For the curve shown in the diagram, the normal to the curve at the point $$\(P\)$$ with coordinates $$\((x, y)\)$$ meets the $$\(x\)$$-axis at $$\(N\)$$. The point $$\(M\)$$ is the foot of the perpendicular from $$\(P\)$$ to the $$\(x\)$$-axis. The curve is such that for all values of $$\(x\)$$ in the interval $$\(0 \leqslant x<\frac{1}{2} \pi\)$$, the area of triangle $$\(P M N\)$$ is equal to $$\(\tan x\)$$. Given that $$\(y=1\)$$ when $$\(x=0\)$$, solve this differential equation to find the equation of the curve,expressing $$\(y\)$$ in terms of $$\(x\)$$. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................
Exam No：9709_s21_qp_33 Year：2021 Question No：7(b)
Answer:
Separate variables and integrate at least one side
Obtain term \(\frac{1}{6} y^{3}\)
Obtain term of the form \(\pm \ln \cos x\)
Evaluate a constant or use \(x=0\) and \(y=1\) in a solution containing terms \(a y^{3}\) and \(\pm \ln \cos x\), or equivalent
Obtain correct answer in any form, e.g. \(\frac{1}{6} y^{3}=-\ln \cos x+\frac{1}{6}\)
Obtain final answer \(y=\sqrt[3]{(1-6 \ln \cos x)}\)
Obtain term \(\frac{1}{6} y^{3}\)
Obtain term of the form \(\pm \ln \cos x\)
Evaluate a constant or use \(x=0\) and \(y=1\) in a solution containing terms \(a y^{3}\) and \(\pm \ln \cos x\), or equivalent
Obtain correct answer in any form, e.g. \(\frac{1}{6} y^{3}=-\ln \cos x+\frac{1}{6}\)
Obtain final answer \(y=\sqrt[3]{(1-6 \ln \cos x)}\)
Knowledge points:
3.5.4 recognise an integrand of the form , and integrate such functions
3.8.2 find by integration a general form of solution for a first order differential equation in which the variables are separable (Including any of the integration techniques from topic 3.5 above.)
3.8.3 use an initial condition to find a particular solution
Solution:
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