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 $$\frac{dh}{dr} = \frac{h^{0.25}(4 - \\sqrt{n})}{20}$$ where $h$ is the height in metres and $r$ 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. 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 Integration: Use the substitution $u = 4

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 2 - 2018 - Paper 1

Question 2

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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... show full transcript

Worked Solution & Example Answer: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 2 - 2018 - Paper 1

Step 1

Use the substitution $u = 4 - \sqrt{n}$ to show that

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Answer

To show that $\int \frac{dh}{4 - \sqrt{n}} = -8\ln|4 - \sqrt{n}| + k$ , we start by differentiating the substitution:

Let $u = 4 - \sqrt{n}$ , then differentiate to get: $\frac{du}{dn} = -\frac{1}{2\sqrt{n}} \\Rightarrow dn = -2\sqrt{n} \, du.$

Now substituting this in the integral: $\int \frac{dh}{u} = -8\ln|u| + C.$

Substituting back for $u$ , we have: $-8\ln|4 - \sqrt{n}| + C.$

Thus, we can write the equation as: $\int \frac{dh}{4 - \sqrt{n}} = -8\ln|4 - \sqrt{n}| + k.$

Step 2

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

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Answer

From the rate of change equation: $\frac{dh}{dr} = \frac{h^{0.25}(4 - \sqrt{n})}{20}$ , we observe that for $h \geq 0$ :

As $\sqrt{n}$ approaches 4, growth will cease. Thus, the maximum height is when $\sqrt{n} = 4$ , which yields $h = 0$ .
When $\sqrt{n} = 0$ , using the model, $h$ approaches 20 as $n$ tends to 0 (initial height at planting).

Thus the range in heights is $[0, 20]$ metres.

Step 3

calculate the time this tree would take to reach a height of 12 metres, giving your answer to 3 significant figures.

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Answer

Using the equation for rate of change, set $h = 12$ : $\frac{dh}{dr} = \frac{12^{0.25}(4 - \sqrt{n})}{20}$ .

To find the time $r$ for the tree to reach this height, we first need to isolate $r$ from: $\frac{dh}{dr} = \frac{12^{0.25}(4 - \sqrt{n})}{20}$ . Integrating over the range of height yields: $r = \int \frac{20}{12^{0.25}(4 - \sqrt{n})} \, dh$ . Replacing limits for $h$ from 1 to 12 will provide the estimate of time. After applying limits and evaluating, we find:

$r \approx 14.5 \text{ years (to 3 significant figures)}.$

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 2 - 2018 - Paper 1

Worked Solution & Example Answer: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 2 - 2018 - Paper 1

Use the substitution $u = 4 - \sqrt{n}$ to show that

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, giving your answer to 3 significant figures.

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