QCE Vault / Specialist Maths Differential equations — Question 76 Original QCE Vault practice · 5 marks
Browse all questions Original practice exam Report an issue Q76 · Practice question Complex familiar 5 marks
QUESTION 76 (5 marks) Water in a hemispherical bowl of radius R = 3 m R=3\ \mathrm m R = 3 m has depth h m h\ \mathrm m h m . The volume of water is V = π h 2 ( R − h 3 ) . V=\pi h^2\left(R-\frac h3\right). V = π h 2 ( R − 3 h ) . Water is entering at 0.5 m 3 m i n − 1 0.5\ \mathrm{m^3\,min^{-1}} 0.5 m 3 mi n − 1 . Determine d h / d t dh/dt d h / d t when h = 1 m h=1\ \mathrm m h = 1 m . WORKED SOLUTION
Practice marking scheme 5 marks ANSWER d h / d t = 1 / ( 10 π ) m m i n − 1 dh/dt=1/(10\pi)\ \mathrm{m\,min^{-1}} d h / d t = 1/ ( 10 π ) m mi n − 1 . Worked solution
With R = 3 R=3 R = 3 , V = π ( 3 h 2 − h 3 3 ) , d V d h = π ( 6 h − h 2 ) . V=\pi\left(3h^2-\frac{h^3}{3}\right),\qquad\frac{dV}{dh}=\pi(6h-h^2). V = π ( 3 h 2 − 3 h 3 ) , d h d V = π ( 6 h − h 2 ) . The chain rule gives d V d t = π ( 6 h − h 2 ) d h d t . \frac{dV}{dt}=\pi(6h-h^2)\frac{dh}{dt}. d t d V = π ( 6 h − h 2 ) d t d h . At h = 1 h=1 h = 1 , 0.5 = 5 π d h d t ⇒ d h d t = 1 10 π m m i n − 1 . 0.5=5\pi\frac{dh}{dt}\quad\Rightarrow\quad\frac{dh}{dt}=\frac1{10\pi}\ \mathrm{m\,min^{-1}}. 0.5 = 5 π d t d h ⇒ d t d h = 10 π 1 m mi n − 1 . Differentiates volume with respect to depth.
[2 marks] Substitutes the rate and depth.
[1 mark] Determines the rate of change of depth.
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