The volume of water in a tank, \(V\) litres, at time \(t\) minutes is given by:
\(V(t)=2 t^3-15 t^2+24 t+50\) for \(0 \leqslant t \leqslant 6\)
- Find an expression for the rate at which water is flowing at time \(t\). (1 mark)
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- Calculate the rate of flow at \(t=4\) minutes and interpret this value. (1 mark)
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- Deduce when the water level in the tank is increasing. (2 marks)
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a. \(\dfrac{dV}{d t}=6 t^2-30 t+24\)
b. \(\text{At} \ \ t=4:\ \ \dfrac{dV}{d t}=0 \ \text{litres/minute}\)
\(\text{Interpretation: At \(\ t=4 \ \), water has stopped flowing either into or}\)
\(\text{out of the tank.}\)
c. \(\text{If water level is increasing} \ \Rightarrow \ \dfrac{dV}{d t}>0:\)
\(\text {Find \(t\) when}\ \dfrac{dV}{d t}>0:\)
| \(6 t^2-30t+24\) | \(\gt 0\) | |
| \(6\left(t^2-5 t+4\right)\) | \(\gt 0\) | |
| \((t-4)(t-1)\) | \(\gt 0\) |
\(\therefore \ \text{Water level increases for}\ \ t \in [0,1) \cup (4, 6]\)
a. \(V=2 t^3-15 t^2+24 t+50\)
\(\dfrac{dV}{d t}=6 t^2-30 t+24\)
b. \(\text{At} \ \ t=4:\)
\(\dfrac{dV}{d t}=6 \times 4^2-30 \times 4+24=0 \ \text{litres/minute}\)
\(\text{Interpretation: At \(\ t=4 \ \), water has stopped flowing either into or}\)
\(\text{out of the tank.}\)
c. \(\text{If water level is increasing} \ \Rightarrow \ \dfrac{dV}{d t}>0:\)
\(\text {Find \(t\) when}\ \dfrac{dV}{d t}>0:\)
| \(6 t^2-30t+24\) | \(\gt 0\) | |
| \(6\left(t^2-5 t+4\right)\) | \(\gt 0\) | |
| \((t-4)(t-1)\) | \(\gt 0\) |
\(\therefore \ \text{Water level increases for}\ \ t \in [0,1) \cup (4, 6]\)