Parametric curve crosses axis, find gradient there

6 \includegraphics[max width=\textwidth, alt={}, center]{b104e2a7-06c8-4e2e-a4f9-5095ad56897a-10_803_394_269_863} The diagram shows the curve with parametric equations $$x = 3 \ln ( 2 t - 3 ) , \quad y = 4 t \ln t$$ The curve crosses the \(y\)-axis at the point \(A\). At the point \(B\), the gradient of the curve is 12 .

1. Find the exact gradient of the curve at \(A\).
2. Show that the value of the parameter \(t\) at \(B\) satisfies the equation $$t = \frac { 9 } { 1 + \ln t } + \frac { 3 } { 2 }$$
3. Use an iterative formula, based on the equation in (b), to find the value of \(t\) at \(B\), giving your answer correct to 3 significant figures. Use an initial value of 5 and give the result of each iteration to 5 significant figures.

3 The parametric equations of a curve are $$x = \ln ( 2 t + 3 ) , \quad y = \frac { 3 t + 2 } { 2 t + 3 }$$ Find the gradient of the curve at the point where it crosses the \(y\)-axis.

8. \begin{figure}[h] \end{figure} The curve shown in Figure 3 has parametric equations $$x = t - 4 \sin t , y = 1 - 2 \cos t , \quad - \frac { 2 \pi } { 3 } \leqslant t \leqslant \frac { 2 \pi } { 3 }$$ The point \(A\), with coordinates ( \(k , 1\) ), lies on the curve. Given that \(k > 0\)

1. find the exact value of \(k\),
2. find the gradient of the curve at the point \(A\). There is one point on the curve where the gradient is equal to \(- \frac { 1 } { 2 }\)
3. Find the value of \(t\) at this point, showing each step in your working and giving your answer to 4 decimal places.
   [0pt] [Solutions based entirely on graphical or numerical methods are not acceptable.]

10 The diagram shows the curve with equation $$x = ( y - 4 ) \ln ( 2 y + 3 ) .$$ The curve crosses the \(y\)-axis at \(A\) and \(B\). \includegraphics[max width=\textwidth, alt={}, center]{afc8561d-94ae-42c0-bc6c-e9b091938368-3_588_780_1087_680}

1. Find an expression for \(\frac { \mathrm { d } x } { \mathrm {~d} y }\) in terms of \(y\).
2. Find the exact gradient of the curve at each of the points \(A\) and \(B\).

A curve has parametric equations $$x = \frac{e^{2t} - 2}{e^{2t} + 1}, \quad y = e^{3t} + 1.$$

1. Find an expression for \(\frac{dy}{dx}\) in terms of \(t\). [4]
2. Find the exact gradient of the curve at the point where the curve crosses the \(y\)-axis. [3]

\includegraphics{figure_3} Figure 3 shows a sketch of the curve \(C\) with parametric equations $$x = 1 + 3\tan t, \quad y = 2\cos 2t, \quad -\frac{\pi}{6} \leq t \leq \frac{\pi}{3}$$ The curve crosses the \(x\)-axis at point \(P\), as shown in Figure 3.

1. Find the equation of the tangent to \(C\) at \(P\), writing your answer in the form \(y = mx + c\), where \(m\) and \(c\) are constants to be found. [5]

The curve \(C\) has equation \(y = f(x)\), where \(f\) is a function with domain \(\left[k, 1 + 3\sqrt{3}\right]\)

1. Find the exact value of the constant \(k\). [1]
2. Find the range of \(f\). [2]