A closed curve is given in polar coordinates by the equation $$r = a(1 - \cos \theta).$$ Show that the tangent at the point \(\theta\) is inclined at an angle \(\psi = \frac{1}{2}\theta\) to the axis \(\theta = 0\). Find the radius of curvature at the point \(\theta\).
The polynomial \(T_n(x)\), where \(n\) is a non-negative integer, satisfies $$(1-x^2) \frac{d^2 T_n}{dx^2} - x \frac{dT_n}{dx} + n^2 T_n = 0;$$ $$T_0(1) = 1; \quad T_n(x) = (-1)^n T_n(-x).$$ By substituting \(x = \cos \theta\) and solving the transformed equation, obtain \(T_n(x)\) in simple form as a function of \(\theta\) and hence show that $$T_0(x) = 1, \quad T_1(x) = x,$$ and that $$T_{n+1}(x) = 2x T_n(x) - T_{n-1}(x) \quad \text{for } n = 1, 2, 3, \ldots$$
The position vector, \(\mathbf{r}(t)\), of a moving point \(P\) relative to a fixed origin satisfies the equation $$\ddot{\mathbf{r}} = \mathbf{k} \times \mathbf{r},$$ where \(\mathbf{k}\) is a constant vector and \(\dot{\mathbf{r}} = d\mathbf{r}/dt\). Show that the locus of \(P\) is a circle. Describe the motion of \(P\) when \(\mathbf{r}\) satisfies the equation $$\ddot{\mathbf{r}} = \mathbf{k} \times \dot{\mathbf{r}}.$$
Let \(u\) be a function of \(x\) and \(y\). If \(x\) and \(y\) are related by \(u(x,y) = \text{constant}\), prove that $$\frac{dy}{dx} = -\frac{\partial u / \partial x}{\partial u / \partial y}.$$ Deduce that the partial differential equation $$2yu \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = 0$$ has solutions given by $$u = f(x - y^2 u),$$ where \(f\) is an arbitrary function. Find the solution such that \(u = x\) when \(y = 0\).
Prove that, if \(g(x) > 0\), then $$\int_a^b f(x)g(x) \, dx \leq \max_{a \leq x \leq b} \{f(x)\} \int_a^b g(x) \, dx$$ and hence that $$\left| \int_a^b f(x)g(x) \, dx \right| \leq \max_{a \leq x \leq b} \{|f(x)|\} \int_a^b g(x) \, dx.$$ Give an example of functions \(f(x)\) and \(g(x)\) for which $$\left| \int_a^b f(x)g(x) \, dx \right| > \max_{a \leq x \leq b} \{|f(x)|\} \int_a^b g(x) \, dx.$$ The function \(h(x)\) vanishes at \(x = 0\), and possesses a first derivative. Show that $$\int_0^a h(x) \, dx = \int_0^a (a-x)h'(x) \, dx,$$ and deduce that $$\left| \int_0^a h(x) \, dx \right| \leq \frac{1}{2}a^2 M,$$ where $$M = \max_{0 \leq x \leq a} \{|h'(x)|\}.$$