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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
Explain why r = \( \frac{h}{4} \)
Answer
In a conical shape, the radius and height are proportional. Given that the height of the entire cone is 20 cm and the radius is 5 cm, we can express this relationship using similar triangles. The proportion holds that:
[ \frac{r}{h} = \frac{5}{20} \to r = \frac{h}{4} ]
Step 2
Show that \( \frac{dv}{dt} = \frac{\pi}{16} h^2 \)
Answer
Starting with the volume equation ( v = \frac{1}{3} \pi r^2 h ), substituting ( r = \frac{h}{4} ) gives us: [ v = \frac{1}{3} \pi \left( \frac{h}{4} \right)^2 h = \frac{\pi}{48} h^3 ]
Now differentiating with respect to time t: [ \frac{dv}{dt} = \frac{\pi}{48} \cdot 3h^2 \cdot \frac{dh}{dt} = \frac{\pi}{16} h^2 \cdot \frac{dh}{dt} ]
Step 3
Show that \( \frac{dh}{dt} = -\frac{0.32}{rh} \)
Answer
From the area of the top surface, we have: [ A = \pi r^2 \Rightarrow \frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt} ]
Given that area is decreasing at a rate of 0.04 cm²s⁻¹, [ \frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt} ightarrow -0.04 = 2\pi r \cdot \frac{dr}{dt} ]
Substituting ( r = \frac{h}{4} ) yields: [ -0.04 = 2\pi \left( \frac{h}{4} \right) \cdot \frac{dh}{dt}]
Solving for ( \frac{dh}{dt} ) gives: [ \frac{dh}{dt} = -\frac{0.32}{rh} ]
Step 4
What is the rate of change of the volume of the soap, with respect to time, when h = 10?
Answer
Using the equation found in part ii: [ \frac{dv}{dt} = \frac{\pi}{16} h^2 \cdot \frac{dh}{dt} ]
Substituting ( h = 10 ): [ \frac{dv}{dt} = \frac{\pi}{16} \cdot 10^2 \cdot \frac{dh}{dt} ]
Using the rate found in part iii, when h = 10, we can calculate: [ \frac{dh}{dt} = -\frac{0.32}{(\frac{10}{4})(10)} = -0.08 ]
Thus: [ \frac{dv}{dt} = \frac{\pi}{16} \cdot 100 \cdot (-0.08) = -0.2 \text{ cm}^3/s ]
Step 5
Explain why \( \frac{dx}{dt} = 0.004(500 - x) \)
Answer
In the reaction, the mass of X increases due to the mass of Y decreasing. Since the total mass is constant at 500 g, we have: [ y = 500 - x ] and so: [ \frac{dy}{dt} = -\frac{dx}{dt} ]
Since it is stated that the rate at which x is formed is proportional to the mass of Y: [ \frac{dx}{dt} = k(500 - x) \text{ with } k = 0.004 ]
Thus, we have: [ \frac{dx}{dt} = 0.004(500 - x) ]
Step 6
Show that \( x = 500 - A e^{-0.004t} \) satisfies the equation in part (i), and find the value of A.
Answer
Starting from ( x = 500 - A e^{-0.004t} ), we differentiate: [ \frac{dx}{dt} = -(-0.004)A e^{-0.004t} = 0.004A e^{-0.004t} ]
Equating with the expression from part (i): [ 0.004(500 - x) = 0.004A e^{-0.004t} ]
Setting initial condition at t = 0, when ( x(0) = 0 ): [ 0 = 500 - A \rightarrow A = 500 ]
Thus: [ x = 500 - 500 e^{-0.004t} ]