
Motion along a Line
Velocity is the rate of change of position. Taking the derivative of velocity gives the acceleration, the rate of change of velocity. It is also important to introduce speed, which is the magnitude of velocity.
📘 Definition — Rate of change
Let $s(t)$ be a function giving the position of an object at time $t$.
velocity$v(t)=s'(t)$
speed$|v(t)|$
acceleration$a(t)=v'(t)=s''(t)$
🧭 Reading the two signs together
- $v(t)>0$ — moving in the positive direction (left to right); $v(t)<0$ — moving right to left.
- $v$ and $a$ have the same sign $\Rightarrow$ the speed $|v|$ is increasing: the object is speeding up.
- $v$ and $a$ have opposite signs $\Rightarrow$ the speed is decreasing: the object is slowing down.
Note that "negative acceleration" does not mean slowing down — it means slowing down only when the velocity is positive.
✏️ Example 1 — Comparing instantaneous and average velocity
A ball is dropped from a height of $64$ feet. Its height above ground (in feet) $t$ seconds later is $s(t)=-16t^2+64$.
- a What is the instantaneous velocity of the ball when it hits the ground?
- b What is the average velocity during its fall?
💡 Show solution▶
Step 1 First find when the ball reaches the ground by setting $s(t)=0$:
$$-16t^2+64=0\ \Longrightarrow\ t^2=4\ \Longrightarrow\ t=2\ \text{seconds}.$$
(a) The velocity function is $v(t)=s'(t)=-32t$, so the instantaneous velocity on impact is
$$v(2)=-64\ \text{ft/s}.$$
(b) The average velocity during the fall is
$$v_{\text{ave}}=\frac{s(2)-s(0)}{2-0}=\frac{0-64}{2}=-32\ \text{ft/s}.$$
Compare The impact speed is twice the average speed, because the ball starts at rest and accelerates steadily throughout the fall.
✏️ Example 2 — Interpreting the relationship between $v(t)$ and $a(t)$
A particle moves along a coordinate axis, positive direction to the right. Its position at time $t$ is $s(t)=t^3-4t+2$. Find $v(1)$ and $a(1)$, and use them to answer:
- a Is the particle moving left to right or right to left at $t=1$?
- b Is the particle speeding up or slowing down at $t=1$?
💡 Show solution▶
Step 1 Differentiate twice:
$$v(t)=s'(t)=3t^2-4,\qquad a(t)=v'(t)=s''(t)=6t.$$
At $t=1$: $v(1)=-1$ and $a(1)=6$.
(a) Because $v(1)<0$, the particle is moving from right to left.
(b) Because $v(1)<0$ and $a(1)>0$, velocity and acceleration act in opposite directions. The particle is being accelerated opposite to its direction of travel, so $|v(t)|$ is decreasing: it is slowing down.
✏️ Example 3 — Position and velocity
The position of a particle moving along a coordinate axis is $s(t)=t^3-9t^2+24t+4$, $t\ge0$.
- a Find $v(t)$.
- b At what times is the particle at rest? On what intervals does it move from left to right? From right to left?
- c Use the information to sketch the path of the particle along a coordinate axis.
💡 Show solution▶
(a) $v(t)=s'(t)=3t^2-18t+24$.
(b) At rest means $v(t)=0$:
$$3t^2-18t+24=3(t-2)(t-4)=0\ \Longrightarrow\ t=2\ \text{and}\ t=4.$$
Since $3(t-2)(t-4)>0$ on $[0,2)\cup(4,+\infty)$, the particle moves left to right there; since it is negative on $(2,4)$, the particle moves right to left on that interval.
(c) We need the position at the start and at each turning point: $s(0)=4$, $s(2)=24$, $s(4)=20$. So the particle starts at $4$, travels right to $24$, reverses and moves left to $20$, then turns again and continues right forever.
Watch it Select Example 3 in the Motion Lab above: the number line traces exactly that path, $4\to24\to20\to\infty$.
Exercises
I
A particle moves along a coordinate axis. Its position at time $t$ is $s(t)=t^2-5t+1$. Is the particle moving from right to left or from left to right at time $t=3$?
Watch the particle
time $t$—
$s(t)$—
$v(t)$—
$a(t)$—
speed—
💡 Show solution▶
Step 1 $v(t)=s'(t)=2t-5$.
Step 2 At $t=3$: $v(3)=6-5=1>0$, so the particle is moving from left to right.
Note It changes direction at $t=\tfrac52$, where $v=0$; before that it was moving right to left.
II
The given functions represent the position of a particle travelling along a horizontal line. Find the velocity and acceleration functions, and determine the time intervals when the object is slowing down or speeding up.
- 1 $s(t)=2t^3-3t^2-12t+8$
- 2 $s(t)=2t^3-15t^2+36t-10$
- 3 $s(t)=\dfrac{t}{1+t^2}$
Speeding up or slowing down?
time $t$—
$v(t)$—
$a(t)$—
right now—
💡 Show solutions▶
1 $v(t)=6t^2-6t-12=6(t-2)(t+1)$ and $a(t)=12t-6=6(2t-1)$.
For $t\ge0$: $v<0$ on $(0,2)$ and $v>0$ on $(2,\infty)$; $a<0$ on $\left(0,\tfrac12\right)$ and $a>0$ on $\left(\tfrac12,\infty\right)$.
Same signs on $\left[0,\tfrac12\right)$ and $(2,\infty)$ — speeding up; opposite signs on $\left(\tfrac12,2\right)$ — slowing down.
2 $v(t)=6t^2-30t+36=6(t-2)(t-3)$ and $a(t)=12t-30=6(2t-5)$.
$v>0$ on $(0,2)\cup(3,\infty)$, $v<0$ on $(2,3)$; $a<0$ for $t<\tfrac52$, $a>0$ for $t>\tfrac52$.
Speeding up on $\left(2,\tfrac52\right)\cup(3,\infty)$; slowing down on $(0,2)\cup\left(\tfrac52,3\right)$.
3 Quotient rule:
$$v(t)=\frac{1-t^2}{\left(1+t^2\right)^2},\qquad a(t)=\frac{2t\left(t^2-3\right)}{\left(1+t^2\right)^3}.$$
For $t\ge0$: $v>0$ on $(0,1)$, $v<0$ on $(1,\infty)$; $a<0$ on $\left(0,\sqrt3\right)$, $a>0$ on $\left(\sqrt3,\infty\right)$.
Speeding up on $\left(1,\sqrt3\right)$; slowing down on $(0,1)\cup\left(\sqrt3,\infty\right)$.
Rule Speeding up $\iff v(t)\,a(t)>0$. Comparing signs is faster than reasoning about the words.
III
A rocket is fired vertically upward from the ground. The distance $s$ in feet that the rocket travels from the ground after $t$ seconds is $s(t)=-16t^2+560t$.
- 1 Find the velocity of the rocket $3$ seconds after being fired.
- 2 Find the acceleration of the rocket $3$ seconds after being fired.
- 1 Determine how long it takes for the ball to hit the ground.
- 2 Determine the velocity of the ball when it hits the ground.
Vertical motion lab
time $t$—
height $s(t)$—
$v(t)$—
$a(t)$—
speed—
💡 Show solutions▶
III.1 $v(t)=s'(t)=-32t+560$, so $v(3)=-96+560=464$ ft/s.
III.2 $a(t)=v'(t)=-32$, so $a(3)=-32$ ft/s² — gravity, constant throughout the flight.
VII.1 The ball hits the ground when $s(t)=0$:
$$-16t^2-8t+64=0\ \Longrightarrow\ 2t^2+t-8=0\ \Longrightarrow\ t=\frac{-1+\sqrt{65}}{4}\approx1.7656\ \text{seconds}.$$
(The negative root is discarded.)
VII.2 $v(t)=-32t-8$, so
$$v\!\left(\frac{-1+\sqrt{65}}{4}\right)=-8\sqrt{65}\approx-64.50\ \text{ft/s}.$$
The sign is negative because the ball is moving downward.
IV
The position function $s(t)=t^3-8t$ gives the position in miles of a freight train, where east is the positive direction and $t$ is measured in hours.
- 1 Determine the direction the train is travelling when $s(t)=0$.
- 2 Determine the direction the train is travelling when $a(t)=0$.
- 3 Determine the time intervals when the train is slowing down or speeding up.
Speeding up or slowing down?
time $t$—
$v(t)$—
$a(t)$—
right now—
💡 Show solutions▶
Setup $v(t)=3t^2-8$ and $a(t)=6t$.
1 $s(t)=t\left(t^2-8\right)=0$ at $t=0$ and $t=2\sqrt2\approx2.83$ hours.
At $t=0$: $v(0)=-8<0$, so the train is travelling west.
At $t=2\sqrt2$: $v=3(8)-8=16>0$, so it is travelling east.
2 $a(t)=6t=0$ at $t=0$, where $v(0)=-8<0$: the train is travelling west.
3 For $t>0$ we have $a(t)>0$, while $v(t)=3t^2-8$ is negative for $t<\sqrt{8/3}\approx1.63$ and positive after.
So the train is slowing down on $\left(0,\sqrt{8/3}\right)$ and speeding up on $\left(\sqrt{8/3},\infty\right)$.
V
A potato is launched vertically upward with an initial velocity of $100$ ft/s from a potato gun at the top of an $85$-foot-tall building. The distance in feet that the potato travels from the ground after $t$ seconds is $s(t)=-16t^2+100t+85$.
- 1 Find the velocity of the potato after $0.5$ s and $5.75$ s.
- 2 Find the speed of the potato at $0.5$ s and $5.75$ s.
- 3 Determine when the potato reaches its maximum height.
- 4 Find the acceleration of the potato at $0.5$ s and $1.5$ s.
- 5 Determine how long the potato is in the air.
- 6 Determine the velocity of the potato upon hitting the ground.
Potato flight
time $t$—
height $s(t)$—
$v(t)$—
$a(t)$—
speed—
💡 Show solutions▶
Setup $v(t)=-32t+100$ and $a(t)=-32$.
1 $v(0.5)=-16+100=84$ ft/s (rising); $v(5.75)=-184+100=-84$ ft/s (falling).
2 Speed is $|v|$: both are $84$ ft/s. The potato passes the same height twice at the same speed, once going up and once coming down.
3 Maximum height where $v(t)=0$:
$$-32t+100=0\ \Longrightarrow\ t=\frac{100}{32}=3.125\ \text{seconds}.$$
4 $a(0.5)=a(1.5)=-32$ ft/s². Acceleration is constant — gravity does not care which way the potato is moving.
5 In the air until $s(t)=0$:
$$-16t^2+100t+85=0\ \Longrightarrow\ t=\frac{100+\sqrt{10000+5440}}{32}=\frac{100+\sqrt{15440}}{32}\approx7.0081\ \text{seconds}.$$
6 $v(7.0081)=-32(7.0081)+100\approx-124.26$ ft/s. It lands faster than it left, because it started $85$ feet above the ground.
VI
The position of a hummingbird flying along a straight line in $t$ seconds is $s(t)=3t^3-7t$ metres.
- 1 Determine the velocity of the bird at $t=1$ s.
- 2 Determine the acceleration of the bird at $t=1$ s.
- 3 Determine the acceleration of the bird when the velocity equals $0$.
- 1 Determine the time in seconds when the velocity of the car is $0$.
- 2 Determine the acceleration of the car when the velocity is $0$.
Motion lab
time $t$—
$s(t)$—
$v(t)$—
$a(t)$—
speed—
💡 Show solutions▶
VI.1 $v(t)=9t^2-7$, so $v(1)=2$ m/s.
VI.2 $a(t)=18t$, so $a(1)=18$ m/s².
VI.3 $v=0$ when $9t^2=7$, that is $t=\dfrac{\sqrt7}{3}\approx0.8819$ s. Then
$$a\left(\frac{\sqrt7}{3}\right)=18\cdot\frac{\sqrt7}{3}=6\sqrt7\approx15.87\ \text{m/s}^2.$$
VIII.1 $v(t)=3t^2-12t+9=3(t-1)(t-3)$, so $v=0$ at $t=1$ s and $t=3$ s.
VIII.2 $a(t)=6t-12$, so $a(1)=-6$ m/s² and $a(3)=6$ m/s². At $t=1$ the car is momentarily at rest with $a(1)=-6<0$, so its velocity is about to become negative (it reverses in the model); at $t=3$ it is again at rest with $a(3)=6>0$, so it starts moving forward again.
IX
Each function gives the position (in metres) of a particle moving along a horizontal line, positive direction to the right, with $t$ in seconds and $t\ge0$.
- 1 $s(t)=t^3-12t+3$. Find $v(t)$ and $a(t)$, when the particle is at rest, the intervals on which it moves left and right, and when it is speeding up or slowing down.
- 2 $s(t)=2t^3-21t^2+60t$, $0\le t\le6$. Find when the particle is at rest, when it is speeding up or slowing down, and the total distance travelled on $[0,6]$.
- 3 $s(t)=t^2-6t+5$, $0\le t\le5$. Find the displacement, the total distance travelled, the average velocity and the average speed on $[0,5]$.
- 4 Conceptual. For $s(t)=t^3-3t^2$, the acceleration is negative at $t=\tfrac12$. Is the particle slowing down there? Compare with $t=\tfrac32$ and explain why "negative acceleration" does not mean "slowing down".
Motion lab — rest points and total distance
time $t$—
$s(t)$—
$v(t)$—
$a(t)$—
speed—
💡 Show solutions▶
1 $v(t)=3t^2-12=3(t-2)(t+2)$ and $a(t)=6t$.
For $t\ge0$ the particle is at rest only at $t=2$, where $s(2)=8-24+3=-13$.
$v<0$ on $(0,2)$ — moving right to left; $v>0$ on $(2,\infty)$ — moving left to right.
Since $a(t)=6t>0$ for $t>0$, the signs disagree on $(0,2)$ — slowing down — and agree on $(2,\infty)$ — speeding up.
2 $v(t)=6t^2-42t+60=6(t-2)(t-5)$ and $a(t)=12t-42=6(2t-7)$.
At rest at $t=2$ and $t=5$. Signs: $v>0$ on $(0,2)\cup(5,6)$, $v<0$ on $(2,5)$; $a<0$ for $t<\tfrac72$, $a>0$ for $t>\tfrac72$.
Speeding up on $\left(2,\tfrac72\right)\cup(5,6)$; slowing down on $(0,2)\cup\left(\tfrac72,5\right)$.
For the distance, split $[0,6]$ at the turning points:
$$s(0)=0,\quad s(2)=16-84+120=52,\quad s(5)=250-525+300=25,\quad s(6)=432-756+360=36.$$
$$\text{total distance}=|52-0|+|25-52|+|36-25|=52+27+11=90\ \text{m}.$$
(The displacement is only $s(6)-s(0)=36$ m.)
3 $v(t)=2t-6$, so the particle turns at $t=3$. Positions: $s(0)=5$, $s(3)=9-18+5=-4$, $s(5)=25-30+5=0$.
$$\text{displacement}=s(5)-s(0)=-5\ \text{m},\qquad \text{distance}=|{-4}-5|+|0-(-4)|=9+4=13\ \text{m}.$$
Average velocity $=\dfrac{-5}{5}=-1$ m/s; average speed $=\dfrac{13}{5}=2.6$ m/s. Displacement can cancel; distance cannot.
4 $v(t)=3t^2-6t$ and $a(t)=6t-6$.
At $t=\tfrac12$: $v=\tfrac34-3=-\tfrac94$ and $a=-3$. Both negative, so $v\,a>0$: the particle is speeding up (moving left faster and faster).
At $t=\tfrac32$: $v=\tfrac{27}{4}-9=-\tfrac94$ and $a=3$. Opposite signs: slowing down.
The sign of $a$ alone only tells you which way the velocity is changing; whether the speed grows depends on whether $a$ points the same way as $v$.
X
Vertical motion. In each problem $s(t)$ is the height above the ground and upward is positive.
- 1 A ball is thrown upward at $32$ ft/s from a $48$-foot platform, so $s(t)=-16t^2+32t+48$ (feet). Find (a) the maximum height, (b) when the ball hits the ground, (c) its velocity and speed on impact, and (d) the total distance it travels.
- 2 A stone is thrown upward at $19.6$ m/s from the edge of a $24.5$-m cliff: $s(t)=-4.9t^2+19.6t+24.5$ (metres). Find the maximum height, the time and velocity at which it hits the ground, and its velocity when it passes the cliff edge on the way down.
- 3 On the Moon, a rock is thrown upward from the surface with $s(t)=-0.8t^2+24t$ (metres). Find its acceleration, its maximum height, and the two times at which its speed is $8$ m/s. What is its height at each of those times?
- 4 An object is launched straight up from the ground with initial velocity $v_0$ ft/s, so $s(t)=-16t^2+v_0t$. What value of $v_0$ makes the maximum height exactly $144$ ft? How long is the object in the air?
Vertical motion lab — up, over, down
time $t$—
height $s(t)$—
$v(t)$—
$a(t)$—
speed—
💡 Show solutions▶
1 $v(t)=-32t+32$ and $a(t)=-32$ ft/s².
(a) $v(t)=0$ at $t=1$, so the maximum height is $s(1)=-16+32+48=64$ ft.
(b) $-16t^2+32t+48=-16(t-3)(t+1)=0$ gives $t=3$ s.
(c) $v(3)=-96+32=-64$ ft/s; the speed is $64$ ft/s.
(d) Up from $48$ to $64$ ($16$ ft), then down from $64$ to $0$ ($64$ ft): total distance $=80$ ft.
2 $v(t)=-9.8t+19.6$, zero at $t=2$ s, so the maximum height is $s(2)=-19.6+39.2+24.5=44.1$ m.
Ground: $-4.9\left(t^2-4t-5\right)=-4.9(t-5)(t+1)=0$, so $t=5$ s and $v(5)=-49+19.6=-29.4$ m/s.
Cliff edge: $s(t)=24.5\iff t(-4.9t+19.6)=0\iff t=4$, and $v(4)=-39.2+19.6=-19.6$ m/s — the same speed it was thrown with, now downward.
3 $v(t)=24-1.6t$ and $a(t)=-1.6$ m/s² (the Moon's gravity, about one sixth of Earth's).
$v=0$ at $t=15$ s, so the maximum height is $s(15)=-180+360=180$ m.
Speed $8$ m/s means $v=\pm8$: $24-1.6t=8\Rightarrow t=10$ s (rising) and $24-1.6t=-8\Rightarrow t=20$ s (falling).
$s(10)=-80+240=160$ m and $s(20)=-320+480=160$ m — the same height, symmetric about the peak at $t=15$.
4 $v(t)=-32t+v_0=0$ at $t=\dfrac{v_0}{32}$, so the maximum height is
$$s\!\left(\frac{v_0}{32}\right)=-16\cdot\frac{v_0^2}{1024}+\frac{v_0^2}{32}=\frac{v_0^2}{64}.$$
Setting $\dfrac{v_0^2}{64}=144$ gives $v_0^2=9216$, so $v_0=96$ ft/s. Then $s(t)=-16t^2+96t=-16t(t-6)$, so the object is in the air for $6$ s (up for $3$ s, down for $3$ s).
XI
Rates of change in biology and economics.
- 1 A bacteria culture has $P(t)=500+80t+6t^2$ cells after $t$ hours. Find the growth rate $P'(5)$ and compare it with the average growth rate over $[0,5]$.
- 2 A fish population in a lake is modelled by $P(t)=\dfrac{3000t}{t+10}$, $t$ in years. Find $P'(t)$, evaluate $P'(10)$ and $P'(40)$, and interpret what happens to the growth rate.
- 3 The cost (in OMR) of producing $x$ units is $C(x)=1500+20x-0.05x^2+0.0002x^3$. Find the marginal cost $C'(100)$ and compare it with the actual cost $C(101)-C(100)$ of producing the $101$st unit.
- 4 A firm sells $x$ units at price $p=50-0.01x$ OMR each, and its cost is $C(x)=2000+14x$. Find the revenue $R(x)$ and profit $P(x)$, then $R'(1000)$, $P'(1000)$ and $P'(2000)$. Interpret the sign of $P'(2000)$ and find the production level where the marginal profit is zero.
💡 Show solutions▶
1 $P'(t)=80+12t$, so $P'(5)=140$ cells per hour.
The average rate is $\dfrac{P(5)-P(0)}{5-0}=\dfrac{1050-500}{5}=110$ cells per hour. The instantaneous rate at $t=5$ is larger because the growth is accelerating ($P''=12>0$).
2 Quotient rule:
$$P'(t)=\frac{3000(t+10)-3000t\cdot1}{(t+10)^2}=\frac{30000}{(t+10)^2}.$$
$P'(10)=\dfrac{30000}{400}=75$ fish per year and $P'(40)=\dfrac{30000}{2500}=12$ fish per year.
The population is still growing ($P'>0$), but ever more slowly: $P(t)\to3000$, the lake's carrying capacity.
3 $C'(x)=20-0.1x+0.0006x^2$, so $C'(100)=20-10+6=16$ OMR per unit.
The actual cost of the $101$st unit is
$$C(101)-C(100)=20-0.05\,(101^2-100^2)+0.0002\,(101^3-100^3)=20-10.05+6.0602=16.0102\ \text{OMR}.$$
The marginal cost is an excellent estimate of the cost of one more unit.
4 $R(x)=xp=50x-0.01x^2$ and $P(x)=R(x)-C(x)=-0.01x^2+36x-2000$.
$R'(x)=50-0.02x$, so $R'(1000)=30$ OMR per unit; $P'(x)=36-0.02x$, so $P'(1000)=16$ OMR per unit.
$P'(2000)=36-40=-4$: producing and selling one more unit beyond $2000$ would lower the profit by about $4$ OMR.
$P'(x)=0$ at $x=\dfrac{36}{0.02}=1800$ units — the output level beyond which extra units reduce profit.
XII
Geometric rates and a cooling model.
- 1 Find the rate of change of the area of a circle with respect to its radius when $r=5$ cm. What familiar quantity is this?
- 2 Find the rate of change of the volume of a cube with respect to its edge length $x$ when $x=4$ cm. Use it to estimate the change in volume when $x$ grows from $4$ to $4.1$ cm, and compare with the exact change.
- 3 Find $\dfrac{dV}{dr}$ for a sphere, $V=\tfrac43\pi r^3$, when $r=3$ m, and interpret the result.
- 4 A cup of coffee cools according to $T(t)=20+\dfrac{600}{t+8}$ °C, $t$ in minutes. Find $T(0)$, $T'(2)$ and $T'(22)$ with units, explain the sign, and find when the coffee is cooling at $1.5$ °C per minute.
💡 Show solutions▶
1 $A=\pi r^2$, so $\dfrac{dA}{dr}=2\pi r$ and at $r=5$: $10\pi\approx31.42$ cm² per cm. This is the circumference: a thin ring of width $\Delta r$ adds area $\approx2\pi r\,\Delta r$.
2 $V=x^3$, so $\dfrac{dV}{dx}=3x^2$ and at $x=4$: $48$ cm³ per cm.
Estimate: $\Delta V\approx48(0.1)=4.8$ cm³. Exact: $4.1^3-4^3=68.921-64=4.921$ cm³. The estimate is close because $\Delta x$ is small.
3 $\dfrac{dV}{dr}=4\pi r^2$, so at $r=3$: $36\pi\approx113.1$ m³ per m. This equals the surface area of the sphere — each extra layer of thickness $\Delta r$ adds about $4\pi r^2\,\Delta r$ of volume.
4 $T(0)=20+75=95$ °C and
$$T'(t)=-\frac{600}{(t+8)^2}.$$
$T'(2)=-\dfrac{600}{100}=-6$ °C/min and $T'(22)=-\dfrac{600}{900}=-\dfrac23\approx-0.67$ °C/min. The rates are negative because the temperature is decreasing, and they shrink in size as the coffee approaches room temperature ($20$ °C).
Cooling at $1.5$ °C/min means $T'(t)=-1.5$: $(t+8)^2=400$, so $t+8=20$ and $t=12$ minutes.
XIII
Interpreting $f'(a)$: rates of change and units.
- 1 The cost of roasting $q$ kilograms of coffee is $C(q)$ dollars. (a) What are the units of $C'(50)$, and what does $C'(50)=12$ mean? (b) Given $C(50)=900$ and $C(60)=1010$, find the average rate of change of $C$ on $[50,60]$ and explain how it differs from $C'(50)$.
- 2 The population of a town, $P(t)$ in thousands, $t$ years after 2010, is recorded below.
(a) Find the average rate of change of $P$ over $[0,8]$. (b) Estimate $P'(4)$ by averaging the average rates over $[2,4]$ and $[4,6]$, and interpret it with units.
$t$ (years) 0 2 4 6 8 $P(t)$ (thousands) 120 128 138 150 165 - 3 A cup of coffee cools so that its temperature is $T(t)$ °C after $t$ minutes. It is known that $T(10)=60$ and $T'(10)=-1.5$. Interpret $T'(10)$, including its sign, and use it to estimate $T(10.5)$.
- 4 The area of a circle of radius $r$ cm is $A(r)=\pi r^2$ cm². (a) Find the average rate of change of $A$ as $r$ goes from $2$ to $2.1$. (b) Use the definition to find $A'(2)$, give its units, and compare it with (a).
💡 Show solutions▶
1 (a) $C'$ has units of dollars per kilogram. $C'(50)=12$ means that when $50$ kg have been roasted, the cost is increasing at $12$ dollars per kilogram — roasting the $51$st kilogram costs about 12 dollars (the marginal cost). (b) $\dfrac{C(60)-C(50)}{60-50}=\dfrac{1010-900}{10}=11$ dollars per kg. This is an average over the ten extra kilograms; $C'(50)$ is the instantaneous rate at the single production level $q=50$.
2 (a) $\dfrac{P(8)-P(0)}{8-0}=\dfrac{165-120}{8}=5.625$ thousand people per year. (b) Over $[2,4]$: $\dfrac{138-128}{2}=5$; over $[4,6]$: $\dfrac{150-138}{2}=6$. So $P'(4)\approx\dfrac{5+6}{2}=5.5$ thousand people per year: in 2014 the population was growing at roughly $5500$ people per year. (A table gives only average rates; the derivative can only be estimated from it.)
3 $T'(10)=-1.5$ °C/min means that $10$ minutes after being poured, the coffee's temperature is decreasing (negative sign) at $1.5$ degrees per minute. Over the next half minute it drops by about $1.5\times0.5=0.75$ °C, so $T(10.5)\approx60-0.75=59.25$ °C.
4 (a) $\dfrac{A(2.1)-A(2)}{0.1}=\dfrac{\pi(4.41-4)}{0.1}=4.1\pi\approx12.88$ cm² per cm. (b)
$$A'(2)=\lim_{h\to0}\frac{\pi(2+h)^2-4\pi}{h}=\lim_{h\to0}\frac{\pi\left(4h+h^2\right)}{h}=\lim_{h\to0}\pi(4+h)=4\pi\approx12.57\ \text{cm}^2/\text{cm}.$$
The average rate $4.1\pi$ is exactly $\pi(4+h)$ with $h=0.1$ and approaches $4\pi$ as $h\to0$. (Notice $4\pi$ is the circumference of the circle of radius $2$.)
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes
Research question. How fast is Oman's population growing, and how different are the average and instantaneous growth rates?
Why it matters. Government planners use population growth rates to plan schools, hospitals and water supply; demographers study exactly these rates.
What to do.
- Find Oman's total population for at least six years from the Oman National Centre for Statistics and Information (NCSI) website or data portal.
- Compute the average rate of change (people per year) between each pair of neighbouring years.
- Estimate the instantaneous rate at a middle year using a central difference.
- Divide by the population to get a percentage growth rate. Note the units.
Tools: NCSI website, spreadsheet
Hand in: A half-page note: your table, the growth rate for one year in two forms (people/year and %/year), and one limitation of the data.
Going further. Anywhere: repeat with your own country's statistics office or with tourism arrivals data.
Instructor note: Students should notice that yearly data only give average rates; the instantaneous rate is an estimate, and sudden changes (e.g. 2020) break smooth models.
🧪 Self-check quiz
0/5 correct
Q1. If $v(t)<0$ and $a(t)<0$, the object is:
Q2. A ball dropped from $64$ ft has $s(t)=-16t^2+64$. Its speed on impact is:
Q3. For $s(t)=t^3-9t^2+24t+4$, the particle is at rest at:
Q4. A projectile reaches its maximum height at the moment when:
Q5. For the potato, $v(0.5)=84$ and $v(5.75)=-84$ ft/s. This means: