Problems

1. A particle moves in two-dimensional space. Its position is given by coordinates \((x, y)\) which satisfy \[\frac{\mathrm{d}x}{\mathrm{d}t} = -x + 3y + u\] \[\frac{\mathrm{d}y}{\mathrm{d}t} = x + y + u\] where \(t\) is the time and \(u\) is a function of time. At time \(t = 0\), the particle has position \((x_0,\, y_0)\).
   1. By considering \(\dfrac{\mathrm{d}x}{\mathrm{d}t} - \dfrac{\mathrm{d}y}{\mathrm{d}t}\), show that if the particle is at the origin \((0,0)\) at some time \(t > 0\), then it is necessary that \(x_0 = y_0\).
   2. Given that \(x_0 = y_0\), find a constant value of \(u\) that ensures that the particle is at the origin at a time \(t = T\), where \(T > 0\).
2. A particle whose position in three-dimensional space is given by co-ordinates \((x, y, z)\) moves with time \(t\) such that \[\frac{\mathrm{d}x}{\mathrm{d}t} = 4y - 5z + u\] \[\frac{\mathrm{d}y}{\mathrm{d}t} = x - 2z + u\] \[\frac{\mathrm{d}z}{\mathrm{d}t} = x - 2y + u\] where \(u\) is a function of time. At time \(t = 0\), the particle has position \((x_0,\, y_0,\, z_0)\).
   1. Show that, if the particle is at the origin \((0,0,0)\) at some time \(t > 0\), it is necessary that \(y_0\) is the mean of \(x_0\) and \(z_0\).
   2. Show further that, if the particle is at the origin \((0,0,0)\) at some time \(t > 0\), it is necessary that \(x_0 = y_0 = z_0\).
   3. Given that \(x_0 = y_0 = z_0\), find a constant value of \(u\) that ensures that the particle is at the origin at a time \(t = T\), where \(T > 0\).

Two particles, \(A\) of mass \(m\) and \(B\) of mass \(M\), are fixed to the ends of a light inextensible string \(AB\) of length \(r\) and lie on a smooth horizontal plane. The origin of coordinates and the \(x\)- and \(y\)-axes are in the plane. Initially, \(A\) is at \((0,\,0)\) and \(B\) is at \((r,\,0)\). \(B\) is at rest and \(A\) is given an instantaneous velocity of magnitude \(u\) in the positive \(y\) direction. At a time \(t\) after this, \(A\) has position \((x,\,y)\) and \(B\) has position \((X,\,Y)\). You may assume that, in the subsequent motion, the string remains taut.

1. Explain by means of a diagram why \[X = x + r\cos\theta\] \[Y = y - r\sin\theta\] where \(\theta\) is the angle clockwise from the positive \(x\)-axis of the vector \(\overrightarrow{AB}\).
2. Find expressions for \(\dot{X}\), \(\dot{Y}\), \(\ddot{X}\) and \(\ddot{Y}\) in terms of \(\ddot{x}\), \(\ddot{y}\), \(\dot{x}\), \(\dot{y}\), \(r\), \(\ddot{\theta}\), \(\dot{\theta}\) and \(\theta\), as appropriate. Assume that the tension \(T\) in the string is the only force acting on either particle.
3. Show that \[\ddot{x}\sin\theta + \ddot{y}\cos\theta = 0\] \[\ddot{X}\sin\theta + \ddot{Y}\cos\theta = 0\] and hence that \(\theta = \dfrac{ut}{r}\).
4. Show that \[m\ddot{x} + M\ddot{X} = 0\] \[m\ddot{y} + M\ddot{Y} = 0\] and find \(my + MY\) in terms of \(t\) and \(m, M, u, r\) as appropriate.
5. Show that \[y = \frac{1}{m+M}\left(mut + Mr\sin\!\left(\frac{ut}{r}\right)\right).\]
6. Show that, if \(M > m\), then the \(y\) component of the velocity of particle \(A\) will be negative at some time in the subsequent motion.

In this question, you should consider only points lying in the first quadrant, that is with \(x > 0\) and \(y > 0\).

1. The equation \(x^2 + y^2 = 2ax\) defines a \emph{family} of curves in the first quadrant, one curve for each positive value of \(a\). A second family of curves in the first quadrant is defined by the equation \(x^2 + y^2 = 2by\), where \(b > 0\).
   1. Differentiate the equation \(x^2 + y^2 = 2ax\) implicitly with respect to \(x\), and hence show that every curve in the first family satisfies the differential equation \[2xy\frac{\mathrm{d}y}{\mathrm{d}x} = y^2 - x^2.\] Find similarly a differential equation, independent of \(b\), for the second family of curves.
   2. Hence, or otherwise, show that, at every point with \(y \neq x\) where a curve in the first family meets a curve in the second family, the tangents to the two curves are perpendicular. A curve in the first family meets a curve in the second family at \((c,\,c)\), where \(c > 0\). Find the equations of the tangents to the two curves at this point. Is it true that where a curve in the first family meets a curve in the second family on the line \(y = x\), the tangents to the two curves are perpendicular?
2. Given the family of curves in the first quadrant \(y = c\ln x\), where \(c\) takes any non-zero value, find, by solving an appropriate differential equation, a second family of curves with the property that at every point where a curve in the first family meets a curve in the second family, the tangents to the two curves are perpendicular.
3. A family of curves in the first quadrant is defined by the equation \(y^2 = 4k(x + k)\), where \(k\) takes any non-zero value. Show that, at every point where one curve in this family meets a second curve in the family, the tangents to the two curves are perpendicular.

1. Use the substitution \(y = (x - a)u\), where \(u\) is a function of \(x\), to solve the differential equation \[ (x - a)\frac{dy}{dx} = y - x, \] where \(a\) is a constant.
2. The curve \(C\) with equation \(y = f(x)\) has the property that, for all values of \(t\) except \(t = 1\), the tangent at the point \(\bigl(t,\, f(t)\bigr)\) passes through the point \((1, t)\).
   1. Given that \(f(0) = 0\), find \(f(x)\) for \(x < 1\). Sketch \(C\) for \(x < 1\). You should find the coordinates of any stationary points and consider the gradient of \(C\) as \(x \to 1\). You may assume that \(z\ln|z| \to 0\) as \(z \to 0\).
   2. Given that \(f(2) = 2\), sketch \(C\) for \(x > 1\), giving the coordinates of any stationary points.

The curves \(C_1\) and \(C_2\) both satisfy the differential equation \[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{kxy - y}{x - kxy},\] where \(k = \ln 2\). All points on \(C_1\) have positive \(x\) and \(y\) co-ordinates and \(C_1\) passes through \((1,\,1)\). All points on \(C_2\) have negative \(x\) and \(y\) co-ordinates and \(C_2\) passes through \((-1,\,-1)\).

1. Show that the equation of \(C_1\) can be written as \((x-y)^2 = (x+y)^2 - 2^{x+y}\). Determine a similar result for curve \(C_2\). Hence show that \(y = x\) is a line of symmetry of each curve.
2. Sketch on the same axes the curves \(y = x^2\) and \(y = 2^x\), for \(x \geqslant 0\). Hence show that \(C_1\) lies between the lines \(x + y = 2\) and \(x + y = 4\). Sketch curve \(C_1\).
3. Sketch curve \(C_2\).

1. Given that the variables \(x\), \(y\) and \(u\) are connected by the differential equations \[ \frac{\mathrm{d}u}{\mathrm{d}x} + \mathrm{f}(x)u = \mathrm{h}(x) \quad \text{and} \quad \frac{\mathrm{d}y}{\mathrm{d}x} + \mathrm{g}(x)y = u, \] show that \[ \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + (\mathrm{g}(x) + \mathrm{f}(x))\frac{\mathrm{d}y}{\mathrm{d}x} + (\mathrm{g}'(x) + \mathrm{f}(x)\mathrm{g}(x))y = \mathrm{h}(x). \tag{1} \]
2. Given that the differential equation \[ \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + \left(1 + \frac{4}{x}\right)\frac{\mathrm{d}y}{\mathrm{d}x} + \left(\frac{2}{x} + \frac{2}{x^2}\right)y = 4x + 12 \tag{2} \] can be written in the same form as (1), find a first order differential equation which is satisfied by \(\mathrm{g}(x)\). If \(\mathrm{g}(x) = kx^n\), find a possible value of \(n\) and the corresponding value of \(k\). Hence find a solution of (2) with \(y = 5\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -3\) at \(x = 1\).

1. Differentiate $\displaystyle \; \frac z {(1+z^2)^{\frac12}} \;$ with respect to \(z\).
2. The {\em signed curvature} \(\kappa\) of the curve \(y=\f(x)\) is defined by \[ \kappa = \frac {\f''(x)}{\big({1+ (\f'(x))^2\big)^{\frac32}}} \,.\] Use this definition to determine all curves for which the signed curvature is a non-zero constant. For these curves, what is the geometrical significance of \(\kappa\)?