IB Mathematics AHL 5.15 Slope fields and their diagrams.-AI HL Paper 3- Exam Style Questions- New Syllabus
Question
This question uses differential equations to model the maximum velocity of a skydiver in free fall.
In 2025, Noah Baumgartner leapt from a height of 40000 m. He was attempting to travel at the speed of sound, \(330 \, \text{ms}^{-1}\), whilst free-falling to the Earth.
Before making his attempt, Noah used mathematical models to check how realistic his attempt would be. The simplest model he used suggests that
\[\dfrac{dv}{dt} = g\]
where \(v \, \text{ms}^{-1}\) is Noah’s velocity and \(g \, \text{ms}^{-2}\) is the acceleration due to gravity. The time since Noah began to free-fall is \(t\) seconds and the displacement from his initial position is \(s\) metres.
Throughout this question, the direction towards the centre of the Earth is taken to be positive and \(v\) is a positive quantity.
When \(s = 0\), it is given that Noah jumps with an initial velocity \(v = 10\).
(a)
(i) Use the chain rule to show that \[\dfrac{dv}{dt} = v \dfrac{dv}{ds}\]. [1]
(ii) Assuming that \(g\) is a constant, solve the differential equation \[v \dfrac{dv}{ds} = g\] to find \(v\) as a function of \(s\). [4]
(iii) Using \(g = 9.8\), determine whether the model predicts that Noah will succeed in travelling at the speed of sound at some point before \(s = 40000\). Justify your answer. [3]
(i) Use the chain rule to show that \[\dfrac{dv}{dt} = v \dfrac{dv}{ds}\]. [1]
(ii) Assuming that \(g\) is a constant, solve the differential equation \[v \dfrac{dv}{ds} = g\] to find \(v\) as a function of \(s\). [4]
(iii) Using \(g = 9.8\), determine whether the model predicts that Noah will succeed in travelling at the speed of sound at some point before \(s = 40000\). Justify your answer. [3]
(b)
To test the model
To test the model
\[\dfrac{dv}{dt} = g\],
Noah conducted a trial jump from a lower height, and data for \(v\) against \(t\) was found.
(i) If the model is correct, describe the shape of the graph of \(v\) against \(t\). [2]
Noah’s data are plotted on the following graph.
(ii) Use the plot to comment on the validity of the model in part (a). [1]
(c)
An improved model considers air resistance, using
An improved model considers air resistance, using
\[\dfrac{dv}{dt} = g – k v^2\]
where \(k\) is a positive constant. You are reminded that initially \(s=0\) and \(v=10\).
(i) By using \[\dfrac{dv}{dt} = v \dfrac{dv}{ds}\], solve the differential equation to find \(v\) in terms of \(s, g\) and \(k\). You may assume that \(g – kv^2 > 0\). [5]
Noah uses the graph of \(v\) against \(t\) shown in part (b) to estimate the value of \(k\).
(ii) The gradient is estimated to be 9.672 when \(v=40\). Taking \(g=9.8\), use this information to show that Noah found that \(k = 8 \times 10^{-5}\). [2]
(iii) Hence, find the value of \(v\) predicted by this model, as \(s\) tends to infinity. [1]
(iv) Find the upper bound for the velocity according to this model, given that \(0 < s \leq 40000\). Give your answer to four significant figures. [2]
The assumption that the value of \(g\) is constant is not correct. It can be shown that
\[g = \dfrac{3.98 \times 10^{14}}{(6.41 \times 10^6 – s)^2}.\]
Hence, the new model is given by
\[v \dfrac{dv}{ds} = \dfrac{3.98 \times 10^{14}}{(6.41 \times 10^6 – s)^2} – (8 \times 10^{-5})v^2.\]
When \(s=0\), it is known that \(v=10\).
(d)
Use Euler’s method with a step length of 4000 to estimate the value of \(v\) when \(s = 40000\). [4]
Use Euler’s method with a step length of 4000 to estimate the value of \(v\) when \(s = 40000\). [4]
(e)
After Noah completed his record-breaking jump, Noah found that the answer from part (d) was not supported by data collected during the jump.
After Noah completed his record-breaking jump, Noah found that the answer from part (d) was not supported by data collected during the jump.
(i) Suggest one improvement to the use of Euler’s method which might increase the accuracy of the prediction of the model. [1]
(ii) Suggest one factor not explicitly considered by the model in part (d) which might lead to a difference between the model’s prediction and the data collected. [1]
▶️ Answer/Explanation
Markscheme
(a)(i)
\( \frac{dv}{dt} = \frac{dv}{ds} \times \frac{ds}{dt} \)
\( v = \frac{ds}{dt} \)
\( \frac{dv}{dt} = v \frac{dv}{ds} \)
Chain rule derivation A1
[1 mark]
\( \frac{dv}{dt} = \frac{dv}{ds} \times \frac{ds}{dt} \)
\( v = \frac{ds}{dt} \)
\( \frac{dv}{dt} = v \frac{dv}{ds} \)
Chain rule derivation A1
[1 mark]
(a)(ii)
\( v \frac{dv}{ds} = g \)
Attempt to separate variables:
\( \int v \, dv = \int g \, ds \)
\( \frac{v^2}{2} = gs + c \)
Integration correct A1
Using initial conditions (\( s = 0, v = 10 \)):
\( \frac{10^2}{2} = g(0) + c \)
\( 50 = c \)
Constant found A1
\( \frac{v^2}{2} = gs + 50 \)
\( v = \sqrt{2gs + 100} \)
Solution \( v = \sqrt{2gs + 100} \) A1A1
[4 marks]
\( v \frac{dv}{ds} = g \)
Attempt to separate variables:
\( \int v \, dv = \int g \, ds \)
\( \frac{v^2}{2} = gs + c \)
Integration correct A1
Using initial conditions (\( s = 0, v = 10 \)):
\( \frac{10^2}{2} = g(0) + c \)
\( 50 = c \)
Constant found A1
\( \frac{v^2}{2} = gs + 50 \)
\( v = \sqrt{2gs + 100} \)
Solution \( v = \sqrt{2gs + 100} \) A1A1
[4 marks]
(a)(iii)
EITHER
Attempt to find \( s \) when \( v = 330 \):
\( 330 = \sqrt{2 \times 9.8 \times s + 100} \)
\( 330^2 = 19.6s + 100 \)
\( s = \frac{330^2 – 100}{19.6} \approx 5551.02 \)
Since \( 5551.02 < 40000 \), the model predicts Noah will reach the speed of sound.
Solve for \( s \), compare with 40000 M1 A1
OR
Attempt to find \( v \) when \( s = 40000 \):
\( v = \sqrt{2 \times 9.8 \times 40000 + 100} \)
\( v = \sqrt{784000 + 100} \approx 885.49 \)
Since \( 885 > 330 \), the model predicts Noah will reach the speed of sound before \( s = 40000 \).
Solve for \( v \), compare with 330 M1 A1 A1
[3 marks]
EITHER
Attempt to find \( s \) when \( v = 330 \):
\( 330 = \sqrt{2 \times 9.8 \times s + 100} \)
\( 330^2 = 19.6s + 100 \)
\( s = \frac{330^2 – 100}{19.6} \approx 5551.02 \)
Since \( 5551.02 < 40000 \), the model predicts Noah will reach the speed of sound.
Solve for \( s \), compare with 40000 M1 A1
OR
Attempt to find \( v \) when \( s = 40000 \):
\( v = \sqrt{2 \times 9.8 \times 40000 + 100} \)
\( v = \sqrt{784000 + 100} \approx 885.49 \)
Since \( 885 > 330 \), the model predicts Noah will reach the speed of sound before \( s = 40000 \).
Solve for \( v \), compare with 330 M1 A1 A1
[3 marks]
(b)(i)
\( \frac{dv}{dt} = g \)
\( v = gt + c \)
Gradient is constant, so the graph of \( v \) against \( t \) is a straight line.
Straight line graph M1 A1
[2 marks]
\( \frac{dv}{dt} = g \)
\( v = gt + c \)
Gradient is constant, so the graph of \( v \) against \( t \) is a straight line.
Straight line graph M1 A1
[2 marks]
(b)(ii)
The graph is not a straight line, only approximately straight for small \( t \), so the model does not appear to be valid.
Non-linear graph, model invalid R1
[1 mark]
The graph is not a straight line, only approximately straight for small \( t \), so the model does not appear to be valid.
Non-linear graph, model invalid R1
[1 mark]
(c)(i)
\( v \frac{dv}{ds} = g – k v^2 \)
Separate variables:
\( \int \frac{v}{g – k v^2} dv = \int ds \)
\( -\frac{1}{2k} \ln(g – k v^2) = s + c \)
Integration correct A1 A1
Rearrange:
\( g – k v^2 = A e^{-2ks} \)
\( v = \sqrt{\frac{g – A e^{-2ks}}{k}} \)
Apply initial conditions (\( s = 0, v = 10 \)):
\( 100 = \frac{g – A}{k} \)
\( A = g – 100k \)
Constant found A1
\( v = \sqrt{\frac{g – (g – 100k)e^{-2ks}}{k}} \)
Solution correct A1 A1
[5 marks]
\( v \frac{dv}{ds} = g – k v^2 \)
Separate variables:
\( \int \frac{v}{g – k v^2} dv = \int ds \)
\( -\frac{1}{2k} \ln(g – k v^2) = s + c \)
Integration correct A1 A1
Rearrange:
\( g – k v^2 = A e^{-2ks} \)
\( v = \sqrt{\frac{g – A e^{-2ks}}{k}} \)
Apply initial conditions (\( s = 0, v = 10 \)):
\( 100 = \frac{g – A}{k} \)
\( A = g – 100k \)
Constant found A1
\( v = \sqrt{\frac{g – (g – 100k)e^{-2ks}}{k}} \)
Solution correct A1 A1
[5 marks]
(c)(ii)
\( \frac{dv}{dt} = g – k v^2 \)
At \( v = 40 \), gradient = 9.672, \( g = 9.8 \):
\( 9.672 = 9.8 – k \times 40^2 \)
\( 9.672 = 9.8 – 1600k \)
Substitute values A1
\( k = \frac{9.8 – 9.672}{1600} = 8 \times 10^{-5} \)
\( k = 8 \times 10^{-5} \) A1
[2 marks]
\( \frac{dv}{dt} = g – k v^2 \)
At \( v = 40 \), gradient = 9.672, \( g = 9.8 \):
\( 9.672 = 9.8 – k \times 40^2 \)
\( 9.672 = 9.8 – 1600k \)
Substitute values A1
\( k = \frac{9.8 – 9.672}{1600} = 8 \times 10^{-5} \)
\( k = 8 \times 10^{-5} \) A1
[2 marks]
(c)(iii)
As \( s \to \infty \), \( e^{-2ks} \to 0 \):
\( v = \sqrt{\frac{g}{k}} \)
\( g = 9.8, k = 8 \times 10^{-5} \)
\( v = \sqrt{\frac{9.8}{8 \times 10^{-5}}} \approx 350 \, \text{ms}^{-1} \)
Terminal velocity 350 M1 A1
[2 marks]
As \( s \to \infty \), \( e^{-2ks} \to 0 \):
\( v = \sqrt{\frac{g}{k}} \)
\( g = 9.8, k = 8 \times 10^{-5} \)
\( v = \sqrt{\frac{9.8}{8 \times 10^{-5}}} \approx 350 \, \text{ms}^{-1} \)
Terminal velocity 350 M1 A1
[2 marks]
(c)(iv)
Upper limit at \( s = 40000 \):
\( v = \sqrt{\frac{9.8 – (9.8 – 100 \times 8 \times 10^{-5})e^{-2 \times 8 \times 10^{-5} \times 40000}}{8 \times 10^{-5}}} \)
\( v \approx 349.7 \, \text{ms}^{-1} \) (to 4 s.f.)
Velocity 349.7 M1 A1
[2 marks]
Upper limit at \( s = 40000 \):
\( v = \sqrt{\frac{9.8 – (9.8 – 100 \times 8 \times 10^{-5})e^{-2 \times 8 \times 10^{-5} \times 40000}}{8 \times 10^{-5}}} \)
\( v \approx 349.7 \, \text{ms}^{-1} \) (to 4 s.f.)
Velocity 349.7 M1 A1
[2 marks]
(d)
\( s_{n+1} = s_n + 4000 \)
\( v_{n+1} = v_n + 4000 \left( \frac{3.98 \times 10^{14}}{v_n (6.41 \times 10^6 – s_n)^2} – (8 \times 10^{-5}) v_n \right) \)
Euler’s method setup A1 M1
Initial condition: \( v_0 = 10, s_0 = 0 \).
Iterate 10 steps (\( s = 40000 \)):
\( v_{10} \approx 361 \, \text{ms}^{-1} \) (360.658…)
Calculation and result A1 A1
[4 marks]
\( s_{n+1} = s_n + 4000 \)
\( v_{n+1} = v_n + 4000 \left( \frac{3.98 \times 10^{14}}{v_n (6.41 \times 10^6 – s_n)^2} – (8 \times 10^{-5}) v_n \right) \)
Euler’s method setup A1 M1
Initial condition: \( v_0 = 10, s_0 = 0 \).
Iterate 10 steps (\( s = 40000 \)):
\( v_{10} \approx 361 \, \text{ms}^{-1} \) (360.658…)
Calculation and result A1 A1
[4 marks]
(e)(i)
Use a smaller step length.
Smaller step length R1
[1 mark]
Use a smaller step length.
Smaller step length R1
[1 mark]
(e)(ii)
Ignoring parachute effects.
Parachute not considered R1
[1 mark]
Ignoring parachute effects.
Parachute not considered R1
[1 mark]
Total Marks: 28