1.07d Second derivatives: d^2y/dx^2 notation

7 The volume, \(V \mathrm {~m} ^ { 3 }\), of water in a tank at time \(t\) seconds is given by $$V = \frac { 1 } { 3 } t ^ { 6 } - 2 t ^ { 4 } + 3 t ^ { 2 } , \quad \text { for } t \geqslant 0$$

1. Find:
   1. \(\frac { \mathrm { d } V } { \mathrm {~d} t }\);
      (3 marks)
   2. \(\frac { \mathrm { d } ^ { 2 } V } { \mathrm {~d} t ^ { 2 } }\).
      (2 marks)
2. Find the rate of change of the volume of water in the tank, in \(\mathrm { m } ^ { 3 } \mathrm {~s} ^ { - 1 }\), when \(t = 2\).
3. 1. Verify that \(V\) has a stationary value when \(t = 1\).
   2. Determine whether this is a maximum or minimum value.

6 The diagram below shows a rectangular sheet of metal 24 cm by 9 cm . \includegraphics[max width=\textwidth, alt={}, center]{10bca9b4-5327-4b35-8b75-612b396e8a76-4_512_897_386_561} A square of side \(x \mathrm {~cm}\) is cut from each corner and the metal is then folded along the broken lines to make an open box with a rectangular base and height \(x \mathrm {~cm}\).

1. Show that the volume, \(V \mathrm {~cm} ^ { 3 }\), of liquid the box can hold is given by $$V = 4 x ^ { 3 } - 66 x ^ { 2 } + 216 x$$
2. 1. Find \(\frac { \mathrm { d } V } { \mathrm {~d} x }\).
   2. Show that any stationary values of \(V\) must occur when \(x ^ { 2 } - 11 x + 18 = 0\).
   3. Solve the equation \(x ^ { 2 } - 11 x + 18 = 0\).
   4. Explain why there is only one value of \(x\) for which \(V\) is stationary.
3. 1. Find \(\frac { \mathrm { d } ^ { 2 } V } { \mathrm {~d} x ^ { 2 } }\).
   2. Hence determine whether the stationary value is a maximum or minimum.

7 The volume, \(V \mathrm {~m} ^ { 3 }\), of water in a tank at time \(t\) seconds is given by $$V = \frac { 1 } { 3 } t ^ { 6 } - 2 t ^ { 4 } + 3 t ^ { 2 } , \quad \text { for } t \geqslant 0$$

1. Find:
   1. \(\frac { \mathrm { d } V } { \mathrm {~d} t }\);
      (3 marks)
   2. \(\frac { \mathrm { d } ^ { 2 } V } { \mathrm {~d} t ^ { 2 } }\).
      (2 marks)
2. Find the rate of change of the volume of water in the tank, in \(\mathrm { m } ^ { 3 } \mathrm {~s} ^ { - 1 }\), when \(t = 2\).
   (2 marks)
3. 1. Verify that \(V\) has a stationary value when \(t = 1\).
      (2 marks)
   2. Determine whether this is a maximum or minimum value.
      (2 marks)

5 A model car moves so that its distance, \(x\) centimetres, from a fixed point \(O\) after time \(t\) seconds is given by $$x = \frac { 1 } { 2 } t ^ { 4 } - 20 t ^ { 2 } + 66 t , \quad 0 \leqslant t \leqslant 4$$

1. Find:
   1. \(\frac { \mathrm { d } x } { \mathrm {~d} t }\);
   2. \(\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } }\).
2. Verify that \(x\) has a stationary value when \(t = 3\), and determine whether this stationary value is a maximum value or a minimum value.
3. Find the rate of change of \(x\) with respect to \(t\) when \(t = 1\).
4. Determine whether the distance of the car from \(O\) is increasing or decreasing at the instant when \(t = 2\).

3 The depth of water, \(y\) metres, in a tank after time \(t\) hours is given by $$y = \frac { 1 } { 8 } t ^ { 4 } - 2 t ^ { 2 } + 4 t , \quad 0 \leqslant t \leqslant 4$$

1. Find:
   1. \(\frac { \mathrm { d } y } { \mathrm {~d} t }\);
   2. \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } }\).
2. Verify that \(y\) has a stationary value when \(t = 2\) and determine whether it is a maximum value or a minimum value.
3. 1. Find the rate of change of the depth of water, in metres per hour, when \(t = 1\).
   2. Hence determine, with a reason, whether the depth of water is increasing or decreasing when \(t = 1\).

1 The curve with equation \(y = 13 + 18 x + 3 x ^ { 2 } - 4 x ^ { 3 }\) passes through the point \(P\) where \(x = - 1\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. Show that the point \(P\) is a stationary point of the curve and find the other value of \(x\) where the curve has a stationary point.
3. 1. Find the value of \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) at the point \(P\).
   2. Hence, or otherwise, determine whether \(P\) is a maximum point or a minimum point.
      (l mark)

4 The curve with equation \(y = x ^ { 5 } - 3 x ^ { 2 } + x + 5\) is sketched below. The point \(O\) is at the origin and the curve passes through the points \(A ( - 1,0 )\) and \(B ( 1,4 )\). \includegraphics[max width=\textwidth, alt={}, center]{91170a77-e266-4c81-89ee-1fc29a538485-3_447_752_438_653}

1. Given that \(y = x ^ { 5 } - 3 x ^ { 2 } + x + 5\), find:
   1. \(\frac { \mathrm { d } y } { \mathrm {~d} x }\);
   2. \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\).
2. Find an equation of the tangent to the curve at the point \(A ( - 1,0 )\).
3. Verify that the point \(B\), where \(x = 1\), is a minimum point of the curve.
4. The curve with equation \(y = x ^ { 5 } - 3 x ^ { 2 } + x + 5\) is sketched below. The point \(O\) is at the origin and the curve passes through the points \(A ( - 1,0 )\) and \(B ( 1,4 )\). \includegraphics[max width=\textwidth, alt={}, center]{91170a77-e266-4c81-89ee-1fc29a538485-3_451_757_1736_648}
   1. Find \(\int _ { - 1 } ^ { 1 } \left( x ^ { 5 } - 3 x ^ { 2 } + x + 5 \right) \mathrm { d } x\).
   2. Hence find the area of the shaded region bounded by the curve between \(A\) and \(B\) and the line segments \(A O\) and \(O B\).

2 A bird flies from a tree. At time \(t\) seconds, the bird's height, \(y\) metres, above the horizontal ground is given by $$y = \frac { 1 } { 8 } t ^ { 4 } - t ^ { 2 } + 5 , \quad 0 \leqslant t \leqslant 4$$

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} t }\).
2. 1. Find the rate of change of height of the bird in metres per second when \(t = 1\).
   2. Determine, with a reason, whether the bird's height above the horizontal ground is increasing or decreasing when \(t = 1\).
3. 1. Find the value of \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } }\) when \(t = 2\).
   2. Given that \(y\) has a stationary value when \(t = 2\), state whether this is a maximum value or a minimum value.

4 The curve with equation \(y = x ^ { 3 } - 5 x ^ { 2 } + 7 x - 3\) is sketched below. \includegraphics[max width=\textwidth, alt={}, center]{3729de55-7139-4f41-8584-640f173c0e09-3_444_588_411_717} The curve touches the \(x\)-axis at the point \(A ( 1,0 )\) and cuts the \(x\)-axis at the point \(B\).

1. 1. Use the factor theorem to show that \(x - 3\) is a factor of $$\mathrm { p } ( x ) = x ^ { 3 } - 5 x ^ { 2 } + 7 x - 3$$
   2. Hence find the coordinates of \(B\).
2. The point \(M\), shown on the diagram, is a minimum point of the curve with equation \(y = x ^ { 3 } - 5 x ^ { 2 } + 7 x - 3\).
   1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
   2. Hence determine the \(x\)-coordinate of \(M\).
3. Find the value of \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) when \(x = 1\).
4. 1. Find \(\int \left( x ^ { 3 } - 5 x ^ { 2 } + 7 x - 3 \right) \mathrm { d } x\).
   2. Hence determine the area of the shaded region bounded by the curve and the coordinate axes.

3 A curve has equation \(y = 2 x ^ { 5 } + 5 x ^ { 4 } - 1\).

1. Find:
   1. \(\frac { \mathrm { d } y } { \mathrm {~d} x }\)
   2. \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\)
2. The point on the curve where \(x = - 1\) is \(P\).
   1. Determine whether \(y\) is increasing or decreasing at \(P\), giving a reason for your answer.
   2. Find an equation of the tangent to the curve at \(P\).
3. The point \(Q ( - 2,15 )\) also lies on the curve. Verify that \(Q\) is a maximum point of the curve.
   [0pt] [4 marks]

8 The gradient, \(\frac { \mathrm { d } y } { \mathrm {~d} x }\), at the point \(( x , y )\) on a curve is given by $$\frac { \mathrm { d } y } { \mathrm {~d} x } = 54 + 27 x - 6 x ^ { 2 }$$

1. 1. Find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\).
   2. The curve passes through the point \(P \left( - 1 \frac { 1 } { 2 } , 4 \right)\). Verify that the curve has a minimum point at \(P\).
2. 1. Show that at the points on the curve where \(y\) is decreasing $$2 x ^ { 2 } - 9 x - 18 > 0$$
   2. Solve the inequality \(2 x ^ { 2 } - 9 x - 18 > 0\).

1. Given that

$$y = 2 x ^ { \frac { 3 } { 2 } } - 1$$

1. find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\),
2. show that $$4 x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 3 y = k$$ where \(k\) is an integer to be found,
3. find $$\int y ^ { 2 } \mathrm {~d} x$$

6 A curve \(C\) has the equation $$y = \frac { x ^ { 3 } + \sqrt { x } } { x } , \quad x > 0$$

1. Express \(\frac { x ^ { 3 } + \sqrt { x } } { x }\) in the form \(x ^ { p } + x ^ { q }\).
2. 1. Hence find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
   2. Find an equation of the normal to the curve \(C\) at the point on the curve where \(x = 1\).
3. 1. Find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\).
   2. Hence deduce that the curve \(C\) has no maximum points. \includegraphics[max width=\textwidth, alt={}, center]{f9a7a4dd-f7fd-4135-8872-2c1270d46a14-7_1463_1707_1244_153}

3 A particle is oscillating in a vertical line. At time \(t\) seconds, its displacement above the centre of the oscillations is \(x\) metres, where \(x = A \sin \omega t + B \cos \omega t\) (and \(A , B\) and \(\omega\) are constants).

1. Show that \(\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } = - \omega ^ { 2 } x\). When \(t = 0\), the particle is 2 m above the centre of the oscillations, the velocity is \(1.44 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) downwards, and the acceleration is \(0.18 \mathrm {~m} \mathrm {~s} ^ { - 2 }\) downwards.
2. Find \(A , B\) and \(\omega\).
3. Show that the period of oscillation is 20.9 s (correct to 3 significant figures), and find the amplitude.
4. Find the total distance travelled by the particle between \(t = 12\) and \(t = 24\).

2 The curve with equation \(y = x ^ { 4 } - 32 x + 5\) has a single stationary point, \(M\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. Hence find the \(x\)-coordinate of \(M\).
3. 1. Find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\).
   2. Hence, or otherwise, determine whether \(M\) is a maximum or a minimum point.
4. Determine whether the curve is increasing or decreasing at the point on the curve where \(x = 0\).

4 A model helicopter takes off from a point \(O\) at time \(t = 0\) and moves vertically so that its height, \(y \mathrm {~cm}\), above \(O\) after time \(t\) seconds is given by $$y = \frac { 1 } { 4 } t ^ { 4 } - 26 t ^ { 2 } + 96 t , \quad 0 \leqslant t \leqslant 4$$

1. Find:
   1. \(\frac { \mathrm { d } y } { \mathrm {~d} t }\);
      (3 marks)
   2. \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } }\).
      (2 marks)
2. Verify that \(y\) has a stationary value when \(t = 2\) and determine whether this stationary value is a maximum value or a minimum value.
   (4 marks)
3. Find the rate of change of \(y\) with respect to \(t\) when \(t = 1\).
4. Determine whether the height of the helicopter above \(O\) is increasing or decreasing at the instant when \(t = 3\).

3 Two numbers, \(x\) and \(y\), are such that \(3 x + y = 9\), where \(x \geqslant 0\) and \(y \geqslant 0\). It is given that \(V = x y ^ { 2 }\).

1. Show that \(V = 81 x - 54 x ^ { 2 } + 9 x ^ { 3 }\).
2. 1. Show that \(\frac { \mathrm { d } V } { \mathrm {~d} x } = k \left( x ^ { 2 } - 4 x + 3 \right)\), and state the value of the integer \(k\).
   2. Hence find the two values of \(x\) for which \(\frac { \mathrm { d } V } { \mathrm {~d} x } = 0\).
3. Find \(\frac { \mathrm { d } ^ { 2 } V } { \mathrm {~d} x ^ { 2 } }\).
4. 1. Find the value of \(\frac { \mathrm { d } ^ { 2 } V } { \mathrm {~d} x ^ { 2 } }\) for each of the two values of \(x\) found in part (b)(ii).
   2. Hence determine the value of \(x\) for which \(V\) has a maximum value.
   3. Find the maximum value of \(V\).

3 The curve with equation \(y = x ^ { 5 } + 20 x ^ { 2 } - 8\) passes through the point \(P\), where \(x = - 2\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. Verify that the point \(P\) is a stationary point of the curve.
3. 1. Find the value of \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) at the point \(P\).
   2. Hence, or otherwise, determine whether \(P\) is a maximum point or a minimum point.
4. Find an equation of the tangent to the curve at the point where \(x = 1\).

1 A particle moves under the action of a force, \(\mathbf { F }\) newtons. At time \(t\) seconds, the velocity, \(\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }\), of the particle is given by $$\mathbf { v } = \left( t ^ { 3 } - 15 t - 5 \right) \mathbf { i } + \left( 6 t - t ^ { 2 } \right) \mathbf { j }$$

1. Find an expression for the acceleration of the particle at time \(t\).
2. The mass of the particle is 4 kg .
   1. Show that, at time \(t\), $$\mathbf { F } = \left( 12 t ^ { 2 } - 60 \right) \mathbf { i } + ( 24 - 8 t ) \mathbf { j }$$
   2. Find the magnitude of \(\mathbf { F }\) when \(t = 2\).

3 A curve has equation \(y = k \sqrt { x }\) where \(k\) is a constant. Find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) at the point \(( 4,2 k )\) on the curve, giving your answer as an expression in terms of \(k\).

A curve is such that \(\frac{\mathrm{d}y}{\mathrm{d}x} = \sqrt{(4x + 1)}\) and \((2, 5)\) is a point on the curve.

1. Find the equation of the curve. [4]
2. A point \(P\) moves along the curve in such a way that the \(y\)-coordinate is increasing at a constant rate of 0.06 units per second. Find the rate of change of the \(x\)-coordinate when \(P\) passes through \((2, 5)\). [2]
3. Show that \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \times \frac{\mathrm{d}y}{\mathrm{d}x}\) is constant. [2]

A particle moves in a straight line. The displacement of the particle at time \(t\) s is \(s\) m, where $$s = t^3 - 6t^2 + 4t.$$ Find the velocity of the particle at the instant when its acceleration is zero. [4]

The curve \(C\) has equation \(y = 2x^3 - 5x^2 - 4x + 2\).

1. Find \(\frac{dy}{dx}\). [2]
2. Using the result from part (a), find the coordinates of the turning points of \(C\). [4]
3. Find \(\frac{d^2y}{dx^2}\). [2]
4. Hence, or otherwise, determine the nature of the turning points of \(C\). [2]

\includegraphics{figure_4} Figure 4 shows a solid brick in the shape of a cuboid measuring 2x cm by x cm by y cm. The total surface area of the brick is 600 cm².

1. Show that the volume, \(V\) cm³, of the brick is given by \(V = 200x - \frac{4x^3}{3}\). [4]

Given that \(x\) can vary,

1. use calculus to find the maximum value of \(V\), giving your answer to the nearest cm³. [5]
2. Justify that the value of \(V\) you have found is a maximum. [2]

Given that \(y = \arcsin x\) prove that

1. \(\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}\) [3]
2. \((1-x^2) \frac{d^2 y}{dx^2} - x \frac{dy}{dx} = 0\) [4]

(Total 7 marks)