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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 2 - 2019 - Paper 2
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![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 2-2019-Paper 2.png](https://cdn.simplestudy.io/assets/backend/uploads/question/MathsPure_2019_2_1_QP_14_2019.jpg)
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 scientist... 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 2 - 2019 - Paper 2
Step 1
Use the substitution $u = 4 - \sqrt{n}$ to show that $\int \frac{dh}{4 - \sqrt{n}} = -8\ln|4 - \sqrt{n}| + k$
Answer
To solve the integral, we start with the substitution:
Let . Thus, we have:
[\sqrt{n} = 4 - u \Rightarrow n = (4 - u)^2]
Now, differentiate both sides with respect to :
[\frac{dn}{dt} = -2(4 - u) \frac{du}{dt}]
From the given differential equation, we can express ( \frac{dh}{dt} ) in terms of ( \frac{du}{dt} ):
[\int\frac{dh}{4 - \sqrt{n}} = \int \frac{dh}{u} = -8\ln|u| + k]
Substituting back for gives us:
[\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.
Answer
From the equation ( \frac{dh}{dt} = \frac{0.25(4 - \sqrt{n})}{20} ), we determine the maximum and minimum heights:
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At , the height is when : [ h(0) = 0.25 \cdot 4 = 1 \text{ metre} ]
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As , ( \sqrt{n} \to 4 ): this implies ( h \to 0 )
Hence, the range of heights is ( h \in [1, 4) \text{ metres} )
Step 3
Calculate the time this tree would take to reach a height of 12 metres.
Answer
We set up the equation for height:
From the model, . Setting this to 12 metres and solving for :
[12 = \frac{0.25(4 - \sqrt{n})}{20} \Rightarrow 240 = 4 - \sqrt{n} \Rightarrow \sqrt{n} = 4 - 240 = -236]
Since reaching 12 metres is not in the physical domain of the model, we reformulate the valid range of heights: Therefore, this tree cannot reach 12 metres in this model.