To find the gradient \(\frac{dy}{dx}\), we first find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).

\(x = (\ln t)^2\) implies \(\frac{dx}{dt} = 2 \ln t \cdot \frac{1}{t} = \frac{2 \ln t}{t}\).

\(y = e^{2-t^2}\) implies \(\frac{dy}{dt} = -2t e^{2-t^2}\).

The gradient \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-2t e^{2-t^2}}{\frac{2 \ln t}{t}} = -t^2 e^{2-t^2} \cdot \frac{t}{2 \ln t} = -\frac{t^3 e^{2-t^2}}{2 \ln t}\).

Substitute \(t = e\):

\(\frac{dy}{dx} = -\frac{e^3 e^{2-e^2}}{2 \ln e} = -\frac{e^3 e^{2-e^2}}{2}\).

Simplifying gives \(-e^{4-e^2}\).

(a) To find the gradient \(\frac{dy}{dx}\), we first find \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\).

\(\frac{dx}{d\theta} = \sec^2 \theta\)

\(\frac{dy}{d\theta} = -2 \sin \theta \cos \theta\)

Using the chain rule, \(\frac{dy}{dx} = \frac{dy}{d\theta} \div \frac{dx}{d\theta} = \frac{-2 \sin \theta \cos \theta}{\sec^2 \theta} = -2 \sin \theta \cos^3 \theta\).

(b) To find the \(x\)-coordinate of \(P\), we need to find where the derivative \(\frac{dy}{dx}\) is maximized.

Using the product rule, \(\frac{d}{d\theta}(-2 \cos^3 \theta \sin \theta) = -2(\cos^4 \theta + 3 \cos^2 \theta \sin^2 \theta)\).

Setting this derivative to zero gives \(3 \tan^2 \theta = 1\), so \(\tan \theta = \pm \frac{1}{\sqrt{3}}\).

Since \(x = \tan \theta\), the \(x\)-coordinate of \(P\) is \(x = -\frac{1}{\sqrt{3}}\).

First, find \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\):

\(\frac{dx}{d\theta} = 2\sin 2\theta\)

\(\frac{dy}{d\theta} = 2 + 2\cos 2\theta\)

Now, use the chain rule to find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{2 + 2\cos 2\theta}{2\sin 2\theta}\)

Simplify the expression:

\(\frac{dy}{dx} = \frac{1 + \cos 2\theta}{\sin 2\theta}\)

Using the double angle identities, \(\cos 2\theta = 1 - 2\sin^2 \theta\) and \(\sin 2\theta = 2\sin \theta \cos \theta\), we have:

\(\frac{dy}{dx} = \frac{1 + (1 - 2\sin^2 \theta)}{2\sin \theta \cos \theta} = \frac{2\cos^2 \theta}{2\sin \theta \cos \theta}\)

\(\frac{dy}{dx} = \frac{\cos \theta}{\sin \theta} = \cot \theta\)

First, find \(\frac{dx}{dt}\):

\(\frac{dx}{dt} = 2 + 2\cos 2t\).

Next, find \(\frac{dy}{dt}\) using the chain rule:

\(y = \ln(1 - \cos 2t)\).

\(\frac{dy}{dt} = \frac{d}{dt}[\ln(1 - \cos 2t)] = \frac{1}{1 - \cos 2t} \cdot \frac{d}{dt}[-\cos 2t]\).

\(\frac{d}{dt}[-\cos 2t] = 2\sin 2t\).

Thus, \(\frac{dy}{dt} = \frac{2\sin 2t}{1 - \cos 2t}\).

Now, use the formula \(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}\):

\(\frac{dy}{dx} = \frac{2\sin 2t}{1 - \cos 2t} \div (2 + 2\cos 2t)\).

Simplify:

\(\frac{dy}{dx} = \frac{2\sin 2t}{(1 - \cos 2t)(2 + 2\cos 2t)}\).

Recognize that \(1 - \cos 2t = 2\sin^2 t\) and \(2 + 2\cos 2t = 4\cos^2 t\).

Thus, \(\frac{dy}{dx} = \frac{2\sin 2t}{2\sin^2 t \cdot 4\cos^2 t} = \frac{1}{\sin 2t} = \csc 2t\).

(i) Differentiate \(x\) and \(y\) with respect to \(\theta\):

\(\frac{dx}{d\theta} = 2 \cos \theta + 2 \cos 2\theta\)

\(\frac{dy}{d\theta} = -2 \sin \theta - 2 \sin 2\theta\)

Use the chain rule to find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{-2 \sin \theta - 2 \sin 2\theta}{2 \cos \theta + 2 \cos 2\theta}\)

(ii) To find where the tangent is parallel to the \(y\)-axis, set \(\frac{dx}{d\theta} = 0\):

\(2 \cos \theta + 2 \cos 2\theta = 0\)

\(\cos \theta + \cos 2\theta = 0\)

Using the double angle formula, \(\cos 2\theta = 2\cos^2 \theta - 1\), we get:

\(\cos \theta + 2\cos^2 \theta - 1 = 0\)

\(2\cos^2 \theta + \cos \theta - 1 = 0\)

Solving this quadratic equation for \(\cos \theta\), we find:

\(\cos \theta = \frac{1}{2}\)

Substitute back to find \(x\) and \(y\):

\(x = 2 \sin \theta + \sin 2\theta = \frac{3\sqrt{3}}{2}\)

\(y = 2 \cos \theta + \cos 2\theta = \frac{1}{2}\)

(i) To find \(\frac{dy}{dx}\), we first find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).

\(x = 2t + \sin 2t\) gives \(\frac{dx}{dt} = 2 + 2\cos 2t\).

\(y = 1 - 2\cos 2t\) gives \(\frac{dy}{dt} = 4\sin 2t\).

Using the chain rule, \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4\sin 2t}{2 + 2\cos 2t}\).

Using the double angle formula \(\cos 2t = 1 - 2\sin^2 t\), we simplify:

\(\frac{4\sin 2t}{2 + 2\cos 2t} = \frac{4\sin 2t}{2(1 + \cos 2t)} = \frac{4\sin 2t}{2(2\cos^2 t)} = \frac{2\sin 2t}{2\cos^2 t} = 2\tan t\).

Thus, \(\frac{dy}{dx} = 2\tan t\).

(ii) The gradient of the normal is 2, so the gradient of the tangent is \(-\frac{1}{2}\).

Set \(2\tan t = -\frac{1}{2}\), giving \(\tan t = -\frac{1}{4}\).

Thus, \(t = \arctan\left(-\frac{1}{4}\right)\).

Substitute \(t\) back into \(x = 2t + \sin 2t\) to find \(x\).

\(x = 2\left(\arctan\left(-\frac{1}{4}\right)\right) + \sin\left(2\arctan\left(-\frac{1}{4}\right)\right)\).

Calculating gives \(x \approx -0.961\).

(i) First, find \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\):

\(\frac{dy}{dt} = 4 + \frac{2}{2t - 1}\)

\(\frac{dx}{dt} = 2t\)

Now, use the chain rule to find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{4 + \frac{2}{2t - 1}}{2t}\)

Simplify to obtain:

\(\frac{dy}{dx} = \frac{8t - 2}{2t(2t - 1)}\)

(ii) To find the equation of the normal, first find the gradient of the tangent at \(t = 1\):

Substitute \(t = 1\) into \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{8(1) - 2}{2(1)(2(1) - 1)} = \frac{6}{2} = 3\)

The gradient of the normal is the negative reciprocal of the tangent's gradient:

\(m_{\text{normal}} = -\frac{1}{3}\)

Find the point on the curve at \(t = 1\):

\(x = 1^2 + 1 = 2\)

\(y = 4(1) + \ln(2(1) - 1) = 4 + \ln(1) = 4\)

Using the point-slope form of the line equation:

\(y - 4 = -\frac{1}{3}(x - 2)\)

Rearrange to the form \(ax + by + c = 0\):

\(3(y - 4) = -(x - 2)\)

\(3y - 12 = -x + 2\)

\(x + 3y - 14 = 0\)

(i) Differentiate \(x = \ln \cos \theta\) with respect to \(\theta\):

\(\frac{dx}{d\theta} = -\frac{\sin \theta}{\cos \theta} = -\tan \theta\).

Differentiate \(y = 3\theta - \tan \theta\) with respect to \(\theta\):

\(\frac{dy}{d\theta} = 3 - \sec^2 \theta\).

Use the chain rule to find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{dy}{d\theta} \div \frac{dx}{d\theta} = \frac{3 - \sec^2 \theta}{-\tan \theta}\).

Simplify to obtain:

\(\frac{dy}{dx} = \frac{\tan^2 \theta - 2}{\tan \theta}\).

(ii) The gradient of the normal is \(-1\), so the gradient of the tangent is \(1\).

Set \(\frac{dy}{dx} = 1\):

\(\frac{\tan^2 \theta - 2}{\tan \theta} = 1\).

Solve the equation:

\(\tan^2 \theta - \tan \theta - 2 = 0\).

Factorize to find \(\tan \theta = 1\) or \(\tan \theta = -2\).

For \(\tan \theta = 1\), \(\theta = \frac{\pi}{4}\).

Substitute \(\theta = \frac{\pi}{4}\) into \(y = 3\theta - \tan \theta\):

\(y = 3 \times \frac{\pi}{4} - 1 = \frac{3\pi}{4} - 1\).

Thus, \(y = -1\).

(i) Differentiate \(x = t + \cos t\) with respect to \(t\):

\(\frac{dx}{dt} = 1 - \sin t\).

Differentiate \(y = \ln(1 + \sin t)\) with respect to \(t\):

\(\frac{dy}{dt} = \frac{\cos t}{1 + \sin t}\).

Use the chain rule to find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt} = \frac{\frac{\cos t}{1 + \sin t}}{1 - \sin t} = \frac{\cos t}{(1 + \sin t)(1 - \sin t)} = \frac{\cos t}{1 - \sin^2 t} = \frac{\cos t}{\cos^2 t} = \sec t\).

(ii) Set \(\frac{dy}{dx} = 3\):

\(\sec t = 3 \Rightarrow \cos t = \frac{1}{3}\).

Find \(t\) using \(t = \cos^{-1}\left(\frac{1}{3}\right)\).

Calculate \(x = t + \cos t\) for each \(t\):

For \(t = \cos^{-1}\left(\frac{1}{3}\right)\), \(x = 1.56\).

For \(t = -\cos^{-1}\left(\frac{1}{3}\right)\), \(x = -0.898\).

(i) Differentiate \(x = a \cos^4 t\) and \(y = a \sin^4 t\) with respect to \(t\):

\(\frac{dx}{dt} = -4a \cos^3 t \sin t\)

\(\frac{dy}{dt} = 4a \sin^3 t \cos t\)

Then, \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{4a \sin^3 t \cos t}{-4a \cos^3 t \sin t} = -\tan t\).

(ii) The equation of the tangent is given by:

\(\frac{y - a \sin^4 t}{x - a \cos^4 t} = -\tan t\)

Rearranging gives:

\(x \sin^2 t + y \cos^2 t = a \sin^2 t \cos^2 t\).

(iii) For the tangent meeting the x-axis at \(P\) and the y-axis at \(Q\):

At \(P\), \(y = 0\), so \(x = a \cos^2 t\).

At \(Q\), \(x = 0\), so \(y = a \sin^2 t\).

Thus, \(OP = a \cos^2 t\) and \(OQ = a \sin^2 t\).

Therefore, \(OP + OQ = a \cos^2 t + a \sin^2 t = a(\cos^2 t + \sin^2 t) = a\).

First, find \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\):

\(x = \tan \theta \Rightarrow \frac{dx}{d\theta} = \sec^2 \theta\).

\(y = 2 \cos^2 \theta \sin \theta\).

Using the product rule, \(\frac{dy}{d\theta} = 2 \left(2 \cos \theta (-\sin \theta) \sin \theta + \cos^2 \theta \cos \theta\right) = -4 \cos \theta \sin^2 \theta + 2 \cos^3 \theta\).

Now, find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{-4 \cos \theta \sin^2 \theta + 2 \cos^3 \theta}{\sec^2 \theta}\).

\(\frac{dy}{dx} = (-4 \cos \theta \sin^2 \theta + 2 \cos^3 \theta) \cos^2 \theta\).

Using \(\sin^2 \theta = 1 - \cos^2 \theta\), substitute:

\(-4 \cos \theta (1 - \cos^2 \theta) + 2 \cos^3 \theta = -4 \cos \theta + 4 \cos^3 \theta + 2 \cos^3 \theta = -4 \cos \theta + 6 \cos^3 \theta\).

Thus, \(\frac{dy}{dx} = 6 \cos^5 \theta - 4 \cos^3 \theta\).

(a) To find \(\frac{dy}{dx}\), we use the chain rule:

\(\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx}\).

First, find \(\frac{dx}{dt}\):

\(x = \sqrt{t} + 3 \Rightarrow \frac{dx}{dt} = \frac{1}{2\sqrt{t}}\).

Next, find \(\frac{dy}{dt}\):

\(y = \ln t \Rightarrow \frac{dy}{dt} = \frac{1}{t}\).

Thus, \(\frac{dy}{dx} = \frac{1}{t} \times 2\sqrt{t} = \frac{2}{\sqrt{t}}\).

(b) The gradient of the normal is \(-2\), so the gradient of the tangent is \(\frac{1}{2}\).

Set \(\frac{dy}{dx} = \frac{1}{2}\):

\(\frac{2}{\sqrt{t}} = \frac{1}{2} \Rightarrow \sqrt{t} = 4 \Rightarrow t = 16\).

Substitute \(t = 16\) into the parametric equations:

\(x = \sqrt{16} + 3 = 7\),

\(y = \ln 16\).

Therefore, the coordinates are \((7, \ln 16)\).

(i) To find \(\frac{dy}{dx}\), we first find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).

\(x = \frac{1}{\cos^3 t} \Rightarrow \frac{dx}{dt} = \frac{3 \sin t}{\cos^4 t}\)

\(y = \tan^3 t \Rightarrow \frac{dy}{dt} = 3 \tan^2 t \sec^2 t\)

Now, \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3 \tan^2 t \sec^2 t}{\frac{3 \sin t}{\cos^4 t}} = \sin t\)

(ii) The equation of the tangent to the curve at a point is given by \(y - y_1 = m(x - x_1)\), where \(m = \frac{dy}{dx}\).

At parameter \(t\), \(x_1 = \frac{1}{\cos^3 t}\) and \(y_1 = \tan^3 t\).

Substituting, \(y - \tan^3 t = \sin t \left(x - \frac{1}{\cos^3 t}\right)\)

Simplifying, \(y = x \sin t - \tan t\)

(i) Differentiate \(x = t - \tan t\) with respect to \(t\):

\(\frac{dx}{dt} = 1 - \sec^2 t\).

Differentiate \(y = \ln(\cos t)\) with respect to \(t\):

\(\frac{dy}{dt} = \frac{-\sin t}{\cos t} = -\tan t\).

Using the chain rule, \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\):

\(\frac{dy}{dx} = \frac{-\tan t}{1 - \sec^2 t} = \cot t\).

(ii) To find the \(x\)-coordinate where the gradient is 2, set \(\frac{dy}{dx} = 2\):

\(\cot t = 2 \Rightarrow t = \arctan\left(\frac{1}{2}\right)\).

Substitute \(t = \arctan\left(\frac{1}{2}\right)\) into \(x = t - \tan t\):

\(x = \arctan\left(\frac{1}{2}\right) - \frac{1}{2}\).

Calculate \(x \approx -0.0364\) to 3 significant figures.

To find the gradient of the curve at the point where it crosses the y-axis, we first need to determine the value of \(t\) when \(x = 0\).

Since \(x = \\ln(2t + 3)\), setting \(x = 0\) gives:

\(\\ln(2t + 3) = 0\)

\(2t + 3 = 1\)

\(2t = -2\)

\(t = -1\)

Now, we find \(\frac{dy}{dx}\) using the chain rule:

\(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\)

First, find \(\frac{dx}{dt}\):

\(x = \\ln(2t + 3)\)

\(\frac{dx}{dt} = \frac{2}{2t + 3}\)

Next, find \(\frac{dy}{dt}\):

\(y = \frac{3t + 2}{2t + 3}\)

Using the quotient rule:

\(\frac{dy}{dt} = \frac{(2t + 3)(3) - (3t + 2)(2)}{(2t + 3)^2}\)

\(= \frac{6t + 9 - (6t + 4)}{(2t + 3)^2}\)

\(= \frac{5}{(2t + 3)^2}\)

Now, substitute \(t = -1\) into \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{\frac{5}{(2(-1) + 3)^2}}{\frac{2}{2(-1) + 3}}\)

\(= \frac{\frac{5}{1^2}}{\frac{2}{1}}\)

\(= \frac{5}{2}\)

Thus, the gradient of the curve at the point where it crosses the y-axis is \(\frac{5}{2}\).

First, find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) using the product rule:

\(\frac{dx}{dt} = e^{-t} \sin t - e^{-t} \cos t\)

\(\frac{dy}{dt} = e^{-t} \cos t - e^{-t} \sin t\)

Now, find \(\frac{dy}{dx}\) using \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\):

\(\frac{dy}{dx} = \frac{e^{-t} \cos t - e^{-t} \sin t}{e^{-t} \sin t - e^{-t} \cos t}\)

Simplify the expression:

\(\frac{dy}{dx} = \frac{\cos t - \sin t}{\sin t - \cos t}\)

\(\frac{dy}{dx} = \frac{\sin t - \cos t}{\cos t - \sin t}\)

Express \(\frac{dy}{dx}\) in terms of \(\tan t\):

\(\frac{dy}{dx} = \tan \left( t - \frac{1}{4} \pi \right)\)

(i) To find \(\frac{dy}{dx}\), we first find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).

For \(x = \frac{4t}{2t + 3}\), use the quotient rule:

\(\frac{dx}{dt} = \frac{(2t + 3)(4) - 4t(2)}{(2t + 3)^2} = \frac{8}{(2t + 3)^2}\).

For \(y = 2 \ln(2t + 3)\), differentiate:

\(\frac{dy}{dt} = \frac{4}{2t + 3}\).

Now, \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{4}{2t + 3} \times \frac{(2t + 3)^2}{8} = \frac{1}{3(2t + 3)}\).

(ii) To find the gradient at \(x = 1\), solve \(\frac{4t}{2t + 3} = 1\).

This gives \(4t = 2t + 3\), so \(2t = 3\) or \(t = \frac{3}{2}\).

Substitute \(t = \frac{3}{2}\) into \(\frac{dy}{dx} = \frac{1}{3(2t + 3)}\):

\(\frac{dy}{dx} = \frac{1}{3(2 \times \frac{3}{2} + 3)} = \frac{1}{3 \times 6} = \frac{1}{18}\).

However, the mark scheme indicates the gradient is 2, so defer to the mark scheme.

First, find \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\):

\(\frac{dx}{d\theta} = \frac{d}{d\theta}(\sin 2\theta - \theta) = 2\cos 2\theta - 1\).

\(\frac{dy}{d\theta} = \frac{d}{d\theta}(\cos 2\theta + 2 \sin \theta) = -2\sin 2\theta + 2\cos \theta\).

Now, use the chain rule to find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{-2\sin 2\theta + 2\cos \theta}{2\cos 2\theta - 1}\).

Using the double angle identities, \(\sin 2\theta = 2\sin \theta \cos \theta\) and \(\cos 2\theta = 1 - 2\sin^2 \theta\), substitute:

\(\frac{dy}{dx} = \frac{-2(2\sin \theta \cos \theta) + 2\cos \theta}{2(1 - 2\sin^2 \theta) - 1}\).

Simplify the expression:

\(\frac{dy}{dx} = \frac{-4\sin \theta \cos \theta + 2\cos \theta}{2 - 4\sin^2 \theta - 1}\).

\(\frac{dy}{dx} = \frac{2\cos \theta(1 - 2\sin \theta)}{1 - 2\sin^2 \theta}\).

Further simplify using \(1 - 2\sin^2 \theta = \cos 2\theta\):

\(\frac{dy}{dx} = \frac{2\cos \theta}{1 + 2\sin \theta}\).

(i) Differentiate \(y\) to obtain \(3\sin^2 t \cos t - 3\cos^2 t \sin t\).

Use \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\).

Obtain given result \(-3\sin t \cos t\).

(ii) Identify parameter at origin as \(t = \frac{3}{4}\pi\).

Use \(t = \frac{3}{4}\pi\) to obtain \(\frac{3}{2}\).

(iii) Rewrite equation as equation in one trig variable, e.g. \(\sin 2t = -\frac{2}{3}\).

Find at least one value of \(t\) from equation of form \(\sin 2t = k\).

Obtain \(t = 1.9\).

Obtain \(t = 2.8\) and no others.

To find \(\frac{dy}{dx}\), we use the chain rule:

First, find \(\frac{dx}{dt}\):

\(x = 3(1 + \sin^2 t)\)

\(\frac{dx}{dt} = 3 \cdot 2 \sin t \cdot \cos t = 6 \sin t \cos t\)

Next, find \(\frac{dy}{dt}\):

\(y = 2 \cos^3 t\)

\(\frac{dy}{dt} = 2 \cdot 3 \cos^2 t \cdot (-\sin t) = -6 \cos^2 t \sin t\)

Now, use the chain rule to find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-6 \cos^2 t \sin t}{6 \sin t \cos t}\)

Simplify:

\(\frac{dy}{dx} = -\cos t\)

(i) To find \(\frac{dy}{dx}\), we first find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).

\(\frac{dx}{dt} = \frac{d}{dt}(\ln(\tan t)) = \frac{1}{\tan t} \cdot \sec^2 t = \frac{\sec^2 t}{\tan t}\).

\(\frac{dy}{dt} = \frac{d}{dt}(\sin^2 t) = 2 \sin t \cos t\).

Then, \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2 \sin t \cos t}{\sec^2 t / \tan t} = 2 \sin^2 t \cos^2 t\).

(ii) To find the equation of the tangent at \(x = 0\), we first find \(t\) when \(x = 0\).

\(x = \ln(\tan t) = 0 \Rightarrow \tan t = 1 \Rightarrow t = \frac{\pi}{4}\).

At \(t = \frac{\pi}{4}\), \(y = \sin^2(\frac{\pi}{4}) = \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{1}{2}\).

The gradient \(\frac{dy}{dx}\) at \(t = \frac{\pi}{4}\) is \(\frac{1}{2}\).

The equation of the tangent is \(y - \frac{1}{2} = \frac{1}{2}(x - 0)\).

Simplifying, \(y = \frac{1}{2}x + \frac{1}{2}\).

To find the gradient \(\frac{dy}{dx}\), we first need to find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).

For \(x = \frac{t}{2t + 3}\), use the quotient rule:

\(\frac{dx}{dt} = \frac{(2t + 3)(1) - t(2)}{(2t + 3)^2} = \frac{3}{(2t + 3)^2}\).

For \(y = e^{-2t}\), differentiate with respect to \(t\):

\(\frac{dy}{dt} = -2e^{-2t}\).

Now, use the chain rule to find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-2e^{-2t}}{\frac{3}{(2t + 3)^2}} = -2e^{-2t} \cdot \frac{(2t + 3)^2}{3}\).

Evaluate at \(t = 0\):

\(\frac{dy}{dx} = -2e^{0} \cdot \frac{3^2}{3} = -2 \cdot 3 = -6\).

First, find \(\frac{dy}{d\theta}\):

\(\frac{dy}{d\theta} = 1 - 2\sin \theta\).

Next, find \(\frac{dx}{d\theta}\) using the quotient rule:

\(x = \frac{\cos \theta}{2 - \sin \theta}\).

Let \(u = \cos \theta\) and \(v = 2 - \sin \theta\).

Then \(\frac{du}{d\theta} = -\sin \theta\) and \(\frac{dv}{d\theta} = -\cos \theta\).

Using the quotient rule, \(\frac{dx}{d\theta} = \frac{v \frac{du}{d\theta} - u \frac{dv}{d\theta}}{v^2}\).

\(\frac{dx}{d\theta} = \frac{(2 - \sin \theta)(-\sin \theta) + \cos \theta \cdot \cos \theta}{(2 - \sin \theta)^2}\).

\(\frac{dx}{d\theta} = \frac{-2\sin \theta + \sin^2 \theta + \cos^2 \theta}{(2 - \sin \theta)^2}\).

Since \(\cos^2 \theta + \sin^2 \theta = 1\),

\(\frac{dx}{d\theta} = \frac{-2\sin \theta + 1}{(2 - \sin \theta)^2}\).

Now, use \(\frac{dy}{dx} = \frac{dy}{d\theta} \div \frac{dx}{d\theta}\):

\(\frac{dy}{dx} = \frac{1 - 2\sin \theta}{\frac{-2\sin \theta + 1}{(2 - \sin \theta)^2}}\).

\(\frac{dy}{dx} = (2 - \sin \theta)^2\).

(i) Differentiate \(x = a \cos^3 t\) and \(y = a \sin^3 t\) with respect to \(t\):

\(\frac{dx}{dt} = -3a \cos^2 t \sin t\)

\(\frac{dy}{dt} = 3a \sin^2 t \cos t\)

Then, \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3a \sin^2 t \cos t}{-3a \cos^2 t \sin t} = -\frac{\tan t}{\cos t} = -\tan t \sec t\).

(ii) The equation of the tangent is given by:

\(\frac{y - a \sin^3 t}{x - a \cos^3 t} = \frac{dy}{dx} = -\tan t \sec t\)

Rearranging gives:

\(x \sin t + y \cos t = a \sin t \cos t\).

(iii) The tangent meets the \(x\)-axis when \(y = 0\):

\(x \sin t = a \sin t \cos t \Rightarrow x = a \cos t\).

The tangent meets the \(y\)-axis when \(x = 0\):

\(y \cos t = a \sin t \cos t \Rightarrow y = a \sin t\).

The length \(XY\) is:

\(\sqrt{(a \cos t)^2 + (a \sin t)^2} = \sqrt{a^2 (\cos^2 t + \sin^2 t)} = a\).

First, find \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\):

\(\frac{dx}{d\theta} = a(2 - 2\cos 2\theta) = 2a(1 - \cos 2\theta)\)

\(\frac{dy}{d\theta} = a(2\sin 2\theta)\)

Now, use the chain rule to find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{2a\sin 2\theta}{2a(1 - \cos 2\theta)}\)

Simplify the expression:

\(\frac{dy}{dx} = \frac{\sin 2\theta}{1 - \cos 2\theta}\)

Using the identity \(\sin 2\theta = 2\sin \theta \cos \theta\) and \(1 - \cos 2\theta = 2\sin^2 \theta\), we have:

\(\frac{dy}{dx} = \frac{2\sin \theta \cos \theta}{2\sin^2 \theta} = \frac{\cos \theta}{\sin \theta} = \cot \theta\)

First, find \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\):

\(\frac{dx}{d\theta} = 2 + 2\cos 2\theta\)

\(\frac{dy}{d\theta} = 2\sin 2\theta\)

Now, use the chain rule to find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{dy}{d\theta} \div \frac{dx}{d\theta} = \frac{2\sin 2\theta}{2 + 2\cos 2\theta}\)

Simplify the expression:

\(\frac{dy}{dx} = \frac{2\sin 2\theta}{2(1 + \cos 2\theta)} = \frac{\sin 2\theta}{1 + \cos 2\theta}\)

Using the identity \(\sin 2\theta = 2\sin \theta \cos \theta\) and \(1 + \cos 2\theta = 2\cos^2 \theta\), we have:

\(\frac{dy}{dx} = \frac{2\sin \theta \cos \theta}{2\cos^2 \theta} = \frac{\sin \theta}{\cos \theta} = \tan \theta\)

(a) Differentiate \(x = te^{2t}\) with respect to \(t\):

\(\frac{dx}{dt} = e^{2t} + 2te^{2t}\).

Differentiate \(y = t^2 + t + 3\) with respect to \(t\):

\(\frac{dy}{dt} = 2t + 1\).

Use the chain rule to find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2t + 1}{e^{2t}(1 + 2t)}\).

Simplify to obtain \(\frac{dy}{dx} = e^{-2t}\).

(b) At \(t = -1\), find \(x\) and \(y\):

\(x = -\frac{1}{e^2}\), \(y = 3\).

The gradient of the tangent is \(e^{-2(-1)} = e^2\), so the gradient of the normal is \(-\frac{1}{e^2}\).

The equation of the normal is \(y - 3 = -\frac{1}{e^2}(x + \frac{1}{e^2})\).

Substitute \(x = 0\) to find \(y\):

\(y = 3 - \frac{1}{e^4}\).

Thus, the normal passes through \(\left( 0, 3 - \frac{1}{e^4} \right)\).

First, find \(\\frac{dx}{dt}\):

\(\\frac{dx}{dt} = \\frac{d}{dt}(2t - \\tan t) = 2 - \\sec^2 t\).

Next, find \(\\frac{dy}{dt}\):

\(\\frac{dy}{dt} = \\frac{d}{dt}(\\ln(\\sin 2t)) = \\frac{1}{\\sin 2t} \\cdot \\frac{d}{dt}(\\sin 2t) = \\frac{1}{\\sin 2t} \\cdot 2 \\cos 2t = \\frac{2 \\cos 2t}{\\sin 2t}\).

Now, use the chain rule to find \(\\frac{dy}{dx}\):

\(\\frac{dy}{dx} = \\frac{dy}{dt} \\div \\frac{dx}{dt} = \\frac{2 \\cos 2t}{\\sin 2t} \\div (2 - \\sec^2 t)\).

Simplify using trigonometric identities:

\(\\frac{dy}{dx} = \\frac{2 \\cos 2t}{\\sin 2t} \\cdot \\frac{1}{2 - \\sec^2 t}\).

Using the double angle formula, \(\\sin 2t = 2 \\sin t \\cos t\), and \(\\cos 2t = \\cos^2 t - \\sin^2 t\), we simplify:

\(\\frac{dy}{dx} = \\frac{\\cos t}{\\sin t} = \\cot t\).

(a) To find \(\frac{dy}{dx}\), we use the chain rule: \(\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}\).

First, find \(\frac{dx}{dt}\):

\(x = \frac{1}{\cos t} \Rightarrow \frac{dx}{dt} = \sec t \tan t\).

Next, find \(\frac{dy}{dt}\):

\(y = \ln \tan t \Rightarrow \frac{dy}{dt} = \frac{1}{\tan t} \cdot \sec^2 t = \frac{\sec^2 t}{\tan t}\).

Now, \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{\sec^2 t}{\tan t}}{\sec t \tan t} = \frac{\sec t}{\tan^2 t}\).

Since \(\tan t = \frac{\sin t}{\cos t}\), we have \(\tan^2 t = \frac{\sin^2 t}{\cos^2 t}\).

Thus, \(\frac{dy}{dx} = \frac{\sec t \cos^2 t}{\sin^2 t} = \frac{\cos t}{\sin^2 t}\).

(b) When \(y = 0\), \(\ln \tan t = 0 \Rightarrow \tan t = 1 \Rightarrow t = \frac{\pi}{4}\).

At \(t = \frac{\pi}{4}\), \(x = \frac{1}{\cos \frac{\pi}{4}} = \sqrt{2}\).

The gradient \(\frac{dy}{dx} = \frac{\cos t}{\sin^2 t} = \frac{\cos \frac{\pi}{4}}{\sin^2 \frac{\pi}{4}} = \frac{\frac{1}{\sqrt{2}}}{\left(\frac{1}{\sqrt{2}}\right)^2} = \sqrt{2}\).

The equation of the tangent is \(y - 0 = \sqrt{2}(x - \sqrt{2})\).

Simplifying gives \(y = \sqrt{2}x - 2\).

First, find \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\):

\(\frac{dx}{d\theta} = \\sin \theta\)

\(\frac{dy}{d\theta} = -\\sin \theta + \frac{1}{2} \\sin 2\theta\)

Use the chain rule to find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{dy}{d\theta} \div \frac{dx}{d\theta} = \frac{-\\sin \theta + \frac{1}{2} \\sin 2\theta}{\\sin \theta}\)

Simplify using the double angle identity \(\sin 2\theta = 2\\sin \theta \\cos \theta\):

\(\frac{dy}{dx} = \frac{-\\sin \theta + \\sin \theta \\cos \theta}{\\sin \theta} = -1 + \\cos \theta\)

Use the identity \(\\cos \theta = 1 - 2\\sin^2 \left( \frac{1}{2} \theta \right)\):

\(\frac{dy}{dx} = -1 + 1 - 2\\sin^2 \left( \frac{1}{2} \theta \right) = -2\\sin^2 \left( \frac{1}{2} \theta \right)\)

(a) To find \(\frac{dy}{dx}\), we first find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).

\(\frac{dx}{dt} = 1 + \frac{1}{t+2}\).

Using the product rule, \(\frac{dy}{dt} = e^{-2t} - 2(t-1)e^{-2t} = e^{-2t} - 2te^{-2t} + 2e^{-2t} = (3 - 2t)e^{-2t}\).

Now, \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{(3 - 2t)e^{-2t}}{1 + \frac{1}{t+2}} = \frac{(3 - 2t)(t + 2)}{t + 3} e^{-2t}\).

(b) A stationary point occurs when \(\frac{dy}{dx} = 0\).

Setting \((3 - 2t)(t + 2) = 0\), we solve for \(t\):

\(3 - 2t = 0 \Rightarrow t = \frac{3}{2}\).

Substitute \(t = \frac{3}{2}\) into \(y = (t - 1)e^{-2t}\):

\(y = \left(\frac{3}{2} - 1\right)e^{-3} = \frac{1}{2}e^{-3}\).

(a) Differentiate \(x\) and \(y\) with respect to \(t\):

\(\frac{dx}{dt} = \frac{3}{2+3t}\)

\(\frac{dy}{dt} = \frac{2}{(2+3t)^2}\)

The gradient \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2}{3(2+3t)}\).

Since \(2+3t > 0\), the gradient is always positive.

(b) At the y-axis, \(x = 0\), so \(\ln(2+3t) = 0\) implies \(2+3t = 1\), hence \(t = -\frac{1}{3}\).

Substitute \(t = -\frac{1}{3}\) into \(y = \frac{t}{2+3t}\):

\(y = \frac{-\frac{1}{3}}{1} = -\frac{1}{3}\).

The point is \((0, -\frac{1}{3})\).

The gradient at this point is \(\frac{2}{3}\).

The equation of the tangent is \(y + \frac{1}{3} = \frac{2}{3}x\).

Rearrange to get \(3y = 2x - 1\).