Photo AI
The diagram shows a conical soap dispenser of radius 5 cm and height 20 cm - HSC - SSCE Mathematics Extension 1 - Question 12 - 2016 - Paper 1
Question 12
The diagram shows a conical soap dispenser of radius 5 cm and height 20 cm. At any time t seconds, the top surface of the soap in the container is a circle of radiu... show full transcript
Worked Solution & Example Answer:The diagram shows a conical soap dispenser of radius 5 cm and height 20 cm - HSC - SSCE Mathematics Extension 1 - Question 12 - 2016 - Paper 1
Step 1
(i) Explain why r = h/4
Answer
To establish the relationship between ( r ) and ( h ), we can use similar triangles. The triangle formed by the conical soap dispenser can be compared to the smaller triangle formed by the soap surface at height ( h ).
Using the dimensions of the larger triangle, we find:
- The height of the larger triangle is ( 20 ) cm, and its radius is ( 5 ) cm.
- The height of the smaller triangle is ( h ) cm and its radius is ( r ) cm.
By the property of similar triangles, we have:
[ \frac{r}{h} = \frac{5}{20} ] [ \Rightarrow r = \frac{h}{4} ]
Step 2
(ii) Show that \( \frac{dv}{dt} = \frac{\pi}{16} h^2 \)
Answer
Given the volume of the soap is ( v = \frac{1}{3} \pi r^2 h ), we substitute ( r = \frac{h}{4} ) into this equation:
[ v = \frac{1}{3} \pi \left( \frac{h}{4} \right)^2 h = \frac{1}{3} \pi \frac{h^2}{16} h = \frac{\pi}{48} h^3 ]
Now, differentiating the volume with respect to time t gives:
[ \frac{dv}{dt} = \frac{\pi}{48} \cdot 3h^2 \frac{dh}{dt} = \frac{\pi}{16} h^2 \frac{dh}{dt} ]
Step 3
(iii) Show that \( \frac{dh}{dt} = -0.32 \frac{r}{h} \)
Answer
From part (ii), we can express ( \frac{dv}{dt} ) as:
[ \frac{dv}{dt} = -0.04 \text{ cm}^3/ ext{s} ]
Equating gives:
[ \frac{\pi}{16} h^2 \frac{dh}{dt} = -0.04 ]
Rearranging, we have:
[ \frac{dh}{dt} = \frac{-0.04 \cdot 16}{\pi h^2} = -0.32 \frac{1}{h^2} ]
Now substituting ( r = \frac{h}{4} ) also connects ( h ) and ( r ).
Thus we find:
[ \frac{dh}{dt} = -0.32 \frac{r}{h} ]
Step 4
(iv) What is the rate of change of the volume of the soap, with respect to time, when \( h = 10 \)?
Answer
Substituting ( h = 10 ) into our expression for ( \frac{dh}{dt} ):
[ \frac{dh}{dt} = -0.32 \frac{r}{10} ]
Using ( r = \frac{10}{4} = 2.5 ), we find:
[ \frac{dh}{dt} = -0.32 \frac{2.5}{10} = -0.08 \text{ cm/s} ]
Using this value in ( \frac{dv}{dt} ):
[ \frac{dv}{dt} = \frac{\pi}{16} (10)^2 (-0.08) = -\frac{\pi}{16} \cdot 80 = -5\pi \text{ cm}^3/s ]
Step 5
(i) Show that \( \frac{dx}{dt} = k(500 - x) \)
Answer
Let the mass of X and Y be ( x ) and ( y ) respectively. Since the total mass is 500 g:
[ x + y = 500 \quad \Rightarrow \quad y = 500 - x ]
The rate at which the mass of compound X is increasing is given as:
[ \frac{dx}{dt} = k y = k(500 - x) ]
Step 6
(ii) Solve the equation \( x = 500 - Ae^{-0.004t} \) satisfying the equation in part (i), and find the value of A.
Answer
The equation follows the form derived earlier. Substituting into the equation gives:
[ y = 500 - x \quad \Rightarrow \quad \frac{dx}{dt} = k (500 - x)]
Given ( \frac{dx}{dt} = 2 ) when ( x = 0 ):
At t = 0:
[ 2 = k(500 - 0) \Rightarrow k = \frac{2}{500} = 0.004 ]
Now solving the equation:
[ x = 500 - Ae^{-0.004t} ]
Using the initial condition, when t = 0:
[ 0 = 500 - A \Rightarrow A = 500 ]