Edexcel IAL October 2022 P4 (WMA14/01) Q10 Related Rates of Change, Differential Equations
A spherical ball of ice of radius 12 cm is placed in a bucket of water.
In a model of the situation,
- the ball remains spherical as it melts
- t minutes after the ball of ice is placed in the bucket, its radius is r cm
- the rate of decrease of the radius of the ball of ice is inversely proportional to the square of the radius
- the radius of the ball of ice is 6 cm after 15 minutes
Using the model and the information given,
(a) find an equation linking r and t.
(5)
(b) find the time taken for the ball of ice to melt completely.
(2)
(c) On Diagram 1 on page 27, sketch a graph of r against t.
(1)
Solution to Part a: Relating r and t
Key Concepts Used:
- Differential equation
- Apply Initial Conditions
Step-by-Step Working:
-
Set Up the Differential Equation
It is given that the rate of decrease of the radius of the ball of ice is inversely proportional to the square of the radius. This can be written mathematically as:
\[ \frac{dr}{dt} \propto -\frac{1}{r^2} \]
\[ \frac{dr}{dt} = -\frac{k}{r^2} \]
where \( k > 0 \) is the constant of proportionality (negative sign indicates decrease).
-
Separate Variables and Integrate
\[ \int r^2 \, dr = -k \int dt \]
Let’s integrate it.
\[ \frac{r^3}{3} = -kt + C \]
-
Apply Initial Conditions
At \( t = 0, \, r = 12 \):
\[ \frac{r^3}{3} = -kt + C \]
\[ \frac{12^3}{3} = -k(0) + C \]
\[ \frac{12^3}{3} = C \Rightarrow C = 576 \]
At \( t = 15, \, r = 6 \):
\[ \frac{6^3}{3} = -15k + 576 \]
\[ 72 = -15k + 576 \]
\[ k = \frac{504}{15} = 33.6 \]
-
Final Equation:
\[ \frac{r^3}{3} = -33.6t + 576 \]
\[ r^3 = -100.8t + 1728 \]
Final Answer
\[ r^3 = 1728 – 100.8t \]
Solution to Part b: Time for Ice Ball to Melt Completely
Key Concepts Used:
- Apply Conditions
Step-by-Step Working:
-
Let’s find \( t \) when \( r = 0 \).
We are substituting \( r \) as 0 because when the ice melts, its radius would become zero as there no shape exists now.
\[ 0 = 1728 – 100.8t \]
\[ t = \frac{1728}{100.8} = \frac{17280}{1008} = \frac{120}{7} \approx 17.14 \, \text{minutes} \]
Final Answer
\[ \frac{120}{7} \approx 17.14 \, \text{minutes} \]
(Exact form preferred; ≈17.14 minutes acceptable)
Hence, it takes 17.1 mins to melt completely.
Solution to Part c: Sketching \( r \) vs \( t \)
Key Concepts Used:
- Sketching graphs
Step-by-Step Working:
-
Key Features of the Sketch:
1. Initial Point: (0,12)
2. Final Point: \(\left(\frac{120}{7}, 0\right)\)
3. Shape: Concave downwards (the function is decreasing or becoming more negative).
4. Smooth curve from \( r = 12 \) to \( r = 0 \).
-
Sketch Description:
– Axes:
• Horizontal (t): \( 0 \) to \( \frac{120}{7} \) minutes.
• Vertical (r): \( 0 \) to \( 12 \) cm.
– Curve: Starts at (0,12), passes through (15,6), ends at \( \left( \frac{120}{7}, 0 \right) \).
-
Why is the shape of curve concave downwards?
We know this when the radius is 12, time is 0. And, when the radius becomes zero, time is 17.1 minutes. This shows us that as the radius decreases dr/dt is more negative or the negative gradient becomes more negative. This could easily be understood from the equation below:
\[ \frac{dr}{dt} = -\frac{k}{r^2} \]
\[ \frac{dr}{dt} \propto -\frac{1}{r^2} \]
So, as \( r \) decreases, the rate of change of \( r \) increases or the negative gradient increases. Therefore, the shape of sketch will be concave downwards.
Final Answer
The graph starts at (0,12), passes through (15,6), and ends at (17.14,0), with a concave downward shape.