Question
Kyungyoon investigates the rate at which a cubical block of sugar dissolves in hot coffee. Initially, the cube has side lengths of 10 mm. This information is illustrated in the following diagram:
(a)
(i) The length of one side of a block of sugar, 20 seconds after it is placed in hot coffee.
(ii) The volume of a block of sugar, 20 seconds after it is placed in hot coffee.
Let the function \( V(t) \) represent the volume of the block of sugar, in \( mm^3 \), at time \( t \) seconds after it is placed in hot coffee.
The function is given by:
\[ V(t) = 1000 – 60t + 1.2t^2 – 0.008t^3, \quad 0 \leq t \leq 50 \]
(b) Find \( V'(t) \).
(c) Find the rate of change of the volume of the block of sugar at \( t = 20 \).
(d) State one reason why the side length of the cube may not always decrease at a constant rate.
▶️Answer/Explanation
Detailed Solution
(a) (i) Finding the side length at \( t = 20 \)
The side length decreases by 0.2 mm every second.
\[ \text{New side length} = 10 – 0.2 \times 20 \]
\[ = 6 \text{ mm} \]
(a) (ii) Finding the volume at \( t = 20 \)
Since the new side length is 6 mm:
\[ V = 6^3 = 216 \text{ mm}^3 \]
(b) Finding \( V'(t) \) (the derivative of \( V(t) \))
Using the power rule to differentiate:
\[ V(t) = 1000 – 60t + 1.2t^2 – 0.008t^3 \]
\[ V'(t) = -60 + 2.4t – 0.024t^2 \]
(c) Finding \( V'(20) \)
Substituting \( t = 20 \) into \( V'(t) \):
\[ V'(20) = -60 + 2.4(20) – 0.024(20)^2 \]
\[ = -60 + 48 – 9.6 \]
\[ = -21.6 \text{ mm}^3 \text{ s}^{-1} \]
Thus, the rate of change of the volume at \( t = 20 \) is \(-21.6\) mm³/s.
(d) Reasons why the side length may not decrease at a constant rate:
- Change in coffee temperature over time.
- Decrease in the surface area of the cube as it dissolves.
- The cube may break apart into smaller pieces.
- The dissolution rate depends on the volume of the cube.
……………………………Markscheme……………………………….
(a) (i)
\( 6 \text{ mm} \)
(ii)
\( 216 \text{ mm}^3 \)
(b)
\( V'(t) = -60 + 2.4t – 0.024t^2 \)
(c)
\( V'(20) = -21.6 \text{ mm}^3 \text{ s}^{-1} \)
(d)
Possible reasons:
- Change in coffee temperature.
- Change in the cube’s surface area.
- Cube breaking into smaller pieces.
- Dissolution rate depends on volume.