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 0.6 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 Modelling with Functions: Figure 2 shows a cylindrical wat

Figure 2 shows a cylindrical water tank - Edexcel - A-Level Maths Pure - Question 2 - 2010 - Paper 6

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 - 2010 - Paper 6

Step 1

Show that t minutes after the tap has been opened

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Answer

To derive the equation, we begin with the volume of water in the tank, given by:

$V = \pi r^2 h$

Since the diameter of the tank is 6 m, the radius r is 3 m. Thus,

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

Taking the rate of change of volume:

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

At the same time, the rate of water flowing into the tank is 0.48 m³ min⁻¹, and the rate of water leaving the tank is 0.6 m³ min⁻¹. Therefore, we equate:

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

This simplifies to:

$\frac{dV}{dt} = -0.12$

Setting the two expressions equal gives:

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

Now substituting the values:

$9\pi \frac{dh}{dt} = -0.12$

This leads to:

$\frac{dh}{dt} = \frac{-0.12}{9\pi}$

Next, equating:

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

After simplification, we can confirm:

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

Step 2

Find the value of t when h = 0.5

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Answer

When we have established the relationship, we can separate variables for the equation derived:

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

This leads us to:

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

To find C, we use the initial condition when t = 0 and h = 0.2:

$-15 \ln(4 - 5(0.2)) = 0 + C$

Calculating:

$-15 \ln(4 - 1) = C \Rightarrow C = -15 \ln(3)$

Thus, the equation becomes:

$-15 \ln(4 - 5h) = t - 15 \ln(3)$

Now, substituting h = 0.5:

$-15 \ln(4 - 5(0.5)) = t - 15 \ln(3)$

This simplifies to:

$t = -15 \ln(4 - 2) + 15 \ln(3)$

Calculating:

$t = -15 \ln(2) + 15 \ln(3)$

Which gives:

$t = 15 \ln \left( \frac{3}{2} \right)$

Hence, the value of t when h = 0.5 can be evaluated numerically, leading to:

$t \approx 15 \cdot 0.405 = 6.075.$

Thus, t takes the value approximately 6.08 minutes.

Figure 2 shows a cylindrical water tank - Edexcel - A-Level Maths Pure - Question 2 - 2010 - Paper 6

Worked Solution & Example Answer:Figure 2 shows a cylindrical water tank - Edexcel - A-Level Maths Pure - Question 2 - 2010 - Paper 6

Show that t minutes after the tap has been opened

Find the value of t when h = 0.5

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