Differentiation - Connected Rates of Change
Differentiation - Connected Rates of Change
TGM Original Questions
A cylindrical water dispenser has a base radius \(10\) cm and is initially full of water. The water is dispensed into an inverted conical cup that has a base radius \(4\) cm and height \(8\) cm and is initially empty. After \(t\) seconds, the depth of water in the conical cup is \(x\) cm.
- Show that the volume of water in the cup is \(\frac{\pi x^3}{12}\) cm\(^3\).
- Given that the depth of water in the cylindrical dispenser decreases at a constant rate of \(0.0025\) cm/s, find the rate of change of the volume of water in the cup in terms of \(\pi\).
Hence find, when the volume of water in the conical cup is \(\frac{2\pi}{3}\) cm\(^3\), the rate of increase of
- the depth of the water in the conical cup, and
- the area of the water surface in the conical cup.
Leave your answers in terms of \(\pi\) where possible.
Solution:
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Answer:(i) \(V=\dfrac{\pi x^3}{12}\text{ cm}^3\) (shown) (ii) \(\dfrac\pi4\text{ cm}^3\text{ s}^{-1}\) (iii) \(\dfrac14\text{ cm s}^{-1}\) (iv) \(\dfrac\pi4\text{ cm}^2\text{ s}^{-1}\).
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