Iterative method for parameter value

7 \includegraphics[max width=\textwidth, alt={}, center]{388d7076-636c-417d-84cb-e6e2a3e9a6a0-10_465_785_260_680} The diagram shows the curve with parametric equations $$x = 4 t + \mathrm { e } ^ { 2 t } , \quad y = 6 t \sin 2 t$$ for \(0 \leqslant t \leqslant 1\). The point \(P\) on the curve has parameter \(p\) and \(y\)-coordinate 3 .

1. Show that \(p = \frac { 1 } { 2 \sin 2 p }\).
2. Show by calculation that the value of \(p\) lies between 0.5 and 0.6 .
3. Use an iterative formula, based on the equation in part (a), to find the value of \(p\) correct to 3 significant figures. Use an initial value of 0.55 and give the result of each iteration to 5 significant figures.
4. Find the gradient of the curve at \(P\).
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

7 \includegraphics[max width=\textwidth, alt={}, center]{61df367d-741f-4906-8ab9-2f32e8711aa6-10_465_785_260_680} The diagram shows the curve with parametric equations $$x = 4 t + \mathrm { e } ^ { 2 t } , \quad y = 6 t \sin 2 t$$ for \(0 \leqslant t \leqslant 1\). The point \(P\) on the curve has parameter \(p\) and \(y\)-coordinate 3 .

1. Show that \(p = \frac { 1 } { 2 \sin 2 p }\).
2. Show by calculation that the value of \(p\) lies between 0.5 and 0.6 .
3. Use an iterative formula, based on the equation in part (a), to find the value of \(p\) correct to 3 significant figures. Use an initial value of 0.55 and give the result of each iteration to 5 significant figures.
4. Find the gradient of the curve at \(P\).
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

6 A curve has parametric equations $$x = \frac { 1 } { ( 2 t + 1 ) ^ { 2 } } , \quad y = \sqrt { } ( t + 2 )$$ The point \(P\) on the curve has parameter \(p\) and it is given that the gradient of the curve at \(P\) is - 1 .

1. Show that \(p = ( p + 2 ) ^ { \frac { 1 } { 6 } } - \frac { 1 } { 2 }\).
2. Use an iterative process based on the equation in part (i) to find the value of \(p\) correct to 3 decimal places. Use a starting value of 0.7 and show the result of each iteration to 5 decimal places.

10 \includegraphics[max width=\textwidth, alt={}, center]{3eefd6c1-924c-4b7e-8d17-a2942fb48234-3_515_508_1105_815} The diagram shows part of the curve with parametric equations $$x = 2 \ln ( t + 2 ) , \quad y = t ^ { 3 } + 2 t + 3$$

1. Find the gradient of the curve at the origin.
2. At the point \(P\) on the curve, the value of the parameter is \(p\). It is given that the gradient of the curve at \(P\) is \(\frac { 1 } { 2 }\).
   1. Show that \(p = \frac { 1 } { 3 p ^ { 2 } + 2 } - 2\).
   2. By first using an iterative formula based on the equation in part (a), determine the coordinates of the point \(P\). Give the result of each iteration to 5 decimal places and each coordinate of \(P\) correct to 2 decimal places.

4 A curve has parametric equations $$x = t ^ { 2 } + 3 t + 1 , \quad y = t ^ { 4 } + 1$$ The point \(P\) on the curve has parameter \(p\). It is given that the gradient of the curve at \(P\) is 4 .

1. Show that \(p = \sqrt [ 3 ] { } ( 2 p + 3 )\).
2. Verify by calculation that the value of \(p\) lies between 1.8 and 2.0.
3. Use an iterative formula based on the equation in part (i) to find the value of \(p\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

6 \includegraphics[max width=\textwidth, alt={}, center]{f5e0b088-73db-405b-a832-aa01d9fcba64-08_396_716_260_712} The diagram shows the curve with parametric equations $$x = 3 t - 6 \mathrm { e } ^ { - 2 t } , \quad y = 4 t ^ { 2 } \mathrm { e } ^ { - t }$$ for \(0 \leqslant t \leqslant 2\). At the point \(P\) on the curve, the \(y\)-coordinate is 1 .

1. Show that the value of \(t\) at the point \(P\) satisfies the equation \(t = \frac { 1 } { 2 } \mathrm { e } ^ { \frac { 1 } { 2 } t }\).
2. Use the iterative formula \(t _ { n + 1 } = \frac { 1 } { 2 } \mathrm { e } ^ { \frac { 1 } { 2 } t _ { n } }\) with \(t _ { 1 } = 0.7\) to find the value of \(t\) at \(P\) correct to 3 significant figures. Give the result of each iteration to 5 significant figures.
3. Find the gradient of the curve at \(P\), giving the answer correct to 2 significant figures.

6 \includegraphics[max width=\textwidth, alt={}, center]{0d15e5a1-d05f-48bc-8613-198804ff605c-08_396_716_260_712} The diagram shows the curve with parametric equations $$x = 3 t - 6 \mathrm { e } ^ { - 2 t } , \quad y = 4 t ^ { 2 } \mathrm { e } ^ { - t }$$ for \(0 \leqslant t \leqslant 2\). At the point \(P\) on the curve, the \(y\)-coordinate is 1 .

1. Show that the value of \(t\) at the point \(P\) satisfies the equation \(t = \frac { 1 } { 2 } \mathrm { e } ^ { \frac { 1 } { 2 } t }\).
2. Use the iterative formula \(t _ { n + 1 } = \frac { 1 } { 2 } \mathrm { e } ^ { \frac { 1 } { 2 } t _ { n } }\) with \(t _ { 1 } = 0.7\) to find the value of \(t\) at \(P\) correct to 3 significant figures. Give the result of each iteration to 5 significant figures.
3. Find the gradient of the curve at \(P\), giving the answer correct to 2 significant figures.

4 The parametric equations of a curve are $$x = \mathrm { e } ^ { 2 t - 3 } , \quad y = 4 \ln t$$ where \(t > 0\). When \(t = a\) the gradient of the curve is 2 .

1. Show that \(a\) satisfies the equation \(a = \frac { 1 } { 2 } ( 3 - \ln a )\).
2. Verify by calculation that this equation has a root between 1 and 2 .
3. Use the iterative formula \(a _ { n + 1 } = \frac { 1 } { 2 } \left( 3 - \ln a _ { n } \right)\) to calculate \(a\) correct to 2 decimal places, showing the result of each iteration to 4 decimal places.