12.2 Answers to 2014 test

1. For the partial fractions, we have $\frac{10}{(x-3)(x^{2}+1)}=\frac{1}{x-3}-\frac{x+3}{x^{2}+1}$. For the indefinite integral, we have $\int\frac{10}{(x-3)(x^{2}+1)}=\frac{1}{2}\log\left(\frac{(x-3)^{2}}{x^{2}+1}\right)-3\tan^{-1}x+c$.

2. a) The quick way: $x^{3}-6x^{2}+9x=x(x-3)^{2}=t^{2}(t^{2}-3)^{2}=(t(t^{2}-3))^{2}=(3y)^{2}$.

b) We have $\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}=(t^{2}+1)^{2}$ and the required arc length is 12.

3. a) False, since $z=0$ is not a root of $(z-1)^{4}=2$. (The diagram shows the roots of $(z-1)^{4}=1$.)

b) False, e.g. $f(x,y)=x^{2}-y^{2}$ has a saddle point at $(0,0)$.

c) True, since by Cavalieri’s slicing principle the volume is $\int_{0}^{1}z^{2}\,dz=\frac{1}{3}$. (NB: this is non-examinable material in 2015/16.)

4. The stationary points are $(0,0)$ and $(-3,9)$. (You are not required to show this, but these points are respectively a saddle point and a local maximum.)

5. $\int_{0}^{1}\int_{0}^{y^{2}}(xy^{2}-x^{2})\,dx\,dy=\int_{0}^{1}\frac{y^{6}}{6}\,dy=\frac{1}{42}$.