OCR C4 2008 June — Question 12

Exam BoardOCR
ModuleC4 (Core Mathematics 4)
Year2008
SessionJune
PaperDownload PDF ↗
TopicVectors 3D & Lines
TypeAngle between two lines
DifficultyModerate -0.3 This is a standard two-part question on 3D vectors requiring showing lines intersect (equating parameters and solving) and finding angle between direction vectors using the dot product formula. Both parts are routine C4 techniques with straightforward calculations and no novel problem-solving required, making it slightly easier than average.
Spec1.03g Parametric equations: of curves and conversion to cartesian1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.07l Derivative of ln(x): and related functions1.07s Parametric and implicit differentiation1.08k Separable differential equations: dy/dx = f(x)g(y)

12
0
5 \end{array} \right) + s \left( \begin{array} { r } 1
- 4
- 2 \end{array} \right) .$$
  1. Show that the lines intersect.
  2. Find the angle between the lines.
  3. Show that, if \(y = \operatorname { cosec } x\), then \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) can be expressed as \(- \operatorname { cosec } x \cot x\).
  4. Solve the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = - \sin x \tan x \cot t$$ given that \(x = \frac { 1 } { 6 } \pi\) when \(t = \frac { 1 } { 2 } \pi\). 8
  5. Given that \(\frac { 2 t } { ( t + 1 ) ^ { 2 } }\) can be expressed in the form \(\frac { A } { t + 1 } + \frac { B } { ( t + 1 ) ^ { 2 } }\), find the values of the constants \(A\) and \(B\).
  6. Show that the substitution \(t = \sqrt { 2 x - 1 }\) transforms \(\int \frac { 1 } { x + \sqrt { 2 x - 1 } } \mathrm {~d} x\) to \(\int \frac { 2 t } { ( t + 1 ) ^ { 2 } } \mathrm {~d} t\).
  7. Hence find the exact value of \(\int _ { 1 } ^ { 5 } \frac { 1 } { x + \sqrt { 2 x - 1 } } \mathrm {~d} x\). 9 The parametric equations of a curve are $$x = 2 \theta + \sin 2 \theta , \quad y = 4 \sin \theta$$ and part of its graph is shown below. \includegraphics[max width=\textwidth, alt={}, center]{b8ba126f-c5fa-4828-9439-e5162a03ca5b-3_646_1150_1050_500}
  8. Find the value of \(\theta\) at \(A\) and the value of \(\theta\) at \(B\).
  9. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \sec \theta\).
  10. At the point \(C\) on the curve, the gradient is 2 . Find the coordinates of \(C\), giving your answer in an exact form.

12 \\
0 \\
5
\end{array} \right) + s \left( \begin{array} { r } 
1 \\
- 4 \\
- 2
\end{array} \right) .$$

(i) Show that the lines intersect.\\
(ii) Find the angle between the lines.\\
(i) Show that, if $y = \operatorname { cosec } x$, then $\frac { \mathrm { d } y } { \mathrm {~d} x }$ can be expressed as $- \operatorname { cosec } x \cot x$.\\
(ii) Solve the differential equation

$$\frac { \mathrm { d } x } { \mathrm {~d} t } = - \sin x \tan x \cot t$$

given that $x = \frac { 1 } { 6 } \pi$ when $t = \frac { 1 } { 2 } \pi$.

8 (i) Given that $\frac { 2 t } { ( t + 1 ) ^ { 2 } }$ can be expressed in the form $\frac { A } { t + 1 } + \frac { B } { ( t + 1 ) ^ { 2 } }$, find the values of the constants $A$ and $B$.\\
(ii) Show that the substitution $t = \sqrt { 2 x - 1 }$ transforms $\int \frac { 1 } { x + \sqrt { 2 x - 1 } } \mathrm {~d} x$ to $\int \frac { 2 t } { ( t + 1 ) ^ { 2 } } \mathrm {~d} t$.\\
(iii) Hence find the exact value of $\int _ { 1 } ^ { 5 } \frac { 1 } { x + \sqrt { 2 x - 1 } } \mathrm {~d} x$.

9 The parametric equations of a curve are

$$x = 2 \theta + \sin 2 \theta , \quad y = 4 \sin \theta$$

and part of its graph is shown below.\\
\includegraphics[max width=\textwidth, alt={}, center]{b8ba126f-c5fa-4828-9439-e5162a03ca5b-3_646_1150_1050_500}\\
(i) Find the value of $\theta$ at $A$ and the value of $\theta$ at $B$.\\
(ii) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \sec \theta$.\\
(iii) At the point $C$ on the curve, the gradient is 2 . Find the coordinates of $C$, giving your answer in an exact form.

\hfill \mbox{\textit{OCR C4 2008 Q12}}