Edexcel C4 2011 June — Question 15

Exam BoardEdexcel
ModuleC4 (Core Mathematics 4)
Year2011
SessionJune
TopicVectors 3D & Lines

15
3 \end{array} \right) + \mu \left( \begin{array} { r } 2
- 3
1 \end{array} \right)$$ where \(\lambda\) and \(\mu\) are scalar parameters.
  1. Show that \(l _ { 1 }\) and \(l _ { 2 }\) meet and find the position vector of their point of intersection \(A\).
  2. Find, to the nearest \(0.1 ^ { \circ }\), the acute angle between \(l _ { 1 }\) and \(l _ { 2 }\). The point \(B\) has position vector \(\left( \begin{array} { r } 5
    - 1
    1 \end{array} \right)\).
  3. Show that \(B\) lies on \(l _ { 1 }\).
  4. Find the shortest distance from \(B\) to the line \(l _ { 2 }\), giving your answer to 3 significant figures.
    7. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{9d513d77-b8f9-4223-832f-f566c5f50457-10_643_999_276_475} \captionsetup{labelformat=empty} \caption{Figure 3}
    \end{figure} Figure 3 shows part of the curve \(C\) with parametric equations $$x = \tan \theta , \quad y = \sin \theta , \quad 0 \leqslant \theta < \frac { \pi } { 2 }$$ The point \(P\) lies on \(C\) and has coordinates \(\left( \sqrt { } 3 , \frac { 1 } { 2 } \sqrt { } 3 \right)\).
  5. Find the value of \(\theta\) at the point \(P\). The line \(l\) is a normal to \(C\) at \(P\). The normal cuts the \(x\)-axis at the point \(Q\).
  6. Show that \(Q\) has coordinates \(( k \sqrt { } 3,0 )\), giving the value of the constant \(k\). The finite shaded region \(S\) shown in Figure 3 is bounded by the curve \(C\), the line \(x = \sqrt { } 3\) and the \(x\)-axis. This shaded region is rotated through \(2 \pi\) radians about the \(x\)-axis to form a solid of revolution.
  7. Find the volume of the solid of revolution, giving your answer in the form \(p \pi \sqrt { } 3 + q \pi ^ { 2 }\), where \(p\) and \(q\) are constants. 8. (a) Find \(\int ( 4 y + 3 ) ^ { - \frac { 1 } { 2 } } \mathrm {~d} y\)
  8. Given that \(y = 1.5\) at \(x = - 2\), solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { \sqrt { } ( 4 y + 3 ) } { x ^ { 2 } }$$ giving your answer in the form \(y = \mathrm { f } ( x )\).