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

6. (a) Use algebraic integration to show that the centre of mass of a uniform solid hemisphere of radius \(r\) is at a distance \(\frac { 3 } { 8 } r\) from the centre of its plane face.
[0pt] [You may assume that the volume of a sphere of radius \(r\) is \(\frac { 4 } { 3 } \pi r ^ { 3 }\) ]
(5) \begin{figure}[h] \end{figure} A uniform solid hemisphere of mass \(m\) and radius \(r\) is joined to a uniform solid right circular cone to form a solid \(S\). The cone has mass \(M\), base radius \(r\) and height \(4 r\). The vertex of the cone is \(O\). The plane face of the cone coincides with the plane face of the hemisphere, as shown in Figure 3.
(b) Find the distance of the centre of mass of \(S\) from \(O\). The point \(A\) lies on the circumference of the base of the cone. The solid is placed on a horizontal table with \(O A\) in contact with the table. The solid remains in equilibrium in this position.
(c) Show that \(M \geqslant \frac { 1 } { 10 } m\)

7. \begin{figure}[h] \end{figure} A particle of mass \(m\) is attached to one end of a light rod of length \(l\). The other end of the rod is attached to a fixed point \(O\). The rod can turn freely in a vertical plane about a horizontal axis through \(O\). The particle is projected with speed \(u\) from a point \(A\), where \(O A\) makes an angle \(\alpha\) with the upward vertical through \(O\), as shown in Figure 4. The particle moves in complete vertical circles. Given that \(\cos \alpha = \frac { 4 } { 5 }\)

1. show that \(u > \sqrt { \frac { 2 g l } { 5 } }\) As the rod rotates, the least tension in the rod is \(T\) and the greatest tension is \(4 T\).
2. Show that \(u = \sqrt { \frac { 17 } { 5 } g l }\)
   \includegraphics[max width=\textwidth, alt={}]{ffe0bc72-3136-48d9-9d5b-4a364d134070-12_2639_1830_121_121}

1. \begin{figure}[h] \end{figure} The shaded region \(R\) is bounded by the curve with equation \(y ^ { 2 } = 9 ( 4 - x )\), the positive \(x\)-axis and the positive \(y\)-axis, as shown in Figure 1. A uniform solid \(S\) is formed by rotating \(R\) through \(360 ^ { \circ }\) about the \(x\)-axis.
Use algebraic integration to find the \(x\) coordinate of the centre of mass of \(S\).

2. A particle \(P\) of mass 0.6 kg is moving along the positive \(x\)-axis in the positive direction. The only force acting on \(P\) acts in the direction of \(x\) increasing and has magnitude \(\left( 3 t + \frac { 1 } { 2 } \right) \mathrm { N }\), where \(t\) seconds is the time after \(P\) leaves the origin \(O\). When \(t = 0 , P\) is at rest at \(O\).

1. Find an expression, in terms of \(t\), for the velocity of \(P\) at time \(t\) seconds. The particle passes through the point \(A\) with speed \(\frac { 10 } { 3 } \mathrm {~ms} ^ { - 1 }\).
2. Find the distance \(O A\).

1. A particle of mass 0.9 kg is attached to one end of a light elastic string, of natural length 1.2 m and modulus of elasticity 29.4 N . The other end of the string is attached to a fixed point \(A\) on a ceiling.

The particle is held at \(A\) and then released from rest. The particle first comes to instantaneous rest at the point \(B\). Find the distance \(A B\).
(5)

5. \begin{figure}[h] \end{figure} Figure 3 shows the finite region \(R\) which is bounded by part of the curve with equation \(y = \sin x\), the \(x\)-axis and the line with equation \(x = \frac { \pi } { 2 }\). A uniform solid \(S\) is formed by rotating \(R\) through \(2 \pi\) radians about the \(x\)-axis. Using algebraic integration,

1. show that the volume of \(S\) is \(\frac { \pi ^ { 2 } } { 4 }\)
2. find, in terms of \(\pi\), the \(x\) coordinate of the centre of mass of \(S\).

5. \begin{figure}[h] \end{figure} The region \(R\), shown shaded in Figure 3, is bounded by the circle with centre \(O\) and radius \(r\), the line with equation \(x = \frac { 3 } { 5 } r\) and the \(x\)-axis. The region is rotated through one complete revolution about the \(x\)-axis to form a uniform solid \(S\).

1. Use algebraic integration to show that the \(x\) coordinate of the centre of mass of \(S\) is \(\frac { 48 } { 65 } r\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{ae189c40-0071-4a6b-91eb-8ffebe082a04-16_394_643_1311_653} \captionsetup{labelformat=empty} \caption{Figure 4}
   \end{figure} A bowl is made from a uniform solid hemisphere of radius 6 cm by removing a hemisphere of radius 5 cm . Both hemispheres have the same centre \(A\) and the same axis of symmetry. The bowl is fixed with its open plane face uppermost and horizontal. Liquid is poured into the bowl. The depth of the liquid is 2 cm , as shown in Figure 4. The mass of the empty bowl is \(5 M \mathrm {~kg}\) and the mass of the liquid is \(2 M \mathrm {~kg}\).
2. Find, to 3 significant figures, the distance from \(A\) to the centre of mass of the bowl with its liquid.

1. \begin{figure}[h] \end{figure} The region \(R\), shown shaded in Figure 1, is bounded by the curve with equation \(y = \frac { 1 } { x }\), the line with equation \(x = 1\), the positive \(x\)-axis and the line with equation \(x = a\) where \(a > 1\) A uniform solid \(S\) is formed by rotating \(R\) through \(2 \pi\) radians about the \(x\)-axis.

1. Show that the volume of \(S\) is $$\pi \left( 1 - \frac { 1 } { a } \right)$$
2. Find the \(x\) coordinate of the centre of mass of \(S\).

2. A particle \(P\) of mass \(m\) is at a distance \(x\) above the surface of the Earth. The Earth exerts a gravitational force on \(P\). This force is directed towards the centre of the Earth. The magnitude of this force is inversely proportional to the square of the distance of \(P\) from the centre of the Earth. At the surface of the Earth the acceleration due to gravity is \(g\). The Earth is modelled as a fixed sphere of radius \(R\).

1. Show that the magnitude of the gravitational force on \(P\) is \(\frac { m g R ^ { 2 } } { ( x + R ) ^ { 2 } }\) A particle is released from rest from a point above the surface of the Earth. When the particle is at a distance \(R\) above the surface of the Earth, the particle has speed \(U\). Air resistance is modelled as being negligible.
2. Find, in terms of \(U , g\) and \(R\), the speed of the particle when it strikes the surface of the Earth.
   

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7 A skydiver drops from a helicopter. Before she opens her parachute, her speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) after time \(t\) seconds is modelled by the differential equation $$\frac { \mathrm { d } v } { \mathrm {~d} t } = 10 \mathrm { e } ^ { - \frac { 1 } { 2 } t }$$ When \(t = 0 , v = 0\).

1. Find \(v\) in terms of \(t\).
2. According to this model, what is the speed of the skydiver in the long term? She opens her parachute when her speed is \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Her speed \(t\) seconds after this is \(w \mathrm {~m} \mathrm {~s} ^ { - 1 }\), and is modelled by the differential equation $$\frac { \mathrm { d } w } { \mathrm {~d} t } = - \frac { 1 } { 2 } ( w - 4 ) ( w + 5 )$$
3. Express \(\frac { 1 } { ( w - 4 ) ( w + 5 ) }\) in partial fractions.
4. Using this result, show that \(\frac { w - 4 } { w + 5 } = 0.4 \mathrm { e } ^ { - 4.5 t }\).
5. According to this model, what is the speed of the skydiver in the long term?

4 A skydiver drops from a helicopter. Before she opens her parachute, her speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) after time \(t\) seconds is modelled by the differential equation $$\frac { \mathrm { d } v } { \mathrm {~d} t } = 10 \mathrm { e } ^ { - \frac { 1 } { 2 } t }$$ When \(t = 0 , v = 0\).

1. Find \(v\) in terms of \(t\).
2. According to this model, what is the speed of the skydiver in the long term? She opens her parachute when her speed is \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Her speed \(t\) seconds after this is \(w \mathrm {~m} \mathrm {~s} ^ { - 1 }\), and is modelled by the differential equation $$\frac { \mathrm { d } w } { \mathrm {~d} t } = - \frac { 1 } { 2 } ( w - 4 ) ( w + 5 )$$
3. Express \(\frac { 1 } { ( w - 4 ) ( w + 5 ) }\) in partial fractions.
4. Using this result, show that \(\frac { w - 4 } { w + 5 } = 0.4 \mathrm { e } ^ { - 4.5 t }\).
5. According to this model, what is the speed of the skydiver in the long term?

4.(a)Use the trapezium rule with 4 strips to find an approximate value for $$\int _ { 0 } ^ { 1 } 16 ^ { x } d x$$ (b)Use the trapezium rule with \(n\) strips to write down an expression that would give an approximate value for $$\int _ { 0 } ^ { 1 } 16 ^ { x } d x$$ (c)Hence show that $$\int _ { 0 } ^ { 1 } 16 ^ { x } \mathrm {~d} x = \lim _ { n \rightarrow \infty } \left( \frac { 1 } { n } \left( 1 + 16 ^ { \frac { 1 } { n } } + \ldots + 16 ^ { \frac { n - 1 } { n } } \right) \right)$$ (d)Use integration to determine the exact value of $$\int _ { 0 } ^ { 1 } 16 ^ { x } d x$$ Given that the limit exists,
(e)use part(c)and the answer to part(d)to determine the exact value of $$\lim _ { x \rightarrow 0 } \frac { 16 ^ { x } - 1 } { x }$$

10. A curve with equation \(y = \mathrm { f } ( x )\) passes through the point (4,25). Given that $$f ^ { \prime } ( x ) = \frac { 3 } { 8 } x ^ { 2 } - 10 x ^ { - \frac { 1 } { 2 } } + 1 , \quad x > 0$$

1. find \(\mathrm { f } ( x )\), simplifying each term.
2. Find an equation of the normal to the curve at the point ( 4,25 ). Give your answer in the form \(a x + b y + c = 0\), where \(a\), \(b\) and \(c\) are integers to be found.

2 The gradient of a curve is given by \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 6 x - 4\). The curve passes through the distinct points ( 2,5 ) and ( \(p , 5\) ).

1. Find the equation of the curve.
2. Find the value of \(p\).

3 A curve has an equation which satisfies \(\frac { \mathrm { d } y } { \mathrm {~d} x } = k x ( 2 x - 1 )\) for all values of \(x\). The point \(P ( 2,7 )\) lies on the curve and the gradient of the curve at \(P\) is 9 .

1. Find the value of the constant \(k\).
2. Find the equation of the curve.

6 The gradient of a curve is given by \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 3 x ^ { 2 } + a\), where \(a\) is a constant. The curve passes through the points \(( - 1,2 )\) and \(( 2,17 )\). Find the equation of the curve.

6 The gradient of a curve is given by \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 6 \sqrt { x } - 2\). Given also that the curve passes through the point \(( 9,4 )\), find the equation of the curve.

10 The gradient of a curve is given by \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 4 x + 3\). The curve passes through the point ( 2,9 ).

1. Find the equation of the tangent to the curve at the point \(( 2,9 )\).
2. Find the equation of the curve and the coordinates of its points of intersection with the \(x\)-axis. Find also the coordinates of the minimum point of this curve.
3. Find the equation of the curve after it has been stretched parallel to the \(x\)-axis with scale factor \(\frac { 1 } { 2 }\). Write down the coordinates of the minimum point of the transformed curve.

4 The acceleration of a particle \(P\) moving in a straight line is \(\left( t ^ { 2 } - 9 t + 18 \right) \mathrm { ms } ^ { - 2 }\), where \(t\) is the time in seconds.

1. Find the values of \(t\) for which the acceleration is zero.
2. It is given that when \(t = 3\) the velocity of \(P\) is \(9 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Find the velocity of \(P\) when \(t = 0\).
3. Show that the direction of motion of \(P\) changes before \(t = 1\).

7 A point P on a piece of machinery is moving in a vertical straight line. The displacement of P above ground level at time \(t\) seconds is \(y\) metres. The displacement-time graph for the motion during the time interval \(0 \leqslant t \leqslant 4\) is shown in Fig. 7 . \begin{figure}[h] \end{figure}

1. Using the graph, determine for the time interval \(0 \leqslant t \leqslant 4\) (A) the greatest displacement of P above its position when \(t = 0\),
   (B) the greatest distance of P from its position when \(t = 0\),
   (C) the time interval in which P is moving downwards,
   (D) the times when P is instantaneously at rest. The displacement of P in the time interval \(0 \leqslant t \leqslant 3\) is given by \(y = - 4 t ^ { 2 } + 8 t + 12\).
2. Use calculus to find expressions in terms of \(t\) for the velocity and for the acceleration of P in the interval \(0 \leqslant t \leqslant 3\).
3. At what times does P have a speed of \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in the interval \(0 \leqslant t \leqslant 3\) ? In the time interval \(3 \leqslant t \leqslant 4 , \mathrm { P }\) has a constant acceleration of \(32 \mathrm {~m} \mathrm {~s} ^ { - 2 }\). There is no sudden change in velocity when \(t = 3\).
4. Find an expression in terms of \(t\) for the displacement of P in the interval \(3 \leqslant t \leqslant 4\).

7 A ring is moving on a straight wire. Its velocity is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at time \(t\) seconds after passing a point Q . Model A for the motion of the ring gives the velocity-time graph for \(0 \leqslant t \leqslant 6\) shown in Fig. 7 . \begin{figure}[h] \end{figure} Use model A to calculate the following.

1. The acceleration of the ring when \(t = 0.5\).
2. The displacement of the ring from Q when
   (A) \(t = 2\),
   (B) \(t = 6\). In an alternative model B , the velocity of the ring is given by \(v = 2 t ^ { 2 } - 14 t + 20\) for \(0 \leqslant t \leqslant 6\).
3. Calculate the acceleration of the ring at \(t = 0.5\) as given by model B .
4. Calculate by how much the models differ in their values for the least \(v\) in the time interval \(0 \leqslant t \leqslant 6\).
5. Calculate the displacement of the ring from Q when \(t = 6\) as given by model B .

2 A particle is moving along a straight line and its position is relative to an origin on the line. At time \(t \mathrm {~s}\), the particle's acceleration, \(a \mathrm {~m} \mathrm {~s} ^ { - 2 }\), is given by $$a = 6 t - 12$$ At \(t = 0\) the velocity of the particle is \(+ 9 \mathrm {~ms} ^ { - 1 }\) and its position is - 2 m .

1. Find an expression for the velocity of the particle at time \(t \mathrm {~s}\) and verify that it is stationary when \(t = 3\).
2. Find the position of the particle when \(t = 2\).

6 A particle moves along a straight line through an origin O . Initially the particle is at O .
At time \(t \mathrm {~s}\), its displacement from O is \(x \mathrm {~m}\) and its velocity, \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), is given by $$v = 24 - 18 t + 3 t ^ { 2 } .$$

1. Find the times, \(T _ { 1 } \mathrm {~s}\) and \(T _ { 2 } \mathrm {~s}\) (where \(T _ { 2 } > T _ { 1 }\) ), at which the particle is stationary.
2. Find an expression for \(x\) at time \(t \mathrm {~s}\). Show that when \(t = T _ { 1 } , x = 20\) and find the value of \(x\) when \(t = T _ { 2 }\). Section B (36 marks) \(7 \quad\) Abi and Bob are standing on the ground and are trying to raise a small object of weight 250 N to the top of a building. They are using long light ropes. Fig. 7.1 shows the initial situation. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{83e69140-4abf-4713-85da-922ce7530e47-4_773_1071_429_497} \captionsetup{labelformat=empty} \caption{Fig. 7.1}
   \end{figure} Abi pulls vertically downwards on the rope A with a force \(F\) N. This rope passes over a small smooth pulley and is then connected to the object. Bob pulls on another rope, B, in order to keep the object away from the side of the building. In this situation, the object is stationary and in equilibrium. The tension in rope B, which is horizontal, is 25 N . The pulley is 30 m above the object. The part of rope A between the pulley and the object makes an angle \(\theta\) with the vertical.

3

1. Given that \(\mathrm { f } ( x ) = x ^ { 2 } + 2 x\), use differentiation from first principles to show that \(\mathrm { f } ^ { \prime } ( x ) = 2 x + 2\).
2. The gradient of a curve is given by \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 x + 2\) and the curve passes through the point \(( - 1,5 )\). Find the equation of the curve.