Question

3
the diagram shows a container in the form of an inverted cone, with base radius 12 cm and
height 6 cm. the container is initially empty. at time ( t = 0 ), water is allowed to flow into the
container. after ( t ) seconds, the height of water in the container is ( (6 - x) ) cm and the volume of
water is ( v mathrm{cm}^{3} ), where ( x mathrm{cm} ) is the difference in height between the cone and the water level.
(i) show that ( v=\frac{4 pi}{3}(6 - x)^{3} ).
given that the rate of flow of the water is ( k(6 - x)^{2} mathrm{cm}^{3} / mathrm{s} ), where ( k ) is a constant,
(ii) find an expression for the rate of change of the difference in height between the cone
and the water level in terms of ( pi ) and ( k ).
the water from the tap was allowed to flow for ( 32 pi ) seconds when the height of the water
reached 1 cm.
(iii) find the value of ( k ).

Explanation:

(i)

Step1: Use similar triangles

Let the radius of the water - surface be \(r\). By similar triangles, \(\frac{r}{12}=\frac{6 - x}{6}\), so \(r = 2(6 - x)\).

Step2: Use the volume formula for a cone

The volume of a cone \(V=\frac{1}{3}\pi r^{2}h\). Substitute \(r = 2(6 - x)\) and \(h=(6 - x)\) into the formula.

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(ii)

Step1: Differentiate \(V\) with respect to \(x\)

We know \(V=\frac{4\pi}{3}(6 - x)^{3}\). Using the chain - rule \(\frac{dV}{dx}=4\pi(6 - x)^{2}\).

Step2: Use the chain - rule \(\frac{dx}{dt}=\frac{dx}{dV}\times\frac{dV}{dt}\)

Given \(\frac{dV}{dt}=k(6 - x)^{2}\). Since \(\frac{dx}{dV}=\frac{1}{\frac{dV}{dx}}\), then \(\frac{dx}{dt}=\frac{k(6 - x)^{2}}{4\pi(6 - x)^{2}}\).

Step3: Simplify the expression

Cancel out \((6 - x)^{2}\) (assuming \(x
eq6\)), we get \(\frac{dx}{dt}=\frac{k}{4\pi}\)

(iii)

Step1: Integrate \(\frac{dx}{dt}\)

We know \(\frac{dx}{dt}=\frac{k}{4\pi}\), so \(x=\frac{k}{4\pi}t + C\). When \(t = 0\), \(x=6\) (since the container is initially empty, the height of water \(h = 0\), so \(6-0=6\)), then \(C = 6\). So \(x=\frac{k}{4\pi}t+6\).
When \(t = 32\pi\), \(h = 1\) cm, then \(x=6 - 1=5\)

Step2: Substitute values into the equation

Substitute \(t = 32\pi\) and \(x = 5\) into \(x=\frac{k}{4\pi}t+6\)

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Another way:

Step1: Use the volume formula

The volume of water when \(h = 1\) (i.e., \(x = 5\)) is \(V=\frac{4\pi}{3}(6 - 5)^{3}=\frac{4\pi}{3}\).

Step2: Use the formula \(V=\int_{0}^{t}\frac{dV}{dt}dt\)

Since \(\frac{dV}{dt}=k(6 - x)^{2}\) and \(x = 6 - h\), and \(V=\int_{0}^{32\pi}k(6-(6 - h))^{2}dt\) (but using \(V=\int_{0}^{t}\frac{dV}{dt}dt\) and \(V=\frac{4\pi}{3}\), \(t = 32\pi\), \(\frac{dV}{dt}=k(6 - x)^{2}\) and from \(V=\frac{4\pi}{3}(6 - x)^{3}\), when \(x = 5\), \(V=\frac{4\pi}{3}\)

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We know \(x=\frac{k}{4\pi}t+6\), when \(t = 32\pi\), \(x = 5\)

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Answer:

(i) Shown as above.
(ii) \(\frac{dx}{dt}=\frac{k}{4\pi}\)
(iii) \(k =-\frac{1}{8}\)