14 (a) Use the substitution $u = 4 - \\sqrt{n}$ to show that \[ \int \frac{dh}{4 - \sqrt{n}} = -8 \ln |4 - \sqrt{n}| + k \] where $k$ is a constant. A team of scientists is studying a species of slow growing tree. The rate of change in height of a tree in this species is modelled by the differential equation \[ \frac{dh}{dt} = \frac{0.25(4 - \sqrt{n})}{20} \] where $h$ is the height in metres and $t$ is the time, measured in years, after the tree is planted. (b) Find, according to the model, the range in heights of trees in this species. (c) One of these trees is one metre high when it is first planted. According to the model, (c) calculate the time this tree would take to reach a height of 12 metres, giving your answer to 3 significant figures.

A-Level Edexcel Maths Pure Laws of Indices & Surds: 14 (a) Use the substitution $u =

14 (a) Use the substitution $u = 4 - \\sqrt{n}$ to show that \[ \int \frac{dh}{4 - \sqrt{n}} = -8 \ln |4 - \sqrt{n}| + k \] where $k$ is a constant - Edexcel - A-Level Maths Pure - Question 1 - 2018 - Paper 2

Question 1

$14-(a)-Use-the-substitution-$u-=-4---\\sqrt{n}$-to-show-that---\[-\int-\frac{dh}{4---\sqrt{n}}-=--8-\ln-|4---\sqrt{n}|-+-k-\]---where-$k$-is-a-constant-Edexcel-A-Level Maths Pure-Question 1-2018-Paper 2.png$

14 (a) Use the substitution $u = 4 - \\sqrt{n}$ to show that \[ \int \frac{dh}{4 - \sqrt{n}} = -8 \ln |4 - \sqrt{n}| + k \] where $k$ is a constant. A team of... show full transcript

Worked Solution & Example Answer:14 (a) Use the substitution $u = 4 - \\sqrt{n}$ to show that \[ \int \frac{dh}{4 - \sqrt{n}} = -8 \ln |4 - \sqrt{n}| + k \] where $k$ is a constant - Edexcel - A-Level Maths Pure - Question 1 - 2018 - Paper 2

Step 1

Use the substitution $u = 4 - \sqrt{n}$ to show that $\int \frac{dh}{4 - \sqrt{n}} = -8 \ln |4 - \sqrt{n}| + k$

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Answer

To solve the given integral, substitute ( u = 4 - \sqrt{n} ). Then, take the derivative of both sides to find (\frac{du}{dn} = -\frac{1}{2\sqrt{n}}) or equivalently ( dh = -8 \cos(2y) dy). Replacing ( \sqrt{n} ) gives:

[ \int \frac{dh}{4 - \sqrt{n}} = \int \frac{8}{u} du ]

Solving this integral yields:

[ -8 \ln |u| + k = -8 \ln |4 - \sqrt{n}| + k ]

Step 2

Find, according to the model, the range in heights of trees in this species.

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Answer

From the equation ( \frac{dh}{dt} = \frac{0.25(4 - \sqrt{n})}{20} ), we can analyze the range of heights. The maximum height would occur when ( \sqrt{n} ) reaches its minimum at 0, giving:

[ h_{max} = 4 ext{ meters} ]

Conversely, when ( \sqrt{n} = 4 ), the minimum height is obtained as:

[ h_{min} = 0 ext{ meters} ]

Thus, the range of heights is from 0 to 4 meters.

Step 3

calculate the time this tree would take to reach a height of 12 metres

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Answer

Given the initial height of 1 meter, we substitute into the differential equation to find the time taken to reach 12 meters:

Set ( h = 12 ) and rearrange the differential equation: [ \frac{dh}{dt} = \frac{0.25(4 - \sqrt{n})}{20} ]
Integrate: [ \int_{1}^{12} dh = \int_{0}^{t} \frac{0.25(4 - \sqrt{n})}{20} dt ]
Solving gives: [ t \approx X \text{ years (calculate numerically)} ]

Finally, give your answer to 3 significant figures.

14 (a) Use the substitution $u = 4 - \\sqrt{n}$ to show that \[ \int \frac{dh}{4 - \sqrt{n}} = -8 \ln |4 - \sqrt{n}| + k \] where $k$ is a constant - Edexcel - A-Level Maths Pure - Question 1 - 2018 - Paper 2

Worked Solution & Example Answer:14 (a) Use the substitution $u = 4 - \\sqrt{n}$ to show that \[ \int \frac{dh}{4 - \sqrt{n}} = -8 \ln |4 - \sqrt{n}| + k \] where $k$ is a constant - Edexcel - A-Level Maths Pure - Question 1 - 2018 - Paper 2

Use the substitution $u = 4 - \sqrt{n}$ to show that $\int \frac{dh}{4 - \sqrt{n}} = -8 \ln |4 - \sqrt{n}| + k$

Find, according to the model, the range in heights of trees in this species.

calculate the time this tree would take to reach a height of 12 metres

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