Average and instantaneous rates of change

(a) $3$ litres/second

1. $$\frac{31-7}{10-2}$$
2. The instantaneous rate of change is estimated by the gradient of the tangent.
3. Use its two stated points: (31 $-$ 7)/(10 $-$ 2) = 24/8 = 3 litres per second.

• P1 Establishing $\frac{31-7}{10-2}$ or an equivalent valid method.
• A1 Correct answer: $3$ litres/second

(a) $5$ litres/s

1. $$\frac{45-15}{8-2}$$
2. The instantaneous rate is estimated by the tangent gradient.
3. Gradient = (45 $-$ 15)/(8 $-$ 2) = 5 litres per second.

• P1 Establishing $\frac{45-15}{8-2}$ or an equivalent valid method.
• A1 Correct answer: $5$ litres/s

(a) $6.55\ ^{\circ}\text{C}$ per minute

1. $T(0) = 100$ and $T(5) = 80 \times 0.9^5 + 20 = 67.2392$.
2. $$\frac{100 - 67.2392}{5} = 6.552\ldots \approx 6.55$$

• M1 Finding $T(0) = 100$ and $T(5) = 67.24$.
• M1 Dividing the fall by 5.
• A1 The correct answer, $6.55$.

(b) At t = 0: the graph is steeper at the start.

1. Each minute the excess temperature $T - 20$ is multiplied by $0.9$, so the fall is $10\%$ of the excess. The excess is largest at the start, so cooling is fastest at $t = 0$.

• C1 At $t = 0$, because the curve is steeper there (the excess temperature is larger).

(a) $30$ per year

1. Gradient of the tangent: $\frac{560 - 200}{16 - 4} = \frac{360}{12} = 30$.

• M1 Finding the tangent's gradient from the two points.
• A1 The correct answer, $30$ per year.

(b) The tangent touches the curve only at t = 10, so its gradient is the rate at that instant; a chord gives an average over 12 years.

1. A tangent has the same direction as the curve at the point of contact, so its gradient is the instantaneous rate. A chord joins two points and gives the average rate between them.

• C1 Tangent: rate at an instant; chord: average rate over an interval.

(a) $18$ litres/s

1. Evaluate both endpoint volumes and calculate their difference.
   $$(3\times 5^{2}+7)-(3+7)$$
2. Divide by the four-second interval.
   $$72/4$$
3. Therefore $18$ litres/s.

• M1 Evaluate both endpoint volumes and calculate their difference.
• M1 Divide by the four-second interval.
• A1 Correct answer: $18$ litres/s

(b) $33.3$%

1. Compare the increase in rate with the average as the reference.
   $$(24-18)/18\times 100$$
2. Therefore $33.3$%.

• M1 Compare the increase in rate with the average as the reference.
• A1 Correct answer: $33.3$%

(a) $1500$ bacteria per hour

1. Divide the change in N by the change in t.
   $$\frac{500\times 2^{3}-500\times 2^{1}}{2}$$
2. Therefore $1500$ bacteria per hour.

• M1 Divide the change in N by the change in t.
• A1 Correct answer: $1500$ bacteria per hour

(b) $1440$ bacteria per hour

1. Find the gradient of the short chord.
   $$(500\times 2^{2}.1-500\times 2^{2})/0.1$$
2. Therefore $1440$ bacteria per hour.

• M1 Find the gradient of the short chord.
• A1 Correct answer: $1440$ bacteria per hour

(c) A shorter chord is closer to the tangent at t = 2.

1. A shorter chord is closer to the tangent at t = 2.

• C1 Correct conclusion with supporting reasoning: A shorter chord is closer to the tangent at t = 2.

(a) $-0.8$ metres per hour

1. Find the gradient of the tangent.
   $$\frac{2.0-5.2}{8-4}$$
2. Therefore $-0.8$ metres per hour.

• P1 Find the gradient of the tangent.
• A1 Correct answer: $-0.8$ metres per hour

(b) It is negative, so at 6 am the water depth is decreasing, by about 0.8 m per hour.

1. It is negative, so at 6 am the water depth is decreasing, by about 0.8 m per hour.

• C1 Correct conclusion with supporting reasoning: It is negative, so at 6 am the water depth is decreasing, by about 0.8 m per hour.

(c) $-0.95$ metres per hour

1. Divide the change in depth by 4 hours.
   $$\frac{1.8-5.6}{4}$$
2. Therefore $-0.95$ metres per hour.

• P1 Divide the change in depth by 4 hours.
• A1 Correct answer: $-0.95$ metres per hour

(a) $0$ m/s

1. $h(1) = 20 - 5 = 15$ and $h(3) = 60 - 45 = 15$.
2. Average rate: $\frac{15 - 15}{2} = 0$.

• M1 Finding $h(1) = 15$ and $h(3) = 15$.
• A1 The correct answer, $0$.

(b) The ball is at the same height at t = 1 and t = 3: it rises and then falls back.

1. The ball goes up and comes back down to the same height, so its overall change in height over those 2 seconds is zero, even though it was moving.

• C1 Saying the ball is at the same height at both times (up then down), so the average rate is zero although it moves.

(c) $-0.5$ m/s

1. $h(2) = 40 - 20 = 20$ and $h(2.1) = 42 - 22.05 = 19.95$.
2. $$\frac{19.95 - 20}{0.1} = -0.5$$
3. (The true rate at $t = 2$ is 0: the top of the flight. A shorter chord gives an estimate closer to 0.)

• P1 Finding $h(2.1) = 19.95$.
• A1 The correct answer, $-0.5$.