Prove that the radius of curvature at any point of a curve is given by \[ \rho = \frac{(x'^2+y'^2)^{\frac{3}{2}}}{x'y''-y'x''}, \] where accents denote differentiations with regard to a parameter. \par Find the curvature at the origin of each of the branches of the curve \[ x^3-x^2y-xy^2+y^3=4xy-2y^2, \] and trace the curve.
If \(z=\frac{xy}{x-y}\), find all the second order differential coefficients of \(z\) with respect to \(x\) and \(y\), and verify that \[ x^2\frac{\partial^2 z}{\partial x^2} + 2xy\frac{\partial^2 z}{\partial x \partial y} + y^2\frac{\partial^2 z}{\partial y^2} = 0. \]
Obtain formulae of reduction for \[ \int x^n\cos mx\,dx, \quad \int x^k(a+bx^n)^p\,dx. \]
Find the area of the surface generated by the revolution of the lemniscate \(r^2=a^2\cos 2\theta\) round the initial line.
\(AB\) and \(AC\) are two fixed straight lines, and \(O\) is a fixed point. Two circles are drawn through \(O\) one of which touches \(AB\) and \(AC\) at \(D\) and \(E\) respectively and the other touches them at \(F\) and \(G\) respectively. Prove that the circles \(ODF\) and \(OEG\) touch at \(O\).
\(PCQ\) is a given diameter of an ellipse whose centre is \(C\), and \(D\) is any other point on the ellipse. If the area of the triangle \(PDQ\) is a maximum, then \(CP\) and \(CD\) are conjugate diameters. If the perimeter of the triangle \(PDQ\) is a maximum, then the tangents at \(P\) and \(D\) are at right angles.
Prove that a tetrahedron can be constructed so as to have four equal acute-angled triangles for its faces; shew also that in such a tetrahedron the centre of the circumscribing sphere, the centre of the inscribed sphere, and the centre of inertia all coincide, and that the lines joining the middle points of pairs of opposite edges bisect each other at right angles at this same point.
If any one of the three quantities \(ax+bz+cy\), \(by+cx+az\), \(cz+ay+bx\) vanishes, prove that the sum of the cubes of the other two is equal to \[ (a^3+b^3+c^3-3abc)(x^3+y^3+z^3-3xyz). \]
Sum to infinity the series \[ \frac{1}{6} + \frac{1\cdot4}{6\cdot12} + \frac{1\cdot4\cdot7}{6\cdot12\cdot18} + \dots. \]
Solution: Notice that \begin{align*} && (1+x)^{-1/3} &= 1 +\frac{-\frac13}{1!}x +\frac{(-\frac13)(-\frac43)}{2!}x^2 +\frac{(-\frac13)(-\frac43)(-\frac73)}{3!}x^3 +\cdots \\ \Rightarrow && (1-x)^{-1/3} &= 1 + \frac{1}{3\cdot 1!}x + \frac{1\cdot4}{3^2\cdot 2!} x^2+\frac{1\cdot4\cdot7}{3^3 \cdot 3!}x^3 + \cdots \\ \Rightarrow && \left(1-\frac12\right)^{-1/3} &= 1 + \frac{1}{6}+\frac{1\cdot4}{6^2 \cdot 2!} + \frac{1 \cdot 4 \cdot 7}{6^3 \cdot 3!} + \cdots \\ \end{align*} Therefore our sum is \(\sqrt[3]{2}-1\)
\(ABCD\) is a quadrilateral circumscribing a circle and \(a,b,c,d\) are the lengths of the tangents from \(A, B, C, D\) respectively; prove that the sum of a pair of opposite angles is \(2\theta\), where \[ (a+b)(b+c)(c+d)(d+a)\cos^2\theta = (ac-bd)^2. \]