Past exam of the mathematics course of the University of Oxford / Undated / Prelims / Mathematics III resit / 2

Put $h(s)=\sin x(s)-\cos y(s)$. The chain rule givesApply the product rule and the chain rule once more: $g^{\prime }=h^{\prime }\cos h=h^{\prime }f$, andConsequently $f^{\prime \prime }=-(h^{\prime }{)}^{2}f-h^{\prime \prime }g$, so the requested coefficients areThis is the second derivative chain rule along the parametric curve $(x(s),y(s))$.

The Jacobian determinant isThe change of variables is defined only for $x\ne 0$. On the interior of the integration region, $x>0$ and $y>0$, so this Jacobian determinant is negative and nonzero.

The inverse change of variables is $x=v/(1+u)$, $y=uv/(1+u)$. Thus the triangular region becomesThe upper limit depends on $u$: the new region is not a rectangle. The absolute inverse Jacobian determinant is $x^2/(x+y)$, so the change of variables formula cancels the rational factor in the integrand:Although the original integrand is undefined at the vertex $x=y=0$, its nonnegative improper integral exists. One can apply the change of variables formula on $x\ge \epsilon$ and let $\epsilon \downarrow 0$ by the monotone convergence theorem; the transformed integrand has the finite integral just computed.

For the graph of a differentiable function, the arc length is $L=\int \sqrt{1+(y^{\prime }{)}^2}dx$. Since $y^{\prime }=2x$,Use the hyperbolic substitution $2x=\sinh t$: then $dx=\frac{1}{2}\cosh tdt$ and $\sqrt{1+4x^2}=\cosh t$. The hyperbolic functions ${\cosh }^{2}t=(1+\cosh 2t)/2$ yieldsHere $\mathrm{arsinh}t=\ln (t+\sqrt{1+t^2})$ is the inverse hyperbolic sine.