 Home

Codex Wiki

List of universities

University of Cambridge

Course of the University of Cambridge

Natural sciences course of the University of Cambridge

Year of the natural sciences course of the University of Cambridge

Year 1 of the natural sciences course of the University of Cambridge

Mathematics course of the year 1 of the natural sciences course of the University of Cambridge

Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge

2018

Paper 2

Write $c=\cos\phi$ and $s=\sin\phi$. The needed coefficient <derivatives> are $c_x=s^2/r$, $s_x=-sc/r$, $c_y=-sc/r$ and $s_y=c^2/r$. Apply the <product rule> to both terms of $f_x=cg_r-(s/r)g_\phi$ and use the <chain rule> for the <derivatives> of $g$. Collecting terms gives
$$
\boxed{f_{xx}=c^2g_{rr}-\frac{2sc}{r}g_{r\phi}+\frac{2sc}{r^2}g_\phi
+\frac{s^2}{r}g_r+\frac{s^2}{r^2}g_{\phi\phi}.}
$$
Similarly, differentiating $f_y=sg_r+(c/r)g_\phi$ gives
$$
\boxed{f_{yy}=s^2g_{rr}+\frac{2sc}{r}g_{r\phi}-\frac{2sc}{r^2}g_\phi
+\frac{c^2}{r}g_r+\frac{c^2}{r^2}g_{\phi\phi}.}
$$
On addition, both the mixed <partial derivative> terms and the first angular <derivative> terms cancel. Using $s^2+c^2=1$, the <Laplacian> in <polar coordinates> is
$$
\boxed{\nabla^2f=g_{rr}+\frac1r g_r+\frac1{r^2}g_{\phi\phi},\qquad r>0.}
$$

Solution

Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2018 / Paper 2 / 12 / d / Solution

Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2018 / Paper 2 / 12 / d / Solution

 Ancestors (13)

 Discussion (0)

 Articles by others on the same topic (0)

Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2018 / Paper 2 / 12 / d / Solution

 Ancestors (13)

 Discussion (0)  Subscribe (1)

 Articles by others on the same topic (0)

 Discussion (0)