Figure 3 shows a vertical cylindrical tank of height 200 cm containing water. Water is leaking from a hole P on the side of the tank. At time t minutes after the leaking starts, the height of water in the tank is h cm. The height h cm of the water in the tank satisfies the differential equation $$\frac{dh}{dr} = k(h - 9)^{-\frac{3}{2}}; \quad 9 < h < 200$$ where k is a constant. Given that, when h = 130, the height of the water is falling at a rate of 1.1 cm per minute, a) find the value of k. Given that the tank was full of water when the leaking started, b) solve the differential equation with your value of k, to find the value of t when h = 50.

A-Level Edexcel Maths Pure Differentiation: Figure 3 shows a vertical cylind

Figure 3 shows a vertical cylindrical tank of height 200 cm containing water - Edexcel - A-Level Maths Pure - Question 1 - 2017 - Paper 5

Question 1

Figure 3 shows a vertical cylindrical tank of height 200 cm containing water. Water is leaking from a hole P on the side of the tank. At time t minutes after the le... show full transcript

Worked Solution & Example Answer:Figure 3 shows a vertical cylindrical tank of height 200 cm containing water - Edexcel - A-Level Maths Pure - Question 1 - 2017 - Paper 5

Step 1

find the value of k.

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Answer

To find the value of k, we start by substituting the given height h = 130 into the differential equation:

$\frac{dh}{dt} = k(h - 9)^{-\frac{3}{2}}$ .

This implies: $-1.1 = k(130 - 9)^{-\frac{3}{2}}$ .

Now, simplifying the expression: $-1.1 = k(121)^{-\frac{3}{2}}$

Calculating $(121)^{-\frac{3}{2}}$ gives: $(121)^{-\frac{3}{2}} = \frac{1}{11^3} = \frac{1}{1331}$ ,

Thus, we have: $-1.1 = k \times \frac{1}{1331},$

Solving for k: $k = -1.1 \times 1331 = -0.1.$

So, the value of k is:

$k = -0.1.$

Step 2

solve the differential equation with your value of k, to find the value of t when h = 50.

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Answer

To solve the differential equation:

Starting with: $\frac{dh}{dt} = -0.1(h - 9)^{-\frac{3}{2}},$

we separate the variables:

$\int (h - 9)^{\frac{3}{2}} dh = -0.1 \int dt$ .

Integrating both sides:

The left-hand side yields: $\frac{2}{5}(h - 9)^{\frac{5}{2}} + C = -0.1t + C.$

Setting limits: When h = 50: $\frac{2}{5}(50 - 9)^{\frac{5}{2}} - \frac{2}{5}(200 - 9)^{\frac{5}{2}} = -0.1t,$

Calculating: $\frac{2}{5}(41)^{\frac{5}{2}} = -0.1t.$

Through simplification: At h = 200, we find: $t = \frac{2}{5} \cdot \frac{(\sqrt{41})^{5}}{-0.1}$ Evaluating gives t as approximately 148 minutes. Therefore, the solution yields: $t \approx 148 minutes.$

Figure 3 shows a vertical cylindrical tank of height 200 cm containing water - Edexcel - A-Level Maths Pure - Question 1 - 2017 - Paper 5

Worked Solution & Example Answer:Figure 3 shows a vertical cylindrical tank of height 200 cm containing water - Edexcel - A-Level Maths Pure - Question 1 - 2017 - Paper 5

find the value of k.

solve the differential equation with your value of k, to find the value of t when h = 50.

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