12.1 Answers to 2015 test

1. i) We make the substitution $x=\cosh t$. The value of the integral is $\sqrt{3}-\frac{1}{2}\cosh^{-1}2$.

ii) We make the substitution $t=\tan x/2$, obtaining the integral $\int_{1}^{\infty}\frac{dt}{(2t-1)(t+2)}$. For the partial fractions step, we obtain $\frac{1}{5}\left(\frac{2}{2t-1}-\frac{1}{t+2}\right)$. The value of the integral is: $\frac{1}{5}\log 6$.

2. i) At $(x(t),y(t))$ we have $\frac{dy}{dx}=\frac{2t}{t^{2}-1}$. The tangent line at $Q$ is $3y=4x+2e^{2}$, or $y=\frac{4}{3}x+\frac{2}{3}e^{2}$. The normal line at $Q$ is $4y+3x=11e^{2}$, or $y=\frac{11}{4}e^{2}-\frac{3}{4}x$.

ii) We have $\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}=(t^{2}+1)^{2}e^{2t}$. The length along the curve is $3e^{2}-3$.

3. i) True, since the auxiliary equation is $s^{2}-3s+2$ which has (distinct) roots $1$ and $2$.

ii) True, by implicit differentiation.

iii) True, since $\frac{\partial u}{\partial t}=u$ and $\frac{\partial^{2}u}{\partial x}^{2}=-u$.

iv) False: it gives an estimate of $\frac{1}{3}(f(0)+4f(1)+f(2))=2$.

4. There are two stationary points: $(1,1)$ and $(-1,-1)$, which are both saddle points (since $\Delta=-8x^{4}-4y^{2}<0$).

5. The integrating factor is $\sqrt{1+x^{2}}$, and multiplying through by this factor produces the integrable equation:

which has general solution

The particular solution with $y(0)=1$ is $y=\sqrt{x^{2}+1}$.