The height $x$ metres, of a column of water in a fountain display satisfies the differential equation $$\frac{dx}{dr} = \frac{8\sin 2t}{3\sqrt{x}}$$, where $t$ is the time in seconds after the display begins. 15 (a) Solve the differential equation, given that initially the column of water has zero height. Express your answer in the form $x = f(t)$. 15 (b) Find the maximum height of the column of water, giving your answer to the nearest cm.

A-Level AQA Maths Pure Differential Equations: The height $x$ metres, of a colu

The height $x$ metres, of a column of water in a fountain display satisfies the differential equation $$\frac{dx}{dr} = \frac{8\sin 2t}{3\sqrt{x}}$$, where $t$ is the time in seconds after the display begins - AQA - A-Level Maths Pure - Question 15 - 2017 - Paper 1

Question 15

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The height $x$ metres, of a column of water in a fountain display satisfies the differential equation $$\frac{dx}{dr} = \frac{8\sin 2t}{3\sqrt{x}}$$, where $t$ is th... show full transcript

Worked Solution & Example Answer:The height $x$ metres, of a column of water in a fountain display satisfies the differential equation $$\frac{dx}{dr} = \frac{8\sin 2t}{3\sqrt{x}}$$, where $t$ is the time in seconds after the display begins - AQA - A-Level Maths Pure - Question 15 - 2017 - Paper 1

Step 1

Solve the differential equation, given that initially the column of water has zero height.

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Answer

To solve the differential equation, we first separate the variables: $\sqrt{x} \, dx = \frac{8\sin 2t}{3} \, dr$ .

Next, we integrate both sides:

The left side becomes: $\int \sqrt{x} \, dx = \frac{2}{3} x^{\frac{3}{2}} + C$
The right side requires integrating:

Equating both integrals, we have: $$\frac{2}{3} x^{\frac{3}{2}} = -\frac{4}{3} \cos 2t + C.$$ Using the initial condition where $x = 0$ when $t = 0$, we find: $$\frac{2}{3} (0)^{\frac{3}{2}} = -\frac{4}{3} \cos 0 + C\Rightarrow C = \frac{4}{3}.$$ Substituting back, we get: $$\frac{2}{3} x^{\frac{3}{2}} = -\frac{4}{3} \cos 2t + \frac{4}{3}.$$ After simplification, the solution for $x$ becomes: $$x = \left(2 - 2\cos 2t\right)^{\frac{2}{3}}.$$

Step 2

Find the maximum height of the column of water, giving your answer to the nearest cm.

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Answer

To find the maximum height of the column of water, we analyze the function: $x = \left(2 - 2\cos 2t\right)^{\frac{2}{3}}.$

The maximum occurs when $, \cos 2t = -1,, \therefore \max, x = \left(2 - 2(-1)\right)^{\frac{2}{3}} = (4)^{\frac{2}{3}} = 4^{\frac{2}{3}} = \frac{4^2}{4} = \frac{16}{4} = 4.$$

Thus, the maximum height is: $\text{Max height} = 4^{\frac{2}{3}} = 252 \, \text{cm}.$

The height $x$ metres, of a column of water in a fountain display satisfies the differential equation $$\frac{dx}{dr} = \frac{8\sin 2t}{3\sqrt{x}}$$, where $t$ is the time in seconds after the display begins - AQA - A-Level Maths Pure - Question 15 - 2017 - Paper 1

Worked Solution & Example Answer:The height $x$ metres, of a column of water in a fountain display satisfies the differential equation $$\frac{dx}{dr} = \frac{8\sin 2t}{3\sqrt{x}}$$, where $t$ is the time in seconds after the display begins - AQA - A-Level Maths Pure - Question 15 - 2017 - Paper 1

Solve the differential equation, given that initially the column of water has zero height.

Find the maximum height of the column of water, giving your answer to the nearest cm.

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