Find second derivative d²y/dx²

5 The curve \(C\) has parametric equations $$\mathrm { x } = \frac { 1 } { 2 } \mathrm { t } ^ { 2 } - \ln \mathrm { t } , \quad \mathrm { y } = 2 \mathrm { t } + 1 , \quad \text { for } \frac { 1 } { 2 } \leqslant t \leqslant 2$$

1. Find the exact length of \(C\).
2. Find \(\frac { \mathrm { d } ^ { 2 } \mathrm { y } } { \mathrm { dx } ^ { 2 } }\) in terms of \(t\), simplifying your answer.

3 It is given that $$\mathrm { x } = \sin ^ { - 1 } \mathrm { t } \quad \text { and } \quad \mathrm { y } = \mathrm { tcos } ^ { - 1 } \mathrm { t } , \quad \text { for } 0 \leqslant t < 1 .$$

1. Show that \(\frac { d y } { d x } = - t + \sqrt { 1 - t ^ { 2 } } \cos ^ { - 1 } t\).
2. Find \(\frac { d ^ { 2 } y } { d x ^ { 2 } }\) in terms of \(t\).

5 The curve \(C\) has parametric equations $$x = 3 t + 2 t ^ { - 1 } + a t ^ { 3 } , \quad y = 4 t - \frac { 3 } { 2 } t ^ { - 1 } + b t ^ { 3 } , \quad \text { for } 1 \leqslant t \leqslant 2$$ where \(a\) and \(b\) are constants.

1. It is given that \(a = \frac { 2 } { 3 }\) and \(b = - \frac { 1 } { 2 }\). Show that \(\left( \frac { d x } { d t } \right) ^ { 2 } + \left( \frac { d y } { d t } \right) ^ { 2 } = \frac { 25 } { 4 } \left( t ^ { 2 } + t ^ { - 2 } \right) ^ { 2 }\) and find the exact length of \(C\).
2. It is given instead that \(\mathrm { a } = \mathrm { b } = 0\). Find the value of \(\frac { d ^ { 2 } y } { d x ^ { 2 } }\) when \(t = 1\).
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2 It is given that $$x = 1 + \frac { 1 } { t } \quad \text { and } \quad y = t e ^ { t }$$

1. Show that \(\frac { d y } { d x } = - e ^ { t } \left( t ^ { 3 } + t ^ { 2 } \right)\).
2. Find \(\frac { \mathrm { d } ^ { 2 } \mathrm { y } } { \mathrm { dx } ^ { 2 } }\) in terms of \(t\).

2 It is given that $$x = 1 + \frac { 1 } { t } \quad \text { and } \quad y = \cos ^ { - 1 } t \quad \text { for } 0 < t < 1$$

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { t ^ { 2 } } { \sqrt { 1 - t ^ { 2 } } }\).
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2. Show that \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = - t ^ { a } \left( 1 - t ^ { 2 } \right) ^ { b } \left( 2 - t ^ { 2 } \right)\) ,where \(a\) and \(b\) are constants to be determined.

2 It is given that $$x = 1 + \frac { 1 } { t } \quad \text { and } \quad y = \cos ^ { - 1 } t \quad \text { for } 0 < t < 1$$

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { t ^ { 2 } } { \sqrt { 1 - t ^ { 2 } } }\).
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2. Show that \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = - t ^ { a } \left( 1 - t ^ { 2 } \right) ^ { b } \left( 2 - t ^ { 2 } \right)\), where \(a\) and \(b\) are constants to be determined.

6 The curve \(C\) is defined parametrically by $$x = 4 t - t ^ { 2 } \quad \text { and } \quad y = 1 - \mathrm { e } ^ { - t }$$ where \(0 \leqslant t < 2\). Show that at all points of \(C\), $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \frac { ( t - 1 ) \mathrm { e } ^ { - t } } { 4 ( 2 - t ) ^ { 3 } }$$ Show that the mean value of \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) with respect to \(x\) over the interval \(0 \leqslant x \leqslant \frac { 7 } { 4 }\) is $$\frac { 4 e ^ { - \frac { 1 } { 2 } } - 3 } { 21 }$$

7 It is given that $$x = t ^ { 2 } \mathrm { e } ^ { - t ^ { 2 } } \quad \text { and } \quad y = t \mathrm { e } ^ { - t ^ { 2 } }$$

1. Show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 - 2 t ^ { 2 } } { 2 t - 2 t ^ { 3 } }$$
2. Find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) in terms of \(t\).

The curve \(C\) has parametric equations $$x = t ^ { 2 } , \quad y = ( 2 - t ) ^ { \frac { 1 } { 2 } } , \quad \text { for } 0 \leqslant t \leqslant 2 .$$ Find

1. \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) in terms of \(t\),
2. the mean value of \(y\) with respect to \(x\) over the interval \(0 \leqslant x \leqslant 4\),
3. the \(y\)-coordinate of the centroid of the region enclosed by \(C\), the \(x\)-axis and the \(y\)-axis.

7 The curve \(C\) has parametric equations $$x = \sin t , \quad y = \sin 2 t , \quad \text { for } 0 \leqslant t \leqslant \pi .$$ Find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) in terms of \(t\). Hence, or otherwise, find the coordinates of the stationary points on \(C\) and determine their nature.

4 It is given that $$x = t + \sin t , \quad y = t ^ { 2 } + 2 \cos t$$ where \(- \pi < t < \pi\). Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\). Show that $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \frac { 2 t \sin t } { ( 1 + \cos t ) ^ { 3 } }$$ Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) increases with \(x\) over the given interval of \(t\).

4 A curve has parametric equations $$x = 2 \sin 2 t , \quad y = 3 \cos 2 t$$ for \(0 < t < \frac { 1 } { 2 } \pi\). For the point on the curve where \(t = \frac { 1 } { 3 } \pi\), find the value of

1. \(\frac { \mathrm { d } y } { \mathrm {~d} x }\),
2. \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\).

4 A curve has parametric equations $$x = 2 \theta - \sin 2 \theta , \quad y = 1 - \cos 2 \theta , \quad \text { for } - 3 \pi \leqslant \theta \leqslant 3 \pi$$ Show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \cot \theta$$ except for certain values of \(\theta\), which should be stated. Find the value of \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) when \(\theta = \frac { 1 } { 4 } \pi\).

4 \end{array} \right) \quad \text { and } \quad \mathbf { d } = \left( \begin{array} { r } 0
- 8
3 \end{array} \right) .$$

1. Show that \(\{ \mathbf { a } , \mathbf { b } , \mathbf { c } \}\) is a basis for \(\mathbb { R } ^ { 3 }\).
2. Express \(\mathbf { d }\) in terms of \(\mathbf { a } , \mathbf { b }\) and \(\mathbf { c }\).\\ 2 The roots of the equation $$x ^ { 3 } + p x ^ { 2 } + q x + r = 0$$ are \(\alpha , 2 \alpha , 4 \alpha\), where \(p , q , r\) and \(\alpha\) are non-zero real constants.
3. Show that $$2 p \alpha + q = 0$$
4. Show that $$p ^ { 3 } r - q ^ { 3 } = 0$$ 3 The sequence of positive numbers \(u _ { 1 } , u _ { 2 } , u _ { 3 } , \ldots\) is such that \(u _ { 1 } < 3\) and, for \(n \geqslant 1\), $$u _ { n + 1 } = \frac { 4 u _ { n } + 9 } { u _ { n } + 4 }$$
5. By considering \(3 - u _ { n + 1 }\), or otherwise, prove by mathematical induction that \(u _ { n } < 3\) for all positive integers \(n\).
6. Show that \(u _ { n + 1 } > u _ { n }\) for \(n \geqslant 1\).\\ 4 A curve is defined parametrically by $$x = t - \frac { 1 } { 2 } \sin 2 t \quad \text { and } \quad y = \sin ^ { 2 } t$$ The arc of the curve joining the point where \(t = 0\) to the point where \(t = \pi\) is rotated through one complete revolution about the \(x\)-axis. The area of the surface generated is denoted by \(S\).
7. Show that $$S = a \pi \int _ { 0 } ^ { \pi } \sin ^ { 3 } t \mathrm {~d} t$$ where the constant \(a\) is to be found.
8. Using the result \(\sin 3 t = 3 \sin t - 4 \sin ^ { 3 } t\), find the exact value of \(S\).

The curve \(C\) is defined parametrically by $$x = 18 t - t ^ { 2 } \quad \text { and } \quad y = 8 t ^ { \frac { 3 } { 2 } }$$ where \(0 < t \leqslant 4\).

1. Show that at all points of \(C\), $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \frac { 3 ( 9 + t ) } { 2 t ^ { \frac { 1 } { 2 } } ( 9 - t ) ^ { 3 } }$$
2. Show that the mean value of \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) with respect to \(x\) over the interval \(0 < x \leqslant 56\) is \(\frac { 3 } { 70 }\).
3. Find the area of the surface generated when \(C\) is rotated through \(2 \pi\) radians about the \(x\)-axis, showing full working.

1 The curve \(C\) is defined parametrically by $$x = 2 \cos ^ { 3 } t \quad \text { and } \quad y = 2 \sin ^ { 3 } t , \quad \text { for } 0 < t < \frac { 1 } { 2 } \pi$$ Show that, at the point with parameter \(t\), $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \frac { 1 } { 6 } \sec ^ { 4 } t \operatorname { cosec } t$$

1 The curve \(C\) is defined parametrically by $$x = 2 \cos ^ { 3 } t \quad \text { and } \quad y = 2 \sin ^ { 3 } t , \quad \text { for } 0 < t < \frac { 1 } { 2 } \pi .$$ Show that, at the point with parameter \(t\), $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \frac { 1 } { 6 } \sec ^ { 4 } t \operatorname { cosec } t$$