9.5 Numerical methods in context — Practise

Work through these on your own. Decide which method you need before you calculate, and write the units with every answer.

The questions and the answers below come from a set that is checked by computer every time the book is built.

A drone's height is $h(t) = 40e^{-t/4} - 2t$ metres. Show that the drone lands between $5$ and $6$ seconds.
For the drone model $h(t) = 40e^{-t/4} - 2t$ , write down $h'(t)$ .
Use Newton–Raphson with $t_0 = 5$ on $h(t) = 40e^{-t/4} - 2t$ . Find $t_1$ to four decimal places.
For the drone, state the landing time to three decimal places.
A river survey gives depths $0, 1.2, 2.5, 3.1, 2.4, 0.9, 0$ metres at distances $0, 2, 4, 6, 8, 10, 12$ metres. Estimate the cross-sectional area.
For the river survey, give a lower limit and an upper limit for the cross-sectional area.
A car's speed is recorded every two seconds: $0, 5, 9, 12, 14, 15$ metres per second at $t = 0, 2, 4, 6, 8, 10$ seconds. Estimate the distance travelled.
For the speed trace in the previous question, give a lower limit and an upper limit for the distance travelled.
A drug's concentration is $C(t) = 6te^{-0.4t}$ milligrams per litre. Show that the concentration reaches $2$ milligrams per litre between $t = 0.35$ and $t = 0.4$ hours.
For $f(t) = 3t - e^{0.4t}$ , use Newton–Raphson with $t_0 = 0.4$ . Find $t_1$ to four decimal places.
State the time from the previous question to two decimal places, in hours.
A cooling model gives $T = 20 + 60e^{-0.05t} - 0.5t$ degrees, $t$ minutes after a switch is pressed. Show that the temperature is $30$ degrees between $t = 21$ and $t = 22$ .
For the cooling model in the previous question, state the time when the temperature is $30$ degrees, to two decimal places.
A second drone has height $h(t) = 30e^{-t/3} - 3t$ metres. Show that it lands between $3$ and $4$ seconds, and state the landing time to two decimal places.

Answers and concise solutions

1. The two heights

Answer: $h(5) = 40e^{-1.25} - 10 = 1.4602 > 0$ and $h(6) = 40e^{-1.5} - 12 = -3.0748 < 0$ . The height changes sign, so the landing time lies in $(5, 6)$ .

2. The derivative

Answer: Differentiate each term: the derivative of $40e^{-t/4}$ is $-10e^{-t/4}$ , and the derivative of $-2t$ is $-2$ . So $h'(t) = -10e^{-t/4} - 2$ .

3. The value of t1

Answer: $h(5) = 1.4602$ and $h'(5) = -4.8650$ , so $t_1 = 5 - \dfrac{1.4602}{-4.8650} = 5.3001$ to four decimal places.

4. The landing time

Answer: The Newton–Raphson values are $t_1 = 5.3001$ , $t_2 = 5.3069$ and $t_3 = 5.3069$ , so the landing time is $5.307$ seconds to three decimal places. In context, the answer is about $5.3$ seconds.

5. The area

Answer: The readings are equally spaced with $h = 2$ , and the interior depths add to $10.1$ . So $A \approx \dfrac{2}{2}\bigl[0 + 0 + 2(10.1)\bigr] = 20.2$ square metres.

6. The limits

Answer: The smaller depth on each strip gives the lower sum $2(0 + 1.2 + 2.5 + 2.4 + 0.9 + 0) = 14.0$ square metres. The larger depth on each strip gives the upper sum $2(1.2 + 2.5 + 3.1 + 3.1 + 2.4 + 0.9) = 26.4$ square metres. So the area lies between $14.0$ and $26.4$ square metres.

7. The distance

Answer: The distance is the area under the speed–time graph. With $h = 2$ , $s \approx \dfrac{2}{2}\bigl[0 + 15 + 2(5 + 9 + 12 + 14)\bigr] = 95$ metres.

8. The limits

Answer: The speed increases, so the lower sum uses the left-hand heights: $2(0 + 5 + 9 + 12 + 14) = 80$ metres. The upper sum uses the right-hand heights: $2(5 + 9 + 12 + 14 + 15) = 110$ metres. The distance lies between $80$ and $110$ metres, with the estimate $95$ in the middle.

9. The two values

Answer: Set $C(t) = 2$ and divide by $2$ : $3te^{-0.4t} = 1$ , so $3t = e^{0.4t}$ . With $f(t) = 3t - e^{0.4t}$ , $f(0.35) = -0.1003 < 0$ and $f(0.4) = 0.0265 > 0$ , so the root lies in $(0.35, 0.4)$ .

10. The value of t1

Answer: $f'(t) = 3 - 0.4e^{0.4t}$ , so $f'(0.4) = 2.5306$ . Then $t_1 = 0.4 - \dfrac{0.0265}{2.5306} = 0.3895$ to four decimal places.

11. The time

Answer: The values settle on $0.3895$ , so the time is $0.39$ hours to two decimal places. That is about $23.4$ minutes.

12. The two values

Answer: Set $T = 30$ and subtract $30$ : $60e^{-0.05t} - 0.5t - 10 = 0$ . At $t = 21$ the value is $0.4963 > 0$ , and at $t = 22$ it is $-1.0277 < 0$ . The sign changes, so the temperature reaches $30$ degrees between $21$ and $22$ minutes.

13. The time

Answer: Newton–Raphson from $t_0 = 21$ gives $21.2925$ , then $21.3219$ and $21.3219$ . The time is $21.32$ minutes to two decimal places.

14. The bracket and the time

Answer: $h(3) = 30e^{-1} - 9 = 2.0364 > 0$ and $h(4) = 30e^{-4/3} - 12 = -4.0921 < 0$ , so the drone lands between $3$ and $4$ seconds. Newton–Raphson gives $3.31$ seconds to two decimal places.