Questions C4 - A-Level Maths

Find the lengths $\mathrm { OA } , \mathrm { OB }$ and AC .
Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $u$. Hence find the angle $\theta$.
Show that the cartesian equation of the curve is $y = \mathrm { e } ^ { \frac { 1 } { 5 } x } + \mathrm { e } ^ { - \frac { 1 } { 5 } x }$. An object is formed by rotating the region OACB through $360 ^ { \circ }$ about $\mathrm { O } x$.
Find the volume of the object.

OCR MEI C4 Q3

3 A curve has parametric equations $$x = 2 \sin \theta , \quad y = \cos 2 \theta$$

Find the exact coordinates and the gradient of the curve at the point with parameter $\theta = \frac { 1 } { 3 } \pi$.
Find $y$ in terms of $x$.

OCR MEI C4 Q4

4 The parametric equations of a curve are $$x = \cos 2 \theta , \quad y = \sin \theta \cos \theta \quad \text { for } 0 \leqslant \theta < \pi$$ Show that the cartesian equation of the curve is $x ^ { 2 } + 4 y ^ { 2 } = 1$.
Sketch the curve.

OCR MEI C4 Q5

5 Part of the track of a roller-coaster is modelled by a curve with the parametric equations $$x = 2 \theta - \sin \theta , \quad y = 4 \cos \theta \quad \text { for } 0 \leqslant \theta \leqslant 2 \pi .$$ This is shown in Fig. 8. B is a minimum point, and BC is vertical. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{c443a5b6-247d-411d-8371-4d6ebd5c3489-3_598_1443_598_385} \captionsetup{labelformat=empty} \caption{Fig. 8}

\end{figure}

Find the values of the parameter at A and B . Hence show that the ratio of the lengths OA and AC is $( \pi - 1 ) : ( \pi + 1 )$.
Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $\theta$. Find the gradient of the track at A .
Show that, when the gradient of the track is $1 , \theta$ satisfies the equation $$\cos \theta - 4 \sin \theta = 2$$
Express $\cos \theta - 4 \sin \theta$ in the form $R \cos ( \theta + \alpha )$. Hence solve the equation $\cos \theta - 4 \sin \theta = 2$ for $0 \leqslant \theta \leqslant 2 \pi$.

OCR MEI C4 Q6

6 A curve has parametric equations $$x = a t ^ { 3 } , \quad y = \frac { a } { 1 + t ^ { 2 } }$$ where $a$ is a constant.
Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { - 2 } { 3 t \left( 1 + t ^ { 2 } \right) ^ { 2 } }$.
Hence find the gradient of the curve at the point $\left( a , \frac { 1 } { 2 } a \right)$.

OCR MEI C4 Q7

7 A curve has parametric equations $x = 1 + u ^ { 2 } , y = 2 u ^ { 3 }$.

Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $u$.
Hence find the gradient of the curve at the point with coordinates $( 5,16 )$.

OCR MEI C4 Q8

8 A curve is defined by parametric equations $$x = \frac { 1 } { t } - 1 , y = \frac { 2 + t } { 1 + t }$$ Show that the cartesian equation of the curve is $y = \frac { 3 + 2 x } { 2 + x }$.

OCR MEI C4 Q2

2 Fig. 6 shows the arch ABCD of a bridge. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{0c0a2fe7-9e69-470a-af2e-fa5fd41e4a27-2_378_1630_397_132} \captionsetup{labelformat=empty} \caption{Fig. 6}

\end{figure} The section from $B$ to $C$ is part of the curve $O B C E$ with parametric equations $$x = a ( \theta - \sin \theta ) , y = a ( 1 - \cos \theta ) \text { for } 0 \leqslant \theta \leqslant 2 \pi \text {, }$$ where $a$ is a constant.

Find, in terms of $a$,
(A) the length of the straight line OE,
(B) the maximum height of the arch.
Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $\theta$. The straight line sections AB and CD are inclined at $30 ^ { \circ }$ to the horizontal, and are tangents to the curve at B and C respectively. BC is parallel to the $x$-axis. BF is parallel to the $y$-axis.
Show that at the point B the parameter $\theta$ satisfies the equation $$\sin \theta = \frac { 1 } { \sqrt { 3 } } ( 1 \quad \cos \theta ) .$$ Verify that $\theta = \frac { 2 } { 3 } \pi$ is a solution of this equation.
Hence show that $\mathrm { BF } = \frac { 3 } { 2 } a$, and find OF in terms of $a$, giving your answer exactly.
Find BC and AF in terms of $a$. Given that the straight line distance AD is 20 metres, calculate the value of $a$.

OCR MEI C4 Q3

3 A curve has carlesian equation $\mathrm { y } ^ { 2 } - \mathrm { x } _ { 2 } = 4$.

Verify that $$\boldsymbol { x } = \boldsymbol { t } - - ^ { 1 } \quad \boldsymbol { t ^ { \prime } } \quad \boldsymbol { y } = \boldsymbol { t } + \frac { 1 } { \boldsymbol { t } ^ { \prime } }$$ are parametric equations of the curve.
(u) Show lhat $\left. \underset { d x } { \mathbf { d y } } = \frac { ( t - I ) ( r } { 12 + 1 } + 1 \right)$. Hence find the coordinates of the staionary points of the curve.

OCR MEI C4 Q4

4 The parametric equations of a curve are $$x = \sin \theta , \quad y = \sin 2 \theta , \quad \text { for } 0 \leqslant \theta \leqslant 2 \pi$$

Find the exact value of the gradient of the curve at the point where $\theta = \frac { 1 } { 6 } \pi$.
Show that the cartesian equation of the curve is $y ^ { 2 } = 4 x ^ { 2 } - 4 x ^ { 4 }$.

OCR MEI C4 Q5

5 A curve is defined parametrically by the equations $$x = \frac { 1 } { 1 + t } , \quad y = \frac { 1 - t } { 1 + 2 t }$$ Find $t$ in terms of $x$. Hence find the cartesian equation of the curve, giving your answer as simply as possible.

OCR MEI C4 Q6

6 A curve has parametric equations $$x = \mathrm { e } ^ { 2 t } , \quad y = \frac { 2 t } { 1 + t }$$

Find the gradient of the curve at the point where $t = 0$.
Find $y$ in terms of $x$.

OCR MEI C4 Q1

1 Fig. 8 illustrates a hot air balloon on its side. The balloon is modelled by the volume of revolution about the $x$-axis of the curve with parametric equations $$x = 2 + 2 \sin \theta , \quad y = 2 \cos \theta + \sin 2 \theta , \quad ( 0 \leqslant \theta \leqslant 2 \pi ) .$$ The curve crosses the $x$-axis at the point $\mathrm { A } ( 4,0 )$. B and C are maximum and minimum points on the curve. Units on the axes are metres. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{1601927c-74d7-4cc2-a7f2-2c2a2e8c2c4c-1_812_809_517_704} \captionsetup{labelformat=empty} \caption{Fig. 8}

\end{figure}

Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $\theta$.
Verify that $\frac { \mathrm { d } y } { \mathrm {~d} x } = 0$ when $\theta = \frac { 1 } { 6 } \pi$, and find the exact coordinates of B . Hence find the maximum width BC of the balloon.
(A) Show that $y = x \cos \theta$.
(B) Find $\sin \theta$ in terms of $x$ and show that $\cos ^ { 2 } \theta = x - \frac { 1 } { 4 } x ^ { 2 }$.
(C) Hence show that the cartesian equation of the curve is $y ^ { 2 } = x ^ { 3 } - \frac { 1 } { 4 } x ^ { 4 }$.
Find the volume of the balloon.

OCR MEI C4 Q2

2 A curve has equation $$x ^ { 2 } + 4 y ^ { 2 } = k ^ { 2 } ,$$ where $k$ is a positive constant.

Verify that $$x = k \cos \theta , \quad y = \frac { 1 } { 2 } k \sin \theta ,$$ are parametric equations for the curve.
Hence or otherwise show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { x } { 4 y }$.
Fig. 8 illustrates the curve for a particular value of $k$. Write down this value of $k$. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{1601927c-74d7-4cc2-a7f2-2c2a2e8c2c4c-2_658_1070_861_567} \captionsetup{labelformat=empty} \caption{Fig. 8}
\end{figure}
Copy Fig. 8 and on the same axes sketch the curves for $k = 1 , k = 3$ and $k = 4$. On a map, the curves represent the contours of a mountain. A stream flows down the mountain. Its path on the map is always at right angles to the contour it is crossing.
Explain why the path of the stream is modelled by the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 4 y } { x } .$$
Solve this differential equation. Given that the path of the stream passes through the point $( 2,1 )$, show that its equation is $y = \frac { x ^ { 4 } } { 16 }$.

OCR MEI C4 Q3

3 A curve is defined parametrically by the equations $$x = t - \ln t , \quad y = t + \ln t \quad ( t > 0 )$$ Find the gradient of the curve at the point where $t = 2$.

OCR MEI C4 Q4

4 Fig. 7a shows the curve with the parametric equations $$x = 2 \cos \theta , \quad y = \sin 2 \theta , \quad - \frac { \pi } { 2 } \leqslant \theta \leqslant \frac { \pi } { 2 } .$$ The curve meets the $x$-axis at O and P . Q and R are turning points on the curve. The scales on the axes are the same. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{1601927c-74d7-4cc2-a7f2-2c2a2e8c2c4c-4_509_660_571_714} \captionsetup{labelformat=empty} \caption{Fig. 7a}

\end{figure}

State, with their coordinates, the points on the curve for which $\theta = - \frac { \pi } { 2 } , \theta = 0$ and $\theta = \frac { \pi } { 2 }$.
Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $\theta$. Hence find the gradient of the curve when $\theta = \frac { \pi } { 2 }$, and verify that the two tangents to the curve at the origin meet at right angles.
Find the exact coordinates of the turning point Q . When the curve is rotated about the $x$-axis, it forms a paperweight shape, as shown in Fig. 7b. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{1601927c-74d7-4cc2-a7f2-2c2a2e8c2c4c-4_324_389_1692_857} \captionsetup{labelformat=empty} \caption{Fig. 7b}
\end{figure}
Express $\sin ^ { 2 } \theta$ in terms of $x$. Hence show that the cartesian equation of the curve is $y ^ { 2 } = x ^ { 2 } \left( 1 - \frac { 1 } { 4 } x ^ { 2 } \right)$.
Find the volume of the paperweight shape.
Express $\frac { 3 } { ( y - 2 ) ( y + 1 ) }$ in partial fractions.
Hence, given that $x$ and $y$ satisfy the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = x ^ { 2 } ( y - 2 ) ( y + 1 )$$ show that $\frac { y - 2 } { y + 1 } = A \mathrm { e } ^ { x ^ { 3 } }$, where $A$ is a constant.

OCR C4 Q1

1 Express $2 \sin \theta - 3 \cos \theta$ in the form $R \sin ( \theta - \alpha )$, where $R$ and $\alpha$ are constants to be determined, and $0 < \alpha < \frac { 1 } { 2 } \pi$. Hence write down the greatest and least possible values of $1 + 2 \sin \theta - 3 \cos \theta$.

OCR C4 Q2

2 Express $4 \cos \theta - \sin \theta$ in the form $R \cos ( \theta + \alpha )$, where $R > 0$ and $0 < \alpha < { } _ { 2 } ^ { 1 } \pi$.-
Hence solve the equation $4 \cos \theta - \sin \theta = 3$, for $0 \leqslant \theta \leqslant 2 \pi$.

OCR C4 Q3

3 Archimedes, about 2200 years ago, used regular polygons inside and outside circles to obtain approximations for $\pi$.

Fig. 8.1 shows a regular 12 -sided polygon inscribed in a circle of radius 1 unit, centre $\mathrm { O } . \mathrm { AB }$ is one of the sides of the polygon. C is the midpoint of AB . Archimedes used the fact that the circumference of the circle is greater than the perimeter of this polygon. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{c0fcd64b-8ca0-4309-9f58-c23cc4208f4d-2_457_422_457_936} \captionsetup{labelformat=empty} \caption{Fig. 8.1}
\end{figure} (A) Show that $\mathrm { AB } = 2 \sin 15 ^ { \circ }$.
(B) Use a double angle formula to express $\cos 30 ^ { \circ }$ in terms of $\sin 15 ^ { \circ }$. Using the exact value of $\cos 30 ^ { \circ }$, show that $\sin 15 ^ { \circ } = \frac { 1 } { 2 } \sqrt { 2 - \sqrt { 3 } }$.
(C) Use this result to find an exact expression for the perimeter of the polygon. Hence show that $\pi > 6 \sqrt { 2 - \sqrt { 3 } }$.
In Fig. 8.2, a regular 12-sided polygon lies outside the circle of radius 1 unit, which touches each side of the polygon. F is the midpoint of DE. Archimedes used the fact that the circumference of the circle is less than the perimeter of this polygon. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{c0fcd64b-8ca0-4309-9f58-c23cc4208f4d-2_450_416_1562_940} \captionsetup{labelformat=empty} \caption{Fig. 8.2}
\end{figure} (A) Show that $\mathrm { DE } = 2 \tan 15 ^ { \circ }$.
(B) Let $t = \tan 15 ^ { \circ }$. Use a double angle formula to express $\tan 30 ^ { \circ }$ in terms of $t$. Hence show that $t ^ { 2 } + 2 \sqrt { 3 } t - 1 = 0$.
(C) Solve this equation, and hence show that $\pi < 12 ( 2 - \sqrt { 3 } )$.
Use the results in parts (i)( $C$ ) and (ii)( $C$ ) to establish upper and lower bounds for the value of $\pi$, giving your answers in decimal form.