Figure 2 shows a cylindrical water tank. The diameter of a circular cross-section of the tank is 6 m. Water is flowing into the tank at a constant rate of 0.48 m³ min⁻¹. At time t minutes, the depth of the water in the tank is h metres. There is a tap at a point T at the bottom of the tank. When the tap is open, water leaves the tank at a rate of 0.6h m³ min⁻¹. (a) Show that t minutes after the tap has been opened $$75 \frac{dh}{dt} = (4 - 5h)$$ When t = 0, h = 0.2 (b) Find the value of t when h = 0.5

A-Level Edexcel Maths Pure Integration: Figure 2 shows a cylindrical wat

Figure 2 shows a cylindrical water tank - Edexcel - A-Level Maths Pure - Question 2 - 2009 - Paper 3

Question 2

Figure 2 shows a cylindrical water tank. The diameter of a circular cross-section of the tank is 6 m. Water is flowing into the tank at a constant rate of 0.48 m³ mi... show full transcript

Worked Solution & Example Answer:Figure 2 shows a cylindrical water tank - Edexcel - A-Level Maths Pure - Question 2 - 2009 - Paper 3

Step 1

Show that t minutes after the tap has been opened

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Answer

To solve this problem, we start by understanding the volume change in the tank. The volume of water in the tank is given by:

$V = \pi r^2 h$

Given the diameter is 6 m, the radius is:

$r = \frac{6}{2} = 3 \text{ m}$

Thus, the volume becomes:

$V = \pi (3)^2 h = 9\pi h$

The rate of change of volume with respect to time can be expressed as:

$\frac{dV}{dt} = 0.48 - 0.6h$

From the formula for volume, we differentiate:

$\frac{dV}{dt} = 9\pi \frac{dh}{dt}$

Equating the two expressions for $\frac{dV}{dt}$ gives:

$9\pi \frac{dh}{dt} = 0.48 - 0.6h$

Rearranging leads to:

$\frac{dh}{dt} = \frac{0.48 - 0.6h}{9\pi}$

Multiplying both sides by 75:

$75 \frac{dh}{dt} = \frac{75(0.48 - 0.6h)}{9\pi}$

Final simplification yields:

$75 \frac{dh}{dt} = (4 - 5h)$

Step 2

Find the value of t when h = 0.5

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Answer

Given the differential equation:

$\frac{75}{(4 - 5h)} dh = dt$

We now separate the variables and integrate:

$\int \frac{75}{(4 - 5h)} dh = \int dt$

This simplifies to:

$-15 \ln |4 - 5h| = t + C$

Using the initial condition when $t = 0$ and $h = 0.2$ , we substitute to find C:

\Rightarrow C = -15 \ln(3)$$ Thus, the equation becomes: $$t = -15 \ln(4 - 5h) + 15 \ln(3)$$ When $$h = 0.5$$: $$t = -15 \ln(4 - 5(0.5)) + 15 \ln(3)\ = -15 \ln(1.5) + 15 \ln(3)\ = 15 (\ln(3) - \ln(1.5))\ = 15 \ln(\frac{3}{1.5})\ = 15 \ln(2)$$ Thus, the final value of t is: $$t = 15 \ln(2) \approx 10.4$$

Figure 2 shows a cylindrical water tank - Edexcel - A-Level Maths Pure - Question 2 - 2009 - Paper 3

Worked Solution & Example Answer:Figure 2 shows a cylindrical water tank - Edexcel - A-Level Maths Pure - Question 2 - 2009 - Paper 3

Show that t minutes after the tap has been opened

Find the value of t when h = 0.5

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