IB Mathematics AHL 5.16 first order differential equations AI HL Paper 3- Exam Style Questions- New Syllabus
Question
This question investigates mathematical models for the water level in a cylindrical container as it drains.
The diagram depicts a cylindrical water tank with a height of 3.2 metres and a base radius of 1 metre. A small circular valve at the base allows water to drain.
Eva seals the valve and fills the tank completely with water.
At time \( t = 0 \), Eva opens the valve. She measures the height, \( h \) metres, of water left in the tank at 5-minute intervals.
| Time, \( t \) (minutes) | Height, \( h \) (metres) |
|---|---|
| 0 | 3.2 |
| 5 | 2.4 |
| 10 | 1.6 |
| 15 | 1.1 |
| 20 | 0.5 |
(a)(i) Determine the equation of the linear regression line for \( h \) as a function of \( t \). [2]
(a)(ii) Explain the significance of parameter \( a \) in the context of this model. [1]
(b)(i) Eva uses the linear regression equation to predict the time required for all water to drain from the tank.
Explain why using the linear regression equation to predict this time might be unreliable. [1]
Explain why using the linear regression equation to predict this time might be unreliable. [1]
(b)(ii) Eva thinks she can improve her model by using a quadratic function, \( h(t) = pt^2 + qt + r \), where \( p, q, r \in \mathbb{R} \).
Determine the equation of the least squares quadratic regression curve. [1]
Determine the equation of the least squares quadratic regression curve. [1]
(b)(iii) Using this equation, Eva predicts the time required for all water to drain from the tank, obtaining \( k \) minutes. Calculate \( k \). [2]
(b)(iv) Hence, state an appropriate domain for Eva’s function \( h(t) = pt^2 + qt + r \). [1]
(c) Let \( V \) be the volume, in cubic metres, of water in the tank at time \( t \). Let \( R \) be the radius, in metres, of the circular valve.
Eva finds a formula for the rate of change of \( V \) through research: \[ \frac{dV}{dt} = -\pi R^2 \sqrt{70560h} \] Show that \(\frac{dh}{dt} = -R^2 \sqrt{70560h}\). [3]
Eva finds a formula for the rate of change of \( V \) through research: \[ \frac{dV}{dt} = -\pi R^2 \sqrt{70560h} \] Show that \(\frac{dh}{dt} = -R^2 \sqrt{70560h}\). [3]
(d) Solve the differential equation \(\frac{dh}{dt} = -R^2 \sqrt{70560h}\), to show that the general solution is given by \[ h = 17640 (c – R^2 t)^2, \quad c \in \mathbb{R}. \] [5]
(e) Eva measures the radius of the valve to be 0.023 metres. Let \( T \) be the time, in minutes, it takes for all the water to drain out of the tank. Using the general solution from part (d) and the initial condition \( h(0) = 3.2 \), calculate the value of \( T \). [4]
(f) Eva plans to use the tank as a timer. She adjusts the initial height of water in the tank so that all the water will drain out in 15 minutes. Find this new height. [3]
Eva has a second water tank identical to the first. She stacks one tank above the other, so water from the upper tank drains into the lower one. The upper tank is filled completely, while the lower tank is filled to a height of 1 metre.
At time \( t = 0 \), Eva opens both valves. Let \( H \) be the height of water, in metres, in the lower tank at time \( t \).
(g)(i) Show that \[ \frac{dH}{dt} \approx 0.2514 – 0.009873t – 0.1405\sqrt{H}, \quad 0 \leq t \leq T \] [4]
(g)(ii) Apply Euler’s method with a step size of 0.5 minutes to estimate the maximum value of \( H \). [3]
▶️ Answer/Explanation
Markscheme
(a)(i)
The equation of the linear regression line for \( h \) as a function of \( t \):
\( h(t) = -0.134t + 3.1 \)
Equation: \( h(t) = -0.134t + 3.1 \) A1A1
[2 marks]
The equation of the linear regression line for \( h \) as a function of \( t \):
\( h(t) = -0.134t + 3.1 \)
Equation: \( h(t) = -0.134t + 3.1 \) A1A1
[2 marks]
(a)(ii)
The parameter \( a \) represents the rate of change of height (of water in metres per minute).
Rate of change = \(-0.134\) m/min A1
[1 mark]
The parameter \( a \) represents the rate of change of height (of water in metres per minute).
Rate of change = \(-0.134\) m/min A1
[1 mark]
(b)(i)
The linear regression equation is unreliable because the rate of change of height might not remain constant as water drains out.
Unreliable due to non-constant rate of change A1
[1 mark]
The linear regression equation is unreliable because the rate of change of height might not remain constant as water drains out.
Unreliable due to non-constant rate of change A1
[1 mark]
(b)(ii)
The equation of the least squares quadratic regression curve is:
\( h(t) = 0.002t^2 – 0.174t + 3.2 \)
Quadratic equation = \( 0.002t^2 – 0.174t + 3.2 \) A1
[1 mark]
The equation of the least squares quadratic regression curve is:
\( h(t) = 0.002t^2 – 0.174t + 3.2 \)
Quadratic equation = \( 0.002t^2 – 0.174t + 3.2 \) A1
[1 mark]
(b)(iii)
Solve: \( 0.002t^2 – 0.174t + 3.2 = 0 \)
\( t \approx 26.4 \) (26.4046…)
\( k = 26.4 \) M1A1
[2 marks]
Solve: \( 0.002t^2 – 0.174t + 3.2 = 0 \)
\( t \approx 26.4 \) (26.4046…)
\( k = 26.4 \) M1A1
[2 marks]
(b)(iv)
A suitable domain for \( h(t) = 0.002t^2 – 0.174t + 3.2 \):
\( 0 \leq t \leq 26.4 \)
Domain = \( [0, 26.4] \) A1
[1 mark]
A suitable domain for \( h(t) = 0.002t^2 – 0.174t + 3.2 \):
\( 0 \leq t \leq 26.4 \)
Domain = \( [0, 26.4] \) A1
[1 mark]
(c)
The volume of water is \( V = \pi (1^2) h = \pi h \).
Given: \( \frac{dV}{dt} = -\pi R^2 \sqrt{70560h} \).
Using the chain rule: \( \frac{dV}{dt} = \pi \frac{dh}{dt} \).
Thus: \( \pi \frac{dh}{dt} = -\pi R^2 \sqrt{70560h} \).
Divide through by \( \pi \):
\( \frac{dh}{dt} = -R^2 \sqrt{70560h} \).
Result: \( \frac{dh}{dt} = -R^2 \sqrt{70560h} \) A1 M1 A1
[3 marks]
The volume of water is \( V = \pi (1^2) h = \pi h \).
Given: \( \frac{dV}{dt} = -\pi R^2 \sqrt{70560h} \).
Using the chain rule: \( \frac{dV}{dt} = \pi \frac{dh}{dt} \).
Thus: \( \pi \frac{dh}{dt} = -\pi R^2 \sqrt{70560h} \).
Divide through by \( \pi \):
\( \frac{dh}{dt} = -R^2 \sqrt{70560h} \).
Result: \( \frac{dh}{dt} = -R^2 \sqrt{70560h} \) A1 M1 A1
[3 marks]
(d)
Solve the differential equation \( \frac{dh}{dt} = -R^2 \sqrt{70560h} \).
Separate variables:
\( \int \frac{1}{\sqrt{70560h}} \, dh = \int -R^2 \, dt \)
Left: \( \frac{2\sqrt{h}}{\sqrt{70560}} = -R^2 t + c \).
Simplify: \( \sqrt{h} = \frac{\sqrt{70560}}{2} (c – R^2 t) \).
Square both sides: \( h = 17640 (c – R^2 t)^2 \).
General solution: \( h = 17640 (c – R^2 t)^2 \) M1A1 A1A1 A1
[5 marks]
Solve the differential equation \( \frac{dh}{dt} = -R^2 \sqrt{70560h} \).
Separate variables:
\( \int \frac{1}{\sqrt{70560h}} \, dh = \int -R^2 \, dt \)
Left: \( \frac{2\sqrt{h}}{\sqrt{70560}} = -R^2 t + c \).
Simplify: \( \sqrt{h} = \frac{\sqrt{70560}}{2} (c – R^2 t) \).
Square both sides: \( h = 17640 (c – R^2 t)^2 \).
General solution: \( h = 17640 (c – R^2 t)^2 \) M1A1 A1A1 A1
[5 marks]
(e)
At \( t = 0 \), \( h = 3.2 \):
\( 3.2 = 17640 c^2 \)
\( c = 0.0134687… \)
When \( h = 0 \):
\( 0 = c – R^2 T \Rightarrow T = \frac{c}{R^2} \).
With \( R = 0.023 \):
\( T = \frac{0.0134687}{0.023^2} \approx 25.5 \) minutes (25.4606…).
\( T \approx 25.5 \) M1 A1 A1
[4 marks]
At \( t = 0 \), \( h = 3.2 \):
\( 3.2 = 17640 c^2 \)
\( c = 0.0134687… \)
When \( h = 0 \):
\( 0 = c – R^2 T \Rightarrow T = \frac{c}{R^2} \).
With \( R = 0.023 \):
\( T = \frac{0.0134687}{0.023^2} \approx 25.5 \) minutes (25.4606…).
\( T \approx 25.5 \) M1 A1 A1
[4 marks]
(f)
For \( t = 15 \):
\( c = R^2 t = 0.023^2 \times 15 = 0.007935 \).
Initial height: \( h = 17640 (0.007935)^2 \approx 1.11 \) metres.
New height = 1.11 m A1 M1 A1
[3 marks]
For \( t = 15 \):
\( c = R^2 t = 0.023^2 \times 15 = 0.007935 \).
Initial height: \( h = 17640 (0.007935)^2 \approx 1.11 \) metres.
New height = 1.11 m A1 M1 A1
[3 marks]
(g)(i)
Let \( h \) be the height of water in the highest container:
\( \frac{dh}{dt} = -35280 R^2 (0.0134687 – R^2 t) \).
For the lowest container:
\( \frac{dH}{dt} = 35280 R^2 (0.0135 – R^2 t) – R^2 \sqrt{70560 H} \).
Simplifies to:
\( \frac{dH}{dt} \approx 0.2514 – 0.009873 t – 0.1405 \sqrt{H} \).
Result: \( \frac{dH}{dt} \approx 0.2514 – 0.009873 t – 0.1405 \sqrt{H} \) M1A1 M1A1
[4 marks]
Let \( h \) be the height of water in the highest container:
\( \frac{dh}{dt} = -35280 R^2 (0.0134687 – R^2 t) \).
For the lowest container:
\( \frac{dH}{dt} = 35280 R^2 (0.0135 – R^2 t) – R^2 \sqrt{70560 H} \).
Simplifies to:
\( \frac{dH}{dt} \approx 0.2514 – 0.009873 t – 0.1405 \sqrt{H} \).
Result: \( \frac{dH}{dt} \approx 0.2514 – 0.009873 t – 0.1405 \sqrt{H} \) M1A1 M1A1
[4 marks]
(g)(ii)
Using Euler’s method with step size 0.5 minutes:
Example: \( y_1 = 1.05545… \).
Maximum \( H = 1.45 \) m (at 8.5 min) (1.44678…).
Maximum \( H = 1.45 \) A1 A2
[3 marks]
Using Euler’s method with step size 0.5 minutes:
Example: \( y_1 = 1.05545… \).
Maximum \( H = 1.45 \) m (at 8.5 min) (1.44678…).
Maximum \( H = 1.45 \) A1 A2
[3 marks]