Show dy/dx equals expression

6 \includegraphics[max width=\textwidth, alt={}, center]{7b39a2ab-305d-43c5-a1e7-9442d6c13886-10_629_620_278_717} The diagram shows the curve with parametric equations $$x = 1 + \sqrt { t } , \quad y = ( \ln t + 2 ) ( \ln t - 3 ) ,$$ for \(0 < t < 25\). The curve crosses the \(x\)-axis at the points \(A\) and \(B\) and has a minimum point \(M\).

1. Show that \(\frac { \mathrm { dy } } { \mathrm { dx } } = \frac { 4 \mathrm { Int } - 2 } { \sqrt { \mathrm { t } } }\).
2. Find the exact gradient of the curve at \(B\).
3. Find the exact coordinates of \(M\).

5 \includegraphics[max width=\textwidth, alt={}, center]{83d0697c-b133-47da-a745-dfdafa7dbf10-08_663_433_260_854} The diagram shows the curve with parametric equations $$x = \ln ( 2 t + 3 ) , \quad y = \frac { 2 t - 3 } { 2 t + 3 } .$$ The curve crosses the \(y\)-axis at the point \(A\) and the \(x\)-axis at the point \(B\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 6 } { 2 t + 3 }\).
2. Find the gradient of the curve at \(A\).
3. Find the gradient of the curve at \(B\).

5 \includegraphics[max width=\textwidth, alt={}, center]{6294c4f4-70a9-4b81-87e0-20e2cc24dd27-08_663_433_260_854} The diagram shows the curve with parametric equations $$x = \ln ( 2 t + 3 ) , \quad y = \frac { 2 t - 3 } { 2 t + 3 }$$ The curve crosses the \(y\)-axis at the point \(A\) and the \(x\)-axis at the point \(B\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 6 } { 2 t + 3 }\).
2. Find the gradient of the curve at \(A\).
3. Find the gradient of the curve at \(B\).

6 Th \(\mathbf { p }\) rametric eq tion \(\mathbf { 6 }\) a cn \(\mathbf { e }\) are $$x = \mathrm { e } ^ { 2 t } , \quad y = 4 t \mathrm { e } ^ { t }$$

1. Stw th \(t \frac { d y } { d x } = \frac { 2 ( t + 1 ) } { e ^ { t } }\).
2. Fid b eq tin th \(\mathbf { n }\) mal to \(\mathbf { b }\) cn at te \(\dot { \mathbf { p } }\) n wh re \(t = 0\) [4]

7 The parametric equations of a curve are $$x = 2 \theta - \sin 2 \theta , \quad y = 2 - \cos 2 \theta$$

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \cot \theta\).
2. Find the equation of the tangent to the curve at the point where \(\theta = \frac { 1 } { 4 } \pi\).
3. For the part of the curve where \(0 < \theta < 2 \pi\), find the coordinates of the points where the tangent is parallel to the \(x\)-axis.

6 The parametric equations of a curve are $$x = 2 t + \ln t , \quad y = t + \frac { 4 } { t }$$ where \(t\) takes all positive values.

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { t ^ { 2 } - 4 } { t ( 2 t + 1 ) }\).
2. Find the equation of the tangent to the curve at the point where \(t = 1\).
3. The curve has one stationary point. Find the \(y\)-coordinate of this point, and determine whether this point is a maximum or a minimum.

4 The parametric equations of a curve are $$x = 4 \sin \theta , \quad y = 3 - 2 \cos 2 \theta$$ where \(- \frac { 1 } { 2 } \pi < \theta < \frac { 1 } { 2 } \pi\). Express \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\), simplifying your answer as far as possible.

5 The parametric equations of a curve are $$x = \mathrm { e } ^ { 2 t } , \quad y = 4 t \mathrm { e } ^ { t }$$

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 ( t + 1 ) } { \mathrm { e } ^ { t } }\).
2. Find the equation of the normal to the curve at the point where \(t = 0\).

5 A curve is defined by the parametric equations $$x = 2 \tan \theta , \quad y = 3 \sin 2 \theta$$ for \(0 \leqslant \theta < \frac { 1 } { 2 } \pi\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 6 \cos ^ { 4 } \theta - 3 \cos ^ { 2 } \theta\).
2. Find the coordinates of the stationary point.
3. Find the gradient of the curve at the point \(\left( 2 \sqrt { } 3 , \frac { 3 } { 2 } \sqrt { } 3 \right)\).

7 \includegraphics[max width=\textwidth, alt={}, center]{a07e6d2f-ded1-4c62-957b-41fb94b46a2d-3_423_837_1352_651} The diagram shows the curve with parametric equations $$x = 2 - \cos t , \quad y = 1 + 3 \cos 2 t$$ for \(0 < t < \pi\). The minimum point is \(M\) and the curve crosses the \(x\)-axis at points \(P\) and \(Q\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - 12 \cos t\).
2. Find the coordinates of \(M\).
3. Find the gradient of the curve at \(P\) and at \(Q\).

7 \includegraphics[max width=\textwidth, alt={}, center]{f85c4010-17b1-441c-ae8a-e77573d1b0c3-3_423_837_1352_651} The diagram shows the curve with parametric equations $$x = 2 - \cos t , \quad y = 1 + 3 \cos 2 t$$ for \(0 < t < \pi\). The minimum point is \(M\) and the curve crosses the \(x\)-axis at points \(P\) and \(Q\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - 12 \cos t\).
2. Find the coordinates of \(M\).
3. Find the gradient of the curve at \(P\) and at \(Q\).

8 \includegraphics[max width=\textwidth, alt={}, center]{de2f8bf3-fd03-4199-9eb2-c9cbac4d4385-10_549_495_258_824} The diagram shows the curve with parametric equations $$x = 2 - \cos 2 t , \quad y = 2 \sin ^ { 3 } t + 3 \cos ^ { 3 } t + 1$$ for \(0 \leqslant t \leqslant \frac { 1 } { 2 } \pi\). The end-points of the curve \(( 1,4 )\) and \(( 3,3 )\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 3 } { 2 } \sin t - \frac { 9 } { 4 } \cos t\).
2. Find the coordinates of the minimum point, giving each coordinate correct to 3 significant figures.
3. Find the exact gradient of the normal to the curve at the point for which \(x = 2\).

8 \includegraphics[max width=\textwidth, alt={}, center]{bdc467f6-105e-4429-95c6-701eaa43deff-10_549_495_258_824} The diagram shows the curve with parametric equations $$x = 2 - \cos 2 t , \quad y = 2 \sin ^ { 3 } t + 3 \cos ^ { 3 } t + 1$$ for \(0 \leqslant t \leqslant \frac { 1 } { 2 } \pi\). The end-points of the curve are \(( 1,4 )\) and \(( 3,3 )\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 3 } { 2 } \sin t - \frac { 9 } { 4 } \cos t\).
2. Find the coordinates of the minimum point, giving each coordinate correct to 3 significant figures.
3. Find the exact gradient of the normal to the curve at the point for which \(x = 2\).

3 The parametric equations of a curve are $$x = 2 \theta + \sin 2 \theta , \quad y = 1 - \cos 2 \theta$$ Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \tan \theta\).

3 The parametric equations of a curve are $$x = \sin 2 \theta - \theta , \quad y = \cos 2 \theta + 2 \sin \theta$$ Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 \cos \theta } { 1 + 2 \sin \theta }\).

4 The parametric equations of a curve are $$x = t - \tan t , \quad y = \ln ( \cos t )$$ for \(- \frac { 1 } { 2 } \pi < t < \frac { 1 } { 2 } \pi\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \cot t\).
2. Hence find the \(x\)-coordinate of the point on the curve at which the gradient is equal to 2 . Give your answer correct to 3 significant figures.
3. The polynomial \(\mathrm { f } ( x )\) is of the form \(( x - 2 ) ^ { 2 } \mathrm {~g} ( x )\), where \(\mathrm { g } ( x )\) is another polynomial. Show that \(( x - 2 )\) is a factor of \(\mathrm { f } ^ { \prime } ( x )\).
4. The polynomial \(x ^ { 5 } + a x ^ { 4 } + 3 x ^ { 3 } + b x ^ { 2 } + a\), where \(a\) and \(b\) are constants, has a factor \(( x - 2 ) ^ { 2 }\). Using the factor theorem and the result of part (i), or otherwise, find the values of \(a\) and \(b\). [5]

4 The parametric equations of a curve are $$x = t + \cos t , \quad y = \ln ( 1 + \sin t )$$ where \(- \frac { 1 } { 2 } \pi < t < \frac { 1 } { 2 } \pi\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \sec t\).
2. Hence find the \(x\)-coordinates of the points on the curve at which the gradient is equal to 3 . Give your answers correct to 3 significant figures.

4 The parametric equations of a curve are $$x = a ( 2 \theta - \sin 2 \theta ) , \quad y = a ( 1 - \cos 2 \theta )$$ Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \cot \theta\).

2 The parametric equations of a curve are $$x = 3 \left( 1 + \sin ^ { 2 } t \right) , \quad y = 2 \cos ^ { 3 } t$$ Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\), simplifying your answer as far as possible.

8 \includegraphics[max width=\textwidth, alt={}, center]{6025cf1d-525e-4f12-9517-f20ef5fff2fa-3_698_1006_758_571} The diagram shows the curve with parametric equations $$x = \sin t + \cos t , \quad y = \sin ^ { 3 } t + \cos ^ { 3 } t$$ for \(\frac { 1 } { 4 } \pi < t < \frac { 5 } { 4 } \pi\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - 3 \sin t \cos t\).
2. Find the gradient of the curve at the origin.
3. Find the values of \(t\) for which the gradient of the curve is 1 , giving your answers correct to 2 significant figures.

4 The parametric equations of a curve are $$x = \mathrm { e } ^ { - t } \cos t , \quad y = \mathrm { e } ^ { - t } \sin t$$ Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \tan \left( t - \frac { 1 } { 4 } \pi \right)\).

4 The parametric equations of a curve are $$x = \frac { 1 } { \cos ^ { 3 } t } , \quad y = \tan ^ { 3 } t$$ where \(0 \leqslant t < \frac { 1 } { 2 } \pi\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \sin t\).
2. Hence show that the equation of the tangent to the curve at the point with parameter \(t\) is \(y = x \sin t - \tan t\).

2 A curve is defined for \(0 < \theta < \frac { 1 } { 2 } \pi\) by the parametric equations $$x = \tan \theta , \quad y = 2 \cos ^ { 2 } \theta \sin \theta$$ Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 6 \cos ^ { 5 } \theta - 4 \cos ^ { 3 } \theta\).

3 The parametric equations of a curve are $$x = 2 t + \sin 2 t , \quad y = \ln ( 1 - \cos 2 t )$$ Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \operatorname { cosec } 2 t\).

4 The parametric equations of a curve are $$x = 1 - \mathrm { e } ^ { - t } , \quad y = \mathrm { e } ^ { t } + \mathrm { e } ^ { - t }$$

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { e } ^ { 2 t } - 1\).
2. Hence find the exact value of \(t\) at the point on the curve at which the gradient is 2 .