12.5 Answers to 2011 test

2. For the partial fractions, we have $\frac{1}{(x+2)(5x+5)}=\frac{1}{6}\left(\frac{5}{5x+4}-\frac{1}{x+2}\right)$. The indefinite integral

Then the integral $\int_{0}^{\infty}\frac{dx}{(x+2)(5x+4)}$ converges to $\frac{1}{6}\log\left(\frac{5}{2}\right)$.

3. The Laplace transform of $\cosh ax$ is $\frac{s}{s^{2}-a^{2}}$.

4. i) $\frac{dy}{dx}=\frac{\sin t}{1-\cos t}=\cot\frac{t}{2}$.

ii) $L=8$. (Note $\frac{dx}{dt}^{2}+\frac{dy}{dt}^{2}=2-2\cos t=4\sin^{2}\frac{t}{2}$.)

5. There are four stationary points at $(2+\sqrt{\frac{7}{2}},2+\sqrt{\frac{7}{2}})$, $(2-\sqrt{\frac{7}{2}},2-\sqrt{\frac{7}{2}})$, $(2+\frac{3\sqrt{2}}{2},-2-\frac{3\sqrt{2}}{2})$, $(2-\frac{3\sqrt{2}}{2},-2+\frac{3\sqrt{2}}{2})$. (Finding these stationary points was covered in the lecture on Friday of week 8, but not determining their nature.) When $x^{2}=y^{2}$ we have $\Delta=8x(2-x)$ and so, since $x>2$ or $x<0$ at each stationary point, they are all saddles except for $(2-\sqrt{\frac{7}{2}},2-\sqrt{\frac{7}{2}})$, which is a local minimum.