1.07s Parametric and implicit differentiation

4 It is given that $$x = - t + \tan ^ { - 1 } t \quad \text { and } \quad y = t + \sinh ^ { - 1 } t$$

1. Show that \(\frac { d y } { d x } = - \frac { t ^ { 2 } + 1 + \sqrt { t ^ { 2 } + 1 } } { t ^ { 2 } }\).
2. Find the value of \(\frac { \mathrm { d } ^ { 2 } \mathrm { y } } { \mathrm { dx } ^ { 2 } }\) when \(t = \frac { 3 } { 4 }\).

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1. Starting from the definitions of sech and tanh in terms of exponentials, prove that $$1 - \operatorname { sech } ^ { 2 } t = \tanh ^ { 2 } t$$ \includegraphics[max width=\textwidth, alt={}]{d3ddf5ce-4399-4438-ab67-7bdb2e1bea6e-14_77_1547_360_347} ......................................................................................................................................... ......................................................................................................................................... . ........................................................................................................................................ ........................................................................................................................................ ....................................................................................................................................... \includegraphics[max width=\textwidth, alt={}, center]{d3ddf5ce-4399-4438-ab67-7bdb2e1bea6e-14_72_1573_911_324} \includegraphics[max width=\textwidth, alt={}, center]{d3ddf5ce-4399-4438-ab67-7bdb2e1bea6e-14_67_1570_1005_324} The curve \(C\) has parametric equations $$\mathrm { x } = \frac { 1 } { 2 } \tanh ^ { 2 } \mathrm { t } + \text { Insecht } , \quad \mathrm { y } = 1 + \tanh ^ { 4 } \mathrm { t } , \quad \text { for } t > 0$$
2. Show that \(\frac { d y } { d x } = - 4 \operatorname { sech } ^ { 2 } t\).
3. Find the coordinates of the point on \(C\) with \(\frac { d ^ { 2 } y } { d x ^ { 2 } } = - \frac { 9 } { 2 }\), giving your answer in the form \(( a + \ln b , c )\) where \(a , b\) and \(c\) are rational numbers.
   If you use the following page to complete the answer to any question, the question number must be clearly shown.

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1. Starting from the definitions of sech and tanh in terms of exponentials, prove that $$1 - \operatorname { sech } ^ { 2 } t = \tanh ^ { 2 } t$$ \includegraphics[max width=\textwidth, alt={}]{482b2236-1f1b-4c53-a1bc-0277cf63dc62-14_77_1547_360_347} ......................................................................................................................................... ......................................................................................................................................... . ........................................................................................................................................ ........................................................................................................................................ ....................................................................................................................................... \includegraphics[max width=\textwidth, alt={}, center]{482b2236-1f1b-4c53-a1bc-0277cf63dc62-14_72_1573_911_324} \includegraphics[max width=\textwidth, alt={}, center]{482b2236-1f1b-4c53-a1bc-0277cf63dc62-14_67_1573_1005_324} The curve \(C\) has parametric equations $$\mathrm { x } = \frac { 1 } { 2 } \tanh ^ { 2 } \mathrm { t } + \text { Insecht } , \quad \mathrm { y } = 1 + \tanh ^ { 4 } \mathrm { t } , \quad \text { for } t > 0$$
2. Show that \(\frac { d y } { d x } = - 4 \operatorname { sech } ^ { 2 } t\).
3. Find the coordinates of the point on \(C\) with \(\frac { d ^ { 2 } y } { d x ^ { 2 } } = - \frac { 9 } { 2 }\), giving your answer in the form \(( a + \ln b , c )\) where \(a , b\) and \(c\) are rational numbers.
   If you use the following page to complete the answer to any question, the question number must be clearly shown.

4 The curve \(C\) has equation $$4 y ^ { 3 } + ( x + y ) ^ { 6 } = 109 .$$

1. Show that, at the point \(( - 4,3 )\) on \(C , \frac { \mathrm { dy } } { \mathrm { dx } } = \frac { 1 } { 17 }\).
2. Find the value of \(\frac { d ^ { 2 } y } { d x ^ { 2 } }\) at the point \(( - 4,3 )\).

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 \(t\).

3 The curve \(C\) has equation $$x ^ { 3 } + 2 x y + 8 y ^ { 3 } = - 12$$

1. Show that, at the point \(( - 2 , - 1 )\) on \(C , \frac { \mathrm {~d} y } { \mathrm {~d} x } = - \frac { 1 } { 2 }\). \includegraphics[max width=\textwidth, alt={}, center]{27485e4a-cd34-43e3-aa92-767820a9f6f9-06_2714_37_143_2008}
2. Find the value of \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) at the point \(( - 2 , - 1 )\).

5 The curve \(C\) has equation $$y ^ { 2 } + ( x y + 1 ) ^ { 2 } = 5$$

1. Show that, at the point \(( 1,1 )\) on \(C , \frac { \mathrm { dy } } { \mathrm { dx } } = - \frac { 2 } { 3 }\).
2. Find the value of \(\frac { d ^ { 2 } y } { d x ^ { 2 } }\) at the point \(( 1,1 )\).

3 The curve \(C\) has equation $$x y ^ { 3 } - 4 x ^ { 3 } y = 3$$

1. Show that, at the point \(( - 1,1 )\) on \(C , \frac { \mathrm { dy } } { \mathrm { dx } } = 11\).
2. Find the value of \(\frac { d ^ { 2 } y } { d x ^ { 2 } }\) at the point \(( - 1,1 )\). \includegraphics[max width=\textwidth, alt={}, center]{37db1c60-0f94-413f-b29b-5872975eee9e-06_535_1584_276_276} The diagram shows the curve with equation \(\mathrm { y } = \frac { \ln \mathrm { x } } { \mathrm { x } ^ { 2 } }\) for \(x \geqslant 2\), together with a set of \(( N - 2 )\) rectangles
   of unit width.

3 The curve \(C\) has equation $$x y ^ { 3 } - 4 x ^ { 3 } y = 3$$

1. Show that, at the point \(( - 1,1 )\) on \(C , \frac { \mathrm { dy } } { \mathrm { dx } } = 11\).
2. Find the value of \(\frac { d ^ { 2 } y } { d x ^ { 2 } }\) at the point \(( - 1,1 )\). \includegraphics[max width=\textwidth, alt={}, center]{59982339-c496-4bd7-8dcd-9b257f3afc02-06_535_1584_276_276} The diagram shows the curve with equation \(\mathrm { y } = \frac { \ln \mathrm { x } } { \mathrm { x } ^ { 2 } }\) for \(x \geqslant 2\), together with a set of \(( N - 2 )\) rectangles
   of unit width.

2 A curve has equation $$( x + 1 ) y + y ^ { 2 } = 2$$

1. Show that \(\frac { \mathrm { dy } } { \mathrm { dx } } = - \frac { 2 } { 3 }\) at the point \(( 0 , - 2 )\).
2. Find the value of \(\frac { d ^ { 2 } y } { d x ^ { 2 } }\) at the point \(( 0 , - 2 )\).

5 The curve \(C\) has parametric equations $$\mathrm { x } = \frac { 2 } { 3 } \mathrm { t } ^ { \frac { 3 } { 2 } } - 2 \mathrm { t } ^ { \frac { 1 } { 2 } } , \quad \mathrm { y } = 2 \mathrm { t } + 5 , \quad \text { for } 0 < t \leqslant 3$$

1. Find the exact length of \(C\).
2. Find the set of values of \(t\) for which \(\frac { d ^ { 2 } y } { d x ^ { 2 } } > 0\).

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 \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { t ^ { 2 } } { \sqrt { 1 - t ^ { 2 } } }\). \includegraphics[max width=\textwidth, alt={}, center]{374b91df-926d-4f7f-a1d3-a54c70e8ff0e-05_2723_33_99_22}
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 \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { t ^ { 2 } } { \sqrt { 1 - t ^ { 2 } } }\). \includegraphics[max width=\textwidth, alt={}, center]{4af32247-c1f9-4c1f-bdf8-bafe17aca1dc-05_2723_33_99_22}
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.

10. The curve \(C\) has equation $$x = 3 \sec ^ { 2 } 2 y \quad x > 3 \quad 0 < y < \frac { \pi } { 4 }$$

1. Find \(\frac { \mathrm { d } x } { \mathrm {~d} y }\) in terms of \(y\).
2. Hence show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { p } { q x \sqrt { x - 3 } }$$ where \(p\) is irrational and \(q\) is an integer, stating the values of \(p\) and \(q\).
3. Find the equation of the normal to \(C\) at the point where \(y = \frac { \pi } { 12 }\), giving your answer in the form \(y = m x + c\), giving \(m\) and \(c\) as exact irrational numbers.
   

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10. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of the curve \(C\) with equation $$x = y \mathrm { e } ^ { 2 y } \quad y \in \mathbb { R }$$

1. Show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y } { x ( 1 + 2 y ) }$$ Given that the straight line with equation \(x = k\), where \(k\) is a constant, cuts \(C\) at exactly two points,
2. find the range of possible values for \(k\).

1. The curve \(C\) has equation

$$x = 3 \tan \left( y - \frac { \pi } { 6 } \right) \quad x \in \mathbb { R } \quad - \frac { \pi } { 3 } < y < \frac { 2 \pi } { 3 }$$

1. Show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { a } { x ^ { 2 } + b }$$ where \(a\) and \(b\) are integers to be found. The point \(P\) with \(y\) coordinate \(\frac { \pi } { 3 }\) lies on \(C\).
   Given that the tangent to \(C\) at \(P\) crosses the \(x\)-axis at the point \(Q\).
2. find, in simplest form, the exact \(x\) coordinate of \(Q\).

7. Given that $$x = 6 \sin ^ { 2 } 2 y \quad 0 < y < \frac { \pi } { 4 }$$ show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { A \sqrt { \left( B x - x ^ { 2 } \right) } }$$ where \(A\) and \(B\) are integers to be found.
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9. \begin{figure}[h] \end{figure} The curve shown in Figure 5 has equation $$x = 4 \sin ^ { 2 } y - 1 \quad 0 \leqslant y \leqslant \frac { \pi } { 2 }$$ The point \(P \left( k , \frac { \pi } { 3 } \right)\) lies on the curve.

1. Verify that \(k = 2\)
2. 1. Find \(\frac { \mathrm { d } x } { \mathrm {~d} y }\) in terms of \(y\)
   2. Hence show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 2 \sqrt { x + 1 } \sqrt { 3 - x } }\) The normal to the curve at \(P\) cuts the \(x\)-axis at the point \(N\).
3. Find the exact area of triangle \(O P N\), where \(O\) is the origin. Give your answer in the form \(a \pi + b \pi ^ { 2 }\) where \(a\) and \(b\) are constants.

1. 1. The curve \(C\) has equation \(y = \mathrm { g } ( x )\) where

$$g ( x ) = e ^ { 3 x } \sec 2 x \quad - \frac { \pi } { 4 } < x < \frac { \pi } { 4 }$$

1. Find \(\mathrm { g } ^ { \prime } ( x )\)
2. Hence find the \(x\) coordinate of the stationary point of \(C\).
   (ii) A different curve has equation $$x = \ln ( \sin y ) \quad 0 < y < \frac { \pi } { 2 }$$ Show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { \mathrm { e } ^ { x } } { \mathrm { f } ( x ) }$$ where \(\mathrm { f } ( x )\) is a function of \(\mathrm { e } ^ { x }\) that should be found.
   

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8. A curve \(C\) has equation \(y = \mathrm { f } ( x )\), where $$f ( x ) = \arcsin \left( \frac { 1 } { 2 } x \right) \quad - 2 \leqslant x \leqslant 2 \quad - \frac { \pi } { 2 } \leqslant y \leqslant \frac { \pi } { 2 }$$

1. Sketch \(C\).
2. Given \(x = 2 \sin y\), show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { \sqrt { A - x ^ { 2 } } }$$ where \(A\) is a constant to be found. The point \(P\) lies on \(C\) and has \(y\) coordinate \(\frac { \pi } { 4 }\)
3. Find the equation of the tangent to \(C\) at \(P\). Write your answer in the form \(y = m x + c\), where \(m\) and \(c\) are constants to be found.
   (3)

1. In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.

A curve \(C\) has equation $$x = \sin ^ { 2 } 4 y \quad 0 \leqslant y \leqslant \frac { \pi } { 8 } \quad 0 \leqslant x \leqslant 1$$ The point \(P\) with \(x\) coordinate \(\frac { 1 } { 4 }\) lies on \(C\)

1. Find the exact \(y\) coordinate of \(P\)
2. Find \(\frac { \mathrm { d } x } { \mathrm {~d} y }\)
3. Hence show that \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) can be written in the form $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { \sqrt { q + r ( x + s ) ^ { 2 } } }$$ where \(q , r\) and \(s\) are constants to be found. Using the answer to part (c),
4. 1. state the \(x\) coordinate of the point where the value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) is a minimum,
   2. state the value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) at this point.

1. The point \(P\) lies on the curve with equation

$$x = ( 4 y - \sin 2 y ) ^ { 2 }$$ Given that \(P\) has \(( x , y )\) coordinates \(\left( p , \frac { \pi } { 2 } \right)\), where \(p\) is a constant,

1. find the exact value of \(p\) The tangent to the curve at \(P\) cuts the \(y\)-axis at the point \(A\).
2. Use calculus to find the coordinates of \(A\).

1. (a) Prove, by using logarithms, that

$$\frac { \mathrm { d } } { \mathrm {~d} x } \left( 2 ^ { x } \right) = 2 ^ { x } \ln 2$$ The curve \(C\) has the equation $$2 x + 3 y ^ { 2 } + 3 x ^ { 2 } y + 12 = 4 \times 2 ^ { x }$$ The point \(P\), with coordinates \(( 2,0 )\), lies on \(C\).
(b) Find an equation of the tangent to \(C\) at \(P\).

11. The curve \(C\) has parametric equations $$x = 10 \cos 2 t , \quad y = 6 \sin t , \quad - \frac { \pi } { 2 } \leqslant t \leqslant \frac { \pi } { 2 }$$ The point \(A\) with coordinates \(( 5,3 )\) lies on \(C\).

1. Find the value of \(t\) at the point \(A\).
2. Show that an equation of the normal to \(C\) at \(A\) is $$3 y = 10 x - 41$$ The normal to \(C\) at \(A\) cuts \(C\) again at the point \(B\).
3. Find the exact coordinates of \(B\).