1.07s Parametric and implicit differentiation

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

1. 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.

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

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

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

1. Find the gradient of the curve at the point where \(t = 0\).
2. Find \(y\) in terms of \(x\).

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] \end{figure}

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\).
2. 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.
3. (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 \(y ^ { 2 } = x ^ { 3 } - \frac { 1 } { 4 } x ^ { 4 }\).
4. Find the volume of the balloon.

2 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]{1601927c-74d7-4cc2-a7f2-2c2a2e8c2c4c-2_658_1070_861_567} \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 \(y = \frac { x ^ { 4 } } { 16 }\).

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 \(t = 2\).

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] \end{figure}

1. State, with their coordinates, the points on the curve for which \(\theta = - \frac { \pi } { 2 } , \theta = 0\) and \(\theta = \frac { \pi } { 2 }\).
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.
3. 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}
4. Express \(\sin ^ { 2 } \theta\) in terms of \(x\). Hence show that the cartesian equation of the curve is \(y ^ { 2 } = x ^ { 2 } \left( 1 - \frac { 1 } { 4 } x ^ { 2 } \right)\).
5. Find the volume of the paperweight shape.
6. Express \(\frac { 3 } { ( y - 2 ) ( y + 1 ) }\) in partial fractions.
7. 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.

4 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 \(0 \leqslant \theta \leqslant 2 \pi\).

1 A curve has parametric equations \(x = \sec \theta , y = 2 \tan \theta\).

1. Given that the derivative of \(\sec \theta\) is \(\sec \theta \tan \theta\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 \operatorname { cosec } \theta\).
2. Verify that the cartesian equation of the curve is \(y ^ { 2 } = 4 x ^ { 2 } - 4\). Fig. 5 shows the region enclosed by the curve and the line \(x = 2\). This region is rotated through \(180 ^ { \circ }\) about the \(x\)-axis. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{1e788cb0-36b0-42a9-9e0c-077022d410ae-1_556_867_580_588} \captionsetup{labelformat=empty} \caption{Fig. 5}
   \end{figure}
3. Find the volume of revolution produced, giving your answer in exact form.

2 In a game of rugby, a kick is to be taken from a point P (see Fig. 7). P is a perpendicular distance \(y\) metres from the line TOA. Other distances and angles are as shown. \begin{figure}[h] \end{figure}

1. Show that \(\theta = \beta - \alpha\), and hence that \(\tan \theta = \frac { 6 y } { 160 + y ^ { 2 } }\). Calculate the angle \(\theta\) when \(y = 6\).
2. By differentiating implicitly, show that \(\frac { \mathrm { d } \theta } { \mathrm { d } y } = \frac { 6 \left( 160 - y ^ { 2 } \right) } { \left( 160 + y ^ { 2 } \right) ^ { 2 } } \cos ^ { 2 } \theta\).
3. Use this result to find the value of \(y\) that maximises the angle \(\theta\). Calculate this maximum value of \(\theta\). [You need not verify that this value is indeed a maximum.]

5 Cartesian coordinates \(( x , y )\) and polar coordinates \(( r , \theta )\) are set up in the usual way, with the pole at the origin and the initial line along the positive \(x\)-axis, so that \(x = r \cos \theta\) and \(y = r \sin \theta\). A curve has polar equation \(r = k + \cos \theta\), where \(k\) is a constant with \(k \geqslant 1\).

1. Use your graphical calculator to obtain sketches of the curve in the three cases $$k = 1 , k = 1.5 \text { and } k = 4$$
2. Name the special feature which the curve has when \(k = 1\).
3. For each of the three cases, state the number of points on the curve at which the tangent is parallel to the \(y\)-axis.
4. Express \(x\) in terms of \(k\) and \(\theta\), and find \(\frac { \mathrm { d } x } { \mathrm {~d} \theta }\). Hence find the range of values of \(k\) for which there are just two points on the curve where the tangent is parallel to the \(y\)-axis. The distance between the point ( \(r , \theta\) ) on the curve and the point ( 1,0 ) on the \(x\)-axis is \(d\).
5. Use the cosine rule to express \(d ^ { 2 }\) in terms of \(k\) and \(\theta\), and deduce that \(k ^ { 2 } \leqslant d ^ { 2 } \leqslant k ^ { 2 } + 1\).
6. Hence show that, when \(k\) is large, the shape of the curve is very nearly circular.

5 A curve has parametric equations \(x = \frac { t ^ { 2 } } { 1 + t ^ { 2 } } , y = t ^ { 3 } - \lambda t\), where \(\lambda\) is a constant.

1. Use your calculator to obtain a sketch of the curve in each of the cases $$\lambda = - 1 , \quad \lambda = 0 \quad \text { and } \quad \lambda = 1 .$$ Name any special features of these curves.
2. By considering the value of \(x\) when \(t\) is large, write down the equation of the asymptote. For the remainder of this question, assume that \(\lambda\) is positive.
3. Find, in terms of \(\lambda\), the coordinates of the point where the curve intersects itself.
4. Show that the two points on the curve where the tangent is parallel to the \(x\)-axis have coordinates $$\left( \frac { \lambda } { 3 + \lambda } , \pm \sqrt { \frac { 4 \lambda ^ { 3 } } { 27 } } \right)$$ Fig. 5 shows a curve which intersects itself at the point ( 2,0 ) and has asymptote \(x = 8\). The stationary points A and B have \(y\)-coordinates 2 and - 2 . \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{43b4c7ed-3556-4d87-8aef-0111fe747982-4_791_609_1482_769} \captionsetup{labelformat=empty} \caption{Fig. 5}
   \end{figure}
5. For the curve sketched in Fig. 5, find parametric equations of the form \(x = \frac { a t ^ { 2 } } { 1 + t ^ { 2 } } , y = b \left( t ^ { 3 } - \lambda t \right)\), where \(a , \lambda\) and \(b\) are to be determined.

3. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of a part of the curve \(C\) with parametric equations $$x = t ^ { 3 } , y = t ^ { 2 } .$$ The tangent at the point \(P ( 8,4 )\) cuts \(C\) at the point \(Q\) .
Find the area of the shaded region between \(P Q\) and \(C\) .

6. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of the curve \(C\) with parametric equations $$x = 2 \sin t , \quad y = \ln ( \sec t ) , \quad 0 \leqslant t < \frac { \pi } { 2 }$$ The tangent to \(C\) at the point \(P\) ,where \(t = \frac { \pi } { 3 }\) ,cuts the \(x\)-axis at \(A\) .

1. Show that the \(x\)-coordinate of \(A\) is \(\frac { \sqrt { } 3 } { 3 } ( 3 - \ln 2 )\) . The shaded region \(R\) lies between \(C\) ,the positive \(x\)-axis and the tangent \(A P\) as shown in Figure 2 .
2. Show that the area of \(R\) is \(\sqrt { 3 } ( 1 + \ln 2 ) - 2 \ln ( 2 + \sqrt { 3 } ) - \frac { \sqrt { 3 } } { 6 } ( \ln 2 ) ^ { 2 }\) .

3.The curve \(C\) has equation $$x ^ { 2 } + y ^ { 2 } + f x y = g ^ { 2 }$$ where \(f\) and \(g\) are constants and \(g \neq 0\) .

1. Find an expression in terms of \(\alpha , \beta\) and \(f\) for the gradient of \(C\) at the point \(( \alpha , \beta )\) . Given that \(f < 2\) and \(f \neq - 2\) and that the gradient of \(C\) at the point \(( \alpha , \beta )\) is 1 ,
2. show that \(\alpha = - \beta = \frac { \pm g } { \sqrt { } ( 2 - f ) }\) . Given that \(f = - 2\) ,
3. sketch \(C\) .

2. The point \(P\) with coordinates \(\left( \frac { \pi } { 2 } , 1 \right)\) lies on the curve with equation $$4 x \sin x = \pi y ^ { 2 } + 2 x , \quad \frac { \pi } { 6 } \leqslant x \leqslant \frac { 5 \pi } { 6 }$$ Find an equation of the normal to the curve at \(P\).

5 A line PQ is of length \(k\) (where \(k > 1\) ) and it passes through the point ( 1,0 ). PQ is inclined at angle \(\theta\) to the positive \(x\)-axis. The end Q moves along the \(y\)-axis. See Fig. 5. The end P traces out a locus. \begin{figure}[h] \end{figure}

1. Show that the locus of P may be expressed parametrically as follows. $$x = k \cos \theta \quad y = k \sin \theta - \tan \theta$$ You are now required to investigate curves with these parametric equations, where \(k\) may take any non-zero value and \(- \frac { 1 } { 2 } \pi < \theta < \frac { 1 } { 2 } \pi\).
2. Use your calculator to sketch the curve in each of the cases \(k = 2 , k = 1 , k = \frac { 1 } { 2 }\) and \(k = - 1\).
3. For what value(s) of \(k\) does the curve have
   (A) an asymptote (you should state what the asymptote is),
   (B) a cusp,
   (C) a loop?
4. For the case \(k = 2\), find the angle at which the curve crosses itself.
5. For the case \(k = 8\), find in an exact form the coordinates of the highest point on the loop.
6. Verify that the cartesian equation of the curve is $$y ^ { 2 } = \frac { ( x - 1 ) ^ { 2 } } { x ^ { 2 } } \left( k ^ { 2 } - x ^ { 2 } \right) .$$

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.

1

1. 1. Differentiate the equation \(\sin y = x\) with respect to \(x\), and hence show that the derivative of \(\arcsin x\) is \(\frac { 1 } { \sqrt { 1 - x ^ { 2 } } }\).
   2. Evaluate the following integrals, giving your answers in exact form.
      (A) \(\int _ { - 1 } ^ { 1 } \frac { 1 } { \sqrt { 2 - x ^ { 2 } } } \mathrm {~d} x\) (B) \(\int _ { - \frac { 1 } { 2 } } ^ { \frac { 1 } { 2 } } \frac { 1 } { \sqrt { 1 - 2 x ^ { 2 } } } \mathrm {~d} x\)
2. A curve has polar equation \(r = \tan \theta , 0 \leqslant \theta < \frac { 1 } { 2 } \pi\). The points on the curve have cartesian coordinates \(( x , y )\). A sketch of the curve is given in Fig. 1. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{99f0c663-bb5b-4456-854c-df177f5d8349-2_493_796_1123_605} \captionsetup{labelformat=empty} \caption{Fig. 1}
   \end{figure} Show that \(x = \sin \theta\) and that \(r ^ { 2 } = \frac { x ^ { 2 } } { 1 - x ^ { 2 } }\).
   Hence show that the cartesian equation of the curve is $$y = \frac { x ^ { 2 } } { \sqrt { 1 - x ^ { 2 } } } .$$ Give the cartesian equation of the asymptote of the curve.

6 \includegraphics[max width=\textwidth, alt={}, center]{89e54367-bb83-483a-add5-0527b71a5cac-4_476_709_251_683} The diagram shows the curve with equation \(x = \ln \left( y ^ { 3 } + 2 y \right)\). At the point \(P\) on the curve, the gradient is 4 and it is given that \(P\) is close to the point with coordinates (7.5,12).

1. Find \(\frac { \mathrm { d } x } { \mathrm {~d} y }\) in terms of \(y\).
2. Show that the \(y\)-coordinate of \(P\) satisfies the equation $$y = \frac { 12 y ^ { 2 } + 8 } { y ^ { 2 } + 2 }$$
3. By first using an iterative process based on the equation in part (ii), find the coordinates of \(P\), giving each coordinate correct to 3 decimal places.

7 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 \(\frac { \mathrm { d } x } { \mathrm {~d} t } = 6\). Section B (36 marks)

3

1. Given that \(y = \sqrt [ 3 ] { 1 + 3 x ^ { 2 } }\), use the chain rule to find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\).
2. Given that \(y ^ { 3 } = 1 + 3 x ^ { 2 }\), use implicit differentiation to find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\). Show that this result is equivalent to the result in part (i).

5 The equation of a curve is given by \(\mathrm { e } ^ { 2 y } = 1 + \sin x\).

1. By differentiating implicitly, find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\).
2. Find an expression for \(y\) in terms of \(x\), and differentiate it to verify the result in part (i).

3 Find the gradient at the point \(( 0 , \ln 2 )\) on the curve with equation \(\mathrm { e } ^ { 2 y } = 5 - \mathrm { e } ^ { - x }\).

9 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 \(x = 4.5\).