Edexcel IAL June 2021 P4 (WMA14/01) Q3 Differentiation, Rates of Change
A bowl with circular cross section and height 20 cm is shown in Figure 2.
The bowl is initially empty and water starts flowing into the bowl.
When the depth of water is h cm, the volume of water in the bowl, \( V \) cm³, is modelled by the equation
\( V = \frac{1}{3}h^{2}(h+4) \qquad 0 \leq h \leq 20 \)
Given that the water flows into the bowl at a constant rate of 160 cm³/s, find, according to the model.
(a) the time taken to fill the bowl,
(2)
(2)
(b) the rate of change of the depth of the water, in \( \text{cm/s} \), when \( h = 5 \)
(5)
(5)
Solution to Question 6(a): Time to Fill the Bowl (\( V = \frac{1}{3}h^{2}(h+4) \))
Key Concepts Used:
- Volume calculation: Substitute \( h = 20 \) cm.
- Rate of change: \( \text{Time} = \frac{\text{Volume}}{\text{Rate}} \)
Step-by-Step Working:
The given formula represents the volume of water in the bowl when the water depth is h cm.
\( V = \frac{1}{3}h^2(h+4) \)
-
Calculate total volume when full.
The bowl is full when \( h = 20 \) cm. Substitute this into the formula:
\( V = \frac{1}{3} \times (20)^2 \times (20 + 4) \)
\( V = \frac{1}{3} \times 400 \times 24 \)
\( V = \frac{1}{3} \times 9600 \)
\( V = 3200 \text{cm}^3 \) -
Determine Time Using Flow Rate
Water flows in at a constant rate of 160 cm³/s.
\( \text{Time} = \frac{\text{Total Volume}}{\text{Flow Rate}} \)
\( \text{Time} = \frac{3200 \text{cm}^3}{160 \text{cm}^3/\text{s}} \)
\( \text{Time} = 20 \text{seconds} \)
Final Answer:
20 seconds
Solution to Part b: Rate of Change of Depth at \( h = 5 \) cm
Key Concepts Used:
- Chain rule: \( \frac{dh}{dt} = \frac{dh}{dV} \cdot \frac{dV}{dt} \)
- Derivative of volume: \( \frac{dV}{dh} = h^2 + \frac{8}{3}h \)
Step-by-Step Working:
Find how fast the water level is rising \( \frac{dh}{dt} \) when the depth is 5 cm.
-
Expand the Volume Formula
First, express V in a form that’s easier to differentiate:
\( V = \frac{1}{3} h^2 (h + 4) \)
\( V = \frac{1}{3} h^3 + \frac{4}{3} h^2 \) -
Differentiate Volume with Respect to Depth
Find \( \frac{dV}{dh} \):
\( \frac{dV}{dh} = h^2 + \frac{8}{3} h \) -
Apply the Chain Rule
We know the chain rule:
\( \frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt} \)
Given \( \frac{dV}{dt} = 160 \text{cm}^3/\text{s} \):
\( 160 = \left( h^2 + \frac{8}{3} h \right) \frac{dh}{dt} \) -
Solve for \( \frac{dh}{dt} \)
\( \frac{dh}{dt} = \frac{160}{h^2 + \frac{8}{3} h} \)
To simplify:
\( \frac{dh}{dt} = \frac{160 \times 3}{3h^2 + 8h} \)
\( \frac{dh}{dt} = \frac{480}{3h^2 + 8h} \) -
Substitute \( h = 5 \text{cm} \)
\( \frac{dh}{dt} = \frac{480}{3(5)^2 + 8(5)} \)
\( \frac{dh}{dt} = \frac{480}{75 + 40} \)
\( \frac{dh}{dt} \approx 4.174 \text{cm/s} \)
Final Answer:
\( \frac{dh}{dt} = \frac{96}{23} \text{cm/s} \) or \( 4.2 \text{cm/s} \)