1.08a Fundamental theorem of calculus: integration as reverse of differentiation

A particle of mass \(m\) is attached to one end \(P\) of a light elastic spring \(PQ\), of natural length \(a\) and modulus of elasticity \(man^2\). At time \(t = 0\), the particle and the spring are at rest on a smooth horizontal table, with the spring straight but unstretched and uncompressed. The end \(Q\) of the spring is then moved in a straight line, in the direction \(PQ\), with constant acceleration \(f\). At time \(t\), the displacement of the particle in the direction \(PQ\) from its initial position is \(x\) and the length of the spring is \((a + y)\).

1. Show that \(x + y = \frac{1}{2}ft^2\). [2]
2. Hence show that $$\frac{d^2x}{dt^2} + n^2x = \frac{1}{2}n^2ft^2.$$ [6]

You are given that the general solution of this differential equation is $$x = A\cos nt + B\sin nt + \frac{1}{2}ft^2 - \frac{f}{n^2},$$ where \(A\) and \(B\) are constants.

1. Find the values of \(A\) and \(B\). [6]
2. Find the maximum tension in the spring. [4]

A particle \(P\) of mass \(m\) is suspended from a fixed point by a light elastic spring. The spring has natural length \(a\) and modulus of elasticity \(2m\omega^2a\), where \(\omega\) is a positive constant. At time \(t = 0\) the particle is projected vertically downwards with speed \(U\) from its equilibrium position. The motion of the particle is resisted by a force of magnitude \(2m\omega v\), where \(v\) is the speed of the particle. At time \(t\), the displacement of \(P\) downwards from its equilibrium position is \(x\).

1. Show that \(\frac{\mathrm{d}^2x}{\mathrm{d}t^2} + 2\omega \frac{\mathrm{d}x}{\mathrm{d}t} + 2\omega^2x = 0\). [5] Given that the solution of this differential equation is \(x = e^{-\omega t}(A \cos \omega t + B \sin \omega t)\), where \(A\) and \(B\) are constants,
2. find \(A\) and \(B\). [4]
3. Find an expression for the time at which \(P\) first comes to rest. [3]

A curve cuts the \(x\)-axis at \((2, 0)\) and has gradient function $$\frac{dy}{dx} = \frac{24}{x^3}$$

1. Find the equation of the curve. [4 marks]
2. Show that the perpendicular bisector of the line joining \(A(-2, 8)\) to \(B(-6, -4)\) is the normal to the curve at \((2, 0)\) [6 marks]

It is given that $$\frac{\mathrm{d}y}{\mathrm{d}x} = (x + 2)(2x - 1)^2$$ and when \(x = 6\), \(y = 900\) Find \(y\) in terms of \(x\) [6 marks]

A function f has domain \(\mathbb{R}\) and range \(\{y \in \mathbb{R} : y \geq c\}\) The graph of \(y = f(x)\) is shown. \includegraphics{figure_2} The gradient of the curve at the point \((x, y)\) is given by \(\frac{dy}{dx} = (x - 1)e^x\) Find an expression for f(x). Fully justify your answer. [8 marks]

An elite athlete runs in a straight line to complete a 100-metre race. During the race, the athlete's velocity, \(v \text{ m s}^{-1}\), may be modelled by $$v = 11.71 - 11.68e^{-0.9t} - 0.03e^{0.3t}$$ where \(t\) is the time, in seconds, after the starting pistol is fired.

1. Find the maximum value of \(v\), giving your answer to one decimal place. Fully justify your answer. [8 marks]
2. Find an expression for the distance run in terms of \(t\). [6 marks]
3. The athlete's actual time for this race is 9.8 seconds. Comment on the accuracy of the model. [2 marks]

At time \(t = 0\), a parachutist jumps out of an airplane that is travelling horizontally. The velocity, \(\mathbf{v}\) m s\(^{-1}\), of the parachutist at time \(t\) seconds is given by: $$\mathbf{v} = (40e^{-0.2t})\mathbf{i} + 50(e^{-0.2t} - 1)\mathbf{j}$$ The unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) are horizontal and vertical respectively. Assume that the parachutist is at the origin when \(t = 0\) Model the parachutist as a particle.

1. Find an expression for the position vector of the parachutist at time \(t\). [4 marks]
2. The parachutist opens her parachute when she has travelled 100 metres horizontally. Find the vertical displacement of the parachutist from the origin when she opens her parachute. [4 marks]
3. Carefully, explaining the steps that you take, deduce the value of \(g\) used in the formulation of this model. [3 marks]

A segment of the line \(y = kx\) is rotated about the \(x\)-axis to generate a cone with vertex \(O\). The distance of \(O\) from the centre of the base of the cone is \(h\). The radius of the base of the cone is \(r\). \includegraphics{figure_15}

1. Find \(k\) in terms of \(r\) and \(h\). [1 mark]
2. Use calculus to prove that the volume of the cone is $$\frac{1}{3}\pi r^2 h$$ [3 marks]

Which one of the expressions below is not equal to zero? Circle your answer. [1 mark] \(\lim_{x \to \infty} (x^2e^{-x})\) \quad \(\lim_{x \to 0} (x^5 \ln x)\) \quad \(\lim_{x \to \infty} \left(\frac{e^x}{x^5}\right)\) \quad \(\lim_{x \to 0^+} (x^3e^x)\)

In this question you must show detailed reasoning. In this question, positions are given relative to a fixed origin, O. The unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) are in the \(x\)- and \(y\)-directions respectively in a horizontal plane. Distances are measured in centimetres and the time, \(t\), is measured in seconds, where \(0 \leq t \leq 5\). A small radio-controlled toy car C moves on a horizontal surface which contains O. The acceleration of C is given by \(2\mathbf{i} + t\mathbf{j} \text{ cm s}^{-2}\). When \(t = 4\), the displacement of C from O is \(16\mathbf{i} + \frac{32}{3}\mathbf{j}\) cm, and the velocity of C is \(8\mathbf{i} \text{ cm s}^{-1}\). Determine a cartesian equation for the path of C for \(0 < t < 5\). You are not required to simplify your answer. [6]

A particle moves along the horizontal \(x\)-axis so that its velocity \(v\) ms\(^{-1}\) at time \(t\) seconds is given by $$v = 6t^2 - 8t - 5.$$ At time \(t = 1\), the particle's displacement from the origin is \(-4\) m. Find an expression for the displacement of the particle at time \(t\) seconds. [3]

The gradient of a curve is given by \(\frac{dy}{dx} = 2x^{-\frac{1}{2}}\), and the curve passes through the point \((4, 5)\). Find the equation of the curve. [6]

A curve has the following properties: • The gradient of the curve is given by \(\frac{dy}{dx} = -2x\). • The curve passes through the point \((4, -13)\). Determine the coordinates of the points where the curve meets the line \(y = 2x\). [7]

A particle \(P\) moves along the \(x\)-axis. At time \(t\) seconds the velocity of \(P\) is \(v\text{m s}^{-1}\), where \(v = 2t^4 + kt^2 - 4\). The acceleration of \(P\) when \(t = 2\) is \(28\text{m s}^{-2}\).

1. Show that \(k = -9\). [3]
2. Show that the velocity of \(P\) has its minimum value when \(t = 1.5\). [3]

When \(t = 1\), \(P\) is at the point \((-6.4125, 0)\).

1. Find the distance of \(P\) from the origin \(O\) when \(P\) is moving with minimum velocity. [4]

A particle \(P\) is free to move along a straight line \(Ox\). It starts from rest at \(O\) and after \(t\) seconds its acceleration \(a\,\mathrm{m}\,\mathrm{s}^{-2}\) is given by \(a = 12 - 6t\).

1. Find an expression in terms of \(t\) for its velocity \(v\,\mathrm{m}\,\mathrm{s}^{-1}\). Hence find the velocity of \(P\) when \(t = 4\). [4]
2. Find the displacement of \(P\) from \(O\) when \(t = 4\). [3]
3. Find the velocity of \(P\) when it returns to \(O\). [3]