1.07s Parametric and implicit differentiation

6 A curve is defined implicitly by the equation $$y ^ { 3 } = 2 x y + x ^ { 2 }$$

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 ( x + y ) } { 3 y ^ { 2 } - 2 x }\).
2. Hence write down \(\frac { \mathrm { d } x } { \mathrm {~d} y }\) in terms of \(x\) and \(y\).

1 Fig. 7 shows the curve \(y = _ { x - 1 }\). It has a minimum at the point P . The line \(l\) is an asymptote to the curve. \begin{figure}[h] \end{figure}

1. Write down the equation of the asymptote \(l\).
2. Find the coordinates of P .
3. Using the substitution \(u = x - 1\), show that the area of the region enclosed by the \(x\)-axis, the curve and the lines \(x = 2\) and \(x = 3\) is given by $$\int _ { 1 } ^ { 2 } \left( u + 2 + \frac { 4 } { u } \right) \mathrm { d } u$$ Evaluate this area exactly.
4. Another curve is defined by the equation \(\mathrm { e } ^ { y } = \frac { x ^ { 2 } + 3 } { x - 1 }\). Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\) by differentiating implicitly. Hence find the gradient of this curve at the point where \(x = 2\).

2 Fig. 7 shows the curve defined implicitly by the equation $$y ^ { 2 } + y = x ^ { 9 } + 2 x$$ together with the line \(x = 2\). \begin{figure}[h] \end{figure} Not to scale Find the coordinates of the points of intersection of the line and the curve.
Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\). Hence find the gradient of the curve at each of these two points.

3 Fig. 8 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { x } { \sqrt { 2 + x _ { 2 } } }\). \begin{figure}[h] \end{figure}

1. Show algebraically that \(\mathrm { f } ( x )\) is an odd function. Interpret this result geometrically.
2. Show that \(\mathrm { f } ^ { \prime } ( x ) = \frac { 2 } { \left( 2 + x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } }\). Hence find the exact gradient of the curve at the origin.
3. Find the exact area of the region bounded by the curve, the \(x\)-axis and the line \(x = 1\).
4. (A) Show that if \(y = \frac { x } { \sqrt { 2 + x ^ { 2 } } }\), then \(\frac { 1 } { y ^ { 2 } } = \frac { 2 } { x ^ { 2 } } + 1\).
   (B) Differentiate \(\frac { 1 } { y ^ { 2 } } = \frac { 2 } { x ^ { 2 } } + 1\) implicitly to show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 y ^ { 3 } } { x ^ { 3 } }\). Explain why this expression cannot be used to find the gradient of the curve at the origin.

1 Fig. 9 shows the curve with equation \(y ^ { 3 } = \frac { x ^ { 3 } } { 2 x - 1 }\). It has an asymptote \(x = a\) and turning point P . \begin{figure}[h] \end{figure}

1. Write down the value of \(a\).
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 4 x ^ { 3 } - 3 x ^ { 2 } } { 3 y ^ { 2 } ( 2 x - 1 ) ^ { 2 } }\). Hence find the coordinates of the turning point P , giving the \(y\)-coordinate to 3 significant figures.
3. Show that the substitution \(u = 2 x - 1\) transforms \(\int \frac { x } { \sqrt [ 3 ] { 2 x - 1 } } \mathrm {~d} x\) to \(\frac { 1 } { 4 } \int \left( u ^ { \frac { 2 } { 3 } } + u ^ { - \frac { 1 } { 3 } } \right) \mathrm { d } u\). Hence find the exact area of the region enclosed by the curve \(y ^ { 3 } = \frac { x ^ { 3 } } { 2 x - 1 }\), the \(x\)-axis and the lines \(x = 1\) and \(x = 4.5\).

2 The variables \(x\) and \(y\) satisfy the equation \(x ^ { \frac { 2 } { 3 } } + y ^ { \frac { 2 } { 3 } } = 5\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \left( \frac { y } { x } \right) ^ { \frac { 1 } { 3 } }\). Both \(x\) and \(y\) are functions of \(t\).
2. Find the value of \(\frac { \mathrm { d } y } { \mathrm {~d} t }\) when \(x = 1 , y = 8\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = 6\).

2 Fig. 8 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { x } { \sqrt { 2 + x ^ { 2 } } }\) \begin{figure}[h] \end{figure}

1. Show algebraically that \(\mathrm { f } ( x )\) is an odd function. Interpret this result geometrically.
2. Show that \(\mathrm { f } ^ { \prime } ( x ) = \frac { 2 } { \left( 2 + x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } }\). Hence find the exact gradient of the curve at the origin.
3. Find the exact area of the region bounded by the curve, the \(x\)-axis and the line \(x = 1\).
4. \(( A )\) Show that if \(y = \frac { x } { \sqrt { 2 + x ^ { 2 } } }\), then \(\frac { 1 } { y ^ { 2 } } = \frac { 2 } { x ^ { 2 } } + 1\).
   (B) Differentiate \(\frac { 1 } { y ^ { 2 } } = \frac { 2 } { x ^ { 2 } } + 1\) implicitly to show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 y ^ { 3 } } { x ^ { 3 } }\). Explain why this expression cannot be used to find the gradient of the curve at the origin.

2 Given that \(\sin y = x y + x ^ { 2 }\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\).

5 A curve is given parametrically by the equations \(x = t ^ { 2 } , y = 2 t\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\), giving your answer in its simplest form.
2. Show that the equation of the tangent to the curve at \(\left( p ^ { 2 } , 2 p \right)\) is $$p y = x + p ^ { 2 } .$$
3. Find the coordinates of the point where the tangent at \(( 9,6 )\) meets the tangent at \(( 25 , - 10 )\).

4 Find the equation of the normal to the curve $$x ^ { 3 } + 4 x ^ { 2 } y + y ^ { 3 } = 6$$ at the point \(( 1,1 )\), giving your answer in the form \(a x + b y + c = 0\), where \(a , b\) and \(c\) are integers.

9 The parametric equations of a curve are \(x = t ^ { 3 } , y = t ^ { 2 }\).

1. Show that the equation of the tangent at the point \(P\) where \(t = p\) is $$3 p y - 2 x = p ^ { 3 } .$$
2. Given that this tangent passes through the point ( \(- 10,7\) ), find the coordinates of each of the three possible positions of \(P\).

5 A curve \(C\) has parametric equations $$x = \cos t , \quad y = 3 + 2 \cos 2 t , \quad \text { where } 0 \leqslant t \leqslant \pi$$

1. Express \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\) and hence show that the gradient at any point on \(C\) cannot exceed 8 .
2. Show that all points on \(C\) satisfy the cartesian equation \(y = 4 x ^ { 2 } + 1\).
3. Sketch the curve \(y = 4 x ^ { 2 } + 1\) and indicate on your sketch the part which represents \(C\).

6 The equation of a curve is \(x ^ { 2 } + 3 x y + 4 y ^ { 2 } = 58\). Find the equation of the normal at the point \(( 2,3 )\) on the curve, giving your answer in the form \(a x + b y + c = 0\), where \(a , b\) and \(c\) are integers.

3 The equation of a curve is \(x ^ { 2 } y - x y ^ { 2 } = 2\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y ^ { 2 } - 2 x y } { x ^ { 2 } - 2 x y }\).
2. (a) Show that, if \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\), then \(y = 2 x\).
   (b) Hence find the coordinates of the point on the curve where the tangent is parallel to the \(x\)-axis.

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.

5

1. For the curve \(2 x ^ { 2 } + x y + y ^ { 2 } = 14\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\).
2. Deduce that there are two points on the curve \(2 x ^ { 2 } + x y + y ^ { 2 } = 14\) at which the tangents are parallel to the \(x\)-axis, and find their coordinates.

6 \includegraphics[max width=\textwidth, alt={}, center]{798da17d-0af5-4aa6-b731-564642dc28d5-3_766_611_251_703} The diagram shows the curve with parametric equations $$x = a \sin \theta , \quad y = a \theta \cos \theta$$ where \(a\) is a positive constant and \(- \pi \leqslant \theta \leqslant \pi\). The curve meets the positive \(y\)-axis at \(A\) and the positive \(x\)-axis at \(B\).

1. Write down the value of \(\theta\) corresponding to the origin, and state the coordinates of \(A\) and \(B\).
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 1 - \theta \tan \theta\), and hence find the equation of the tangent to the curve at the origin.

7 Fig. 7 shows the curve with parametric equations $$x = \cos \theta , y = \sin \theta - \frac { 1 } { 8 } \sin 2 \theta , 0 \leqslant \theta < 2 \pi$$ The curve crosses the \(x\)-axis at points \(\mathrm { A } ( 1,0 )\) and \(\mathrm { B } ( - 1,0 )\), and the positive \(y\)-axis at C . D is the maximum point of the curve, and E is the minimum point. The solid of revolution formed when this curve is rotated through \(360 ^ { \circ }\) about the \(x\)-axis is used to model the shape of an egg. \begin{figure}[h] \end{figure}

1. Show that, at the point \(\mathrm { A } , \theta = 0\). Write down the value of \(\theta\) at the point B , and find the coordinates of C .
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\). Hence show that, at the point D, $$2 \cos ^ { 2 } \theta - 4 \cos \theta - 1 = 0 .$$
3. Solve this equation, and hence find the \(y\)-coordinate of D , giving your answer correct to 2 decimal places. The cartesian equation of the curve (for \(0 \leqslant \theta \leqslant \pi\) ) is $$y = \frac { 1 } { 4 } ( 4 - x ) \sqrt { 1 - x ^ { 2 } } .$$
4. Show that the volume of the solid of revolution of this curve about the \(x\)-axis is given by $$\frac { 1 } { 16 } \pi \int _ { - 1 } ^ { 1 } \left( 16 - 8 x - 15 x ^ { 2 } + 8 x ^ { 3 } - x ^ { 4 } \right) \mathrm { d } x .$$ Evaluate this integral.

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

1. Verify that $$x = k \cos \theta , \quad y = \frac { 1 } { 2 } k \sin \theta ,$$ are parametric equations for the curve.
2. Hence or otherwise show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { x } { 4 y }\).
3. 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]{9a8332ec-2216-4e1f-9768-ef175c9e159b-4_657_1071_938_577} \captionsetup{labelformat=empty} \caption{Fig. 8}
   \end{figure}
4. 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.
5. Explain why the path of the stream is modelled by the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 4 y } { x } .$$
6. Solve this differential equation. Given that the path of the stream passes through the point \(( 2,1 )\), show that its equation is \(y = \frac { x ^ { 4 } } { 16 }\).

8 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] \end{figure}

1. Find the values of the parameter at A and B . Hence show that the ratio of the lengths OA and AC is \(( \pi - 1 ) : ( \pi + 1 )\).
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\). Find the gradient of the track at A .
3. Show that, when the gradient of the track is \(1 , \theta\) satisfies the equation $$\cos \theta - 4 \sin \theta = 2 .$$
4. Express \(\cos \theta - 4 \sin \theta\) in the form \(R \cos ( \theta + \alpha )\). Hence solve the equation \(\cos \theta - 4 \sin \theta = 2\) for \(0 \leqslant \theta \leqslant 2 \pi\). www.ocr.org.uk after the live examination series.
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   Wednesday 9 June 2010 Afternoon \includegraphics[width=\textwidth]{5c149cb5-7392-4219-b285-486f4694aa6f-5_264_456_881_1361} 1 The train journey from Swansea to London is 307 km and that by road is 300 km . Carry out the calculations performed on the First Great Western website to estimate how much lower the carbon dioxide emissions are when travelling by rail rather than road.
   2 The equation of the curve in Fig. 3 is $$y = \frac { 1 } { 10 ^ { 4 } } \left( x ^ { 3 } - 100 x ^ { 2 } - 10000 x + 2100100 \right)$$ Calculate the speed at which the car has its lowest carbon dioxide emissions and the value of its emissions at that speed.
   [0pt] [An answer obtained from the graph will be given no marks.]
   3
5. In line 109 the carbon dioxide emissions for a particular train journey from Exeter to London are estimated to be 3.7 tonnes. Obtain this figure.
6. The text then goes on to state that the emissions per extra passenger on this journey are less than \(\frac { 1 } { 2 } \mathrm {~kg}\). Justify this figure.
7. \(\_\_\_\_\)
8. \(\_\_\_\_\) 4 The daily number of trains, \(n\), on a line in another country may be modelled by the function defined below, where \(P\) is the annual number of passengers. $$\begin{aligned} & n = 10 \text { for } 0 \leqslant P < 10 ^ { 6 } \\ & n = 11 \text { for } 10 ^ { 6 } \leqslant P < 1.5 \times 10 ^ { 6 } \\ & n = 12 \text { for } 1.5 \times 10 ^ { 6 } \leqslant P < 2 \times 10 ^ { 6 } \\ & n = 13 \text { for } 2 \times 10 ^ { 6 } \leqslant P < 2.5 \times 10 ^ { 6 } \\ & n = 14 \text { for } 2.5 \times 10 ^ { 6 } \leqslant P < 3 \times 10 ^ { 6 } \\ & \ldots \text { and so on } \ldots \end{aligned}$$
9. Sketch the graph of \(n\) against \(P\).
10. Describe, in words, the relationship between the daily number of trains and the annual number of passengers.
11. \includegraphics[width=\textwidth]{5c149cb5-7392-4219-b285-486f4694aa6f-7_716_1249_1011_440}
12. \(\_\_\_\_\) 5 The FGW website gives the conversion factor for miles to kilometres to 7 significant figures.
    "We got the distance between the two stations by road from theaa.com. We then converted this distance to kilometres by multiplying it by \(1.609344 . "\) Suppose this conversion factor is applied to a distance of exactly 100 miles.
    State which one of the following best expresses the level of accuracy for the distance in metric units, justifying your answer. A : to the nearest millimetre
    B : to the nearest 10 centimetres
    C : to the nearest metre

6 P is a general point on the curve with parametric equations \(x = 2 t , y = \frac { 2 } { t }\). This is shown in Fig. 6. The tangent at P intersects the \(x\) - and \(y\)-axes at the points Q and R respectively. \begin{figure}[h] \end{figure} Show that the area of the triangle OQR , where O is the origin, is independent of \(t\).

5 A curve is given by the parametric equations \(x = a t ^ { 2 } , y = 2 a\) (where \(a\) is a constant). A point P on the curve has coordinates ( \(a p ^ { 2 }\), 2ap).

1. Find the coordinates of the point, T , where the tangent to the curve at P meets the \(x\)-axis and the coordinates of the point N where the normal to the curve at P meets the \(x\)-axis.
2. Hence show that the area of the triangle PTN is \(2 a ^ { 2 } p \left( p ^ { 2 } + 1 \right)\) square units.

4 A curve is given by the parametric equations \(x = t ^ { 2 } , y = 3 t\) for all values of \(t\). Find the equation of the tangent to the curve at the point where \(t = - 2\).

9 Two astronomers wish to model the path of motion of a particle in two dimensions.
Experimental results show that the position of the particle can be found using the parametric equations $$x = 2 \cos \theta - \sin \theta + 2 \quad y = \cos \theta + 2 \sin \theta - 1 \quad \left( 0 \leq \theta \leq 360 ^ { \circ } \right)$$ One astronomer uses trigonometry.

1. Express \(2 \cos \theta - \sin \theta\) in the form \(R \cos ( \theta + \alpha )\), where \(R\) and \(\alpha\) are constants to be determined. Show also that, for the same values of \(R\) and \(\alpha\), $$\cos \theta + 2 \sin \theta = R \sin ( \theta + \alpha )$$
2. Hence, or otherwise, show that the path of particle may be written in the form $$( x - 2 ) ^ { 2 } + ( y + 1 ) ^ { 2 } = 5$$ Describe the path of the particle. The second astronomer sets up a first order differential equation with the condition that \(x = 4\) when \(y = 0\).
3. Verify that the point with parameter \(\theta = 0\) has coordinates \(( 4,0 )\).
4. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\). Deduce that \(x\) and \(y\) satisfy the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { x - 2 } { y + 1 }$$
5. Solve this differential equation, using the condition that \(y = 0\) when \(x = 4\). Hence show that the two solutions give the same cartesian equation for the path of particle.

2. A curve has the equation $$2 x ^ { 2 } + x y - y ^ { 2 } + 18 = 0$$ Find the coordinates of the points where the tangent to the curve is parallel to the \(x\)-axis.