IB Mathematics AHL 5.11 Definite and indefinite integration-AI HL Paper 2- Exam Style Questions- New Syllabus
The inner structure of a vase is modeled by rotating the area enclosed by the curve \( y = \frac{1}{2}x^2 – 1 \), and the lines \( x = 0 \), \( y = 0 \), and \( y = 15 \), through \( 2\pi \) radians around the \( y \)-axis. Measurements for \( x \) and \( y \) are in centimeters.
The vase is filled with water up to a height of \( h \) cm.
(a) Derive an explicit formula for the water volume in terms of \( h \). [5]
The vase is filled at a rate of \( 20 \, \text{cm}^3 \, \text{s}^{-1} \).
(b) Calculate the time required to fully fill the vase. [2]
(c) Determine the rate of change of the water height when \( h = 10 \, \text{cm} \). [7]
▶️ Answer/Explanation
The vase’s interior is created by rotating \( y = \frac{1}{2} x^2 – 1 \), \( x = 0 \), \( y = 0 \), and \( y = 15 \) around the \( y \)-axis.
Since \( y = \frac{1}{2} x^2 – 1 \iff x^2 = 2(y + 1) \), the radius at height \( y \) is \( r(y) = x = \sqrt{2(y + 1)} \) for \( 0 \le y \le 15 \).
(a) Volume up to height \( h \)
Using washers around the \( y \)-axis, a thin slice at height \( y \) has area \( \pi r(y)^2 = \pi \times [2(y + 1)] \).
Thus,
\( V(h) = \int_0^h \pi r(y)^2 \, dy = \int_0^h \pi \times 2(y + 1) \, dy = 2\pi \big[ \frac{1}{2} y^2 + y \big]_0^h = \pi h^2 + 2\pi h, \quad 0 \le h \le 15. \)
So the explicit volume-height relation is
\( \boxed{ V(h) = \pi h^2 + 2\pi h \, \text{cm}^3 }. \)
(b) Time to completely fill
Total volume (to \( y = 15 \)):
\( V(15) = \pi (15)^2 + 2\pi (15) = 225\pi + 30\pi = 255\pi \, \text{cm}^3. \)
With a constant inflow rate \( \frac{dV}{dt} = 20 \, \text{cm}^3 \, \text{s}^{-1} \),
the fill time is
\( t = \frac{V(15)}{20} = \frac{255\pi}{20} = \frac{51\pi}{4} \, \text{s} \approx 40.1 \, \text{s}. \)
(c) Rate of rise of the water level at \( h = 10 \) cm
Differentiate \( V(h) \):
\( \frac{dV}{dh} = 2\pi h + 2\pi = 2\pi (h + 1). \)
By the chain rule,
\( \frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt} \quad \Longrightarrow \quad \frac{dh}{dt} = \frac{\frac{dV}{dt}}{\frac{dV}{dh}} = \frac{20}{2\pi (h + 1)} = \frac{10}{\pi (h + 1)}. \)
At \( h = 10 \),
\( \boxed{ \frac{dh}{dt} = \frac{10}{11\pi} \, \text{cm s}^{-1} } \approx 0.289 \, \text{cm s}^{-1}. \)