Derivatives of Functions in Parametric Forms Questions in English

(C) Given $x = a(t - \sin t)$ and $y = a(1 - \cos t).$
First,differentiate $x$ with respect to $t$:
$\frac{dx}{dt} = a(1 - \cos t).$
Next,differentiate $y$ with respect to $t$:
$\frac{dy}{dt} = a(0 - (-\sin t)) = a \sin t.$
Now,find $\frac{dy}{dx}$ using the chain rule:
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a \sin t}{a(1 - \cos t)} = \frac{\sin t}{1 - \cos t}.$
Using trigonometric identities $\sin t = 2 \sin \left( \frac{t}{2} \right) \cos \left( \frac{t}{2} \right)$ and $1 - \cos t = 2 \sin^2 \left( \frac{t}{2} \right):$
$\frac{dy}{dx} = \frac{2 \sin (t/2) \cos (t/2)}{2 \sin^2 (t/2)} = \frac{\cos (t/2)}{\sin (t/2)} = \cot \left( \frac{t}{2} \right).$

(A) Given $x = a(\cos t + \log \tan \frac{t}{2})$ and $y = a \sin t$.
Differentiating $y$ with respect to $t$:
$\frac{dy}{dt} = a \cos t$ .....$(i)$
Differentiating $x$ with respect to $t$:
$\frac{dx}{dt} = a [-\sin t + \frac{1}{\tan(t/2)} \cdot \sec^2(t/2) \cdot \frac{1}{2}]$
$= a [-\sin t + \frac{\cos(t/2)}{\sin(t/2)} \cdot \frac{1}{\cos^2(t/2)} \cdot \frac{1}{2}]$
$= a [-\sin t + \frac{1}{2 \sin(t/2) \cos(t/2)}]$
$= a [-\sin t + \frac{1}{\sin t}] = a [\frac{1 - \sin^2 t}{\sin t}] = a \frac{\cos^2 t}{\sin t}$ .....$(ii)$
Now,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a \cos t}{a \cos^2 t / \sin t} = \frac{\sin t}{\cos t} = \tan t$.

(C) Given $\tan y = \frac{2t}{1 - t^2}$ and $\sin x = \frac{2t}{1 + t^2}.$
Let $t = \tan \theta.$
Then $\tan y = \frac{2 \tan \theta}{1 - \tan^2 \theta} = \tan(2\theta),$ which implies $y = 2\theta.$
Also,$\sin x = \frac{2 \tan \theta}{1 + \tan^2 \theta} = \sin(2\theta),$ which implies $x = 2\theta.$
Since $y = 2\theta$ and $x = 2\theta,$ we have $y = x.$
Therefore,$\frac{dy}{dx} = \frac{d}{dx}(x) = 1.$

(C) Given $x = \frac{1 - t^2}{1 + t^2}$ and $y = \frac{2t}{1 + t^2}$.
Substitute $t = \tan \theta$ in both equations:
$x = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \cos 2\theta$
$y = \frac{2 \tan \theta}{1 + \tan^2 \theta} = \sin 2\theta$
Differentiating both with respect to $\theta$:
$\frac{dx}{d\theta} = -2 \sin 2\theta$
$\frac{dy}{d\theta} = 2 \cos 2\theta$
Using the chain rule,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{2 \cos 2\theta}{-2 \sin 2\theta} = -\cot 2\theta$.
Since $x = \cos 2\theta$ and $y = \sin 2\theta$,we have $\frac{dy}{dx} = -\frac{\cos 2\theta}{\sin 2\theta} = -\frac{x}{y}$.

(D) Given $x = at^2$ and $y = 2at$.
First,find $\frac{dx}{dt} = 2at$ and $\frac{dy}{dt} = 2a$.
Then,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a}{2at} = \frac{1}{t}$.
Now,differentiate $\frac{dy}{dx}$ with respect to $x$:
$\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{1}{t}) = \frac{d}{dt}(\frac{1}{t}) \cdot \frac{dt}{dx}$.
Since $\frac{dt}{dx} = \frac{1}{dx/dt} = \frac{1}{2at}$,we have:
$\frac{d^2y}{dx^2} = (-\frac{1}{t^2}) \cdot \frac{1}{2at} = -\frac{1}{2at^3}$.

(A) Given $x = 2\cos t - \cos 2t$ and $y = 2\sin t - \sin 2t$.
Differentiating with respect to $t$:
$\frac{dx}{dt} = -2\sin t + 2\sin 2t$
$\frac{dy}{dt} = 2\cos t - 2\cos 2t$
Using the chain rule,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2\cos t - 2\cos 2t}{-2\sin t + 2\sin 2t} = \frac{\cos t - \cos 2t}{\sin 2t - \sin t}$.
At $t = \frac{\pi}{4}$:
$\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,$\cos \frac{\pi}{2} = 0$,$\sin \frac{\pi}{2} = 1$,$\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$.
Substituting these values:
$\frac{dy}{dx} = \frac{\frac{1}{\sqrt{2}} - 0}{1 - \frac{1}{\sqrt{2}}} = \frac{\frac{1}{\sqrt{2}}}{\frac{\sqrt{2}-1}{\sqrt{2}}} = \frac{1}{\sqrt{2}-1}$.
Rationalizing the denominator:
$\frac{1}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1} = \frac{\sqrt{2}+1}{2-1} = \sqrt{2} + 1$.

(A) Given $x = a \sin 2\theta (1 + \cos 2\theta ) = a(2 \sin \theta \cos \theta)(2 \cos^2 \theta) = 4a \sin \theta \cos^3 \theta$.
Given $y = b \cos 2\theta (1 - \cos 2\theta ) = b(2 \cos^2 \theta - 1)(2 \sin^2 \theta) = 2b \sin^2 \theta (2 \cos^2 \theta - 1)$.
Alternatively,using the chain rule:
$\frac{dx}{d\theta} = a[2 \cos 2\theta (1 + \cos 2\theta) + \sin 2\theta (-2 \sin 2\theta)] = 2a(\cos 2\theta + \cos^2 2\theta - \sin^2 2\theta) = 2a(\cos 2\theta + \cos 4\theta) = 4a \cos 3\theta \cos \theta$.
$\frac{dy}{d\theta} = b[-2 \sin 2\theta (1 - \cos 2\theta) + \cos 2\theta (2 \sin 2\theta)] = 2b(-\sin 2\theta + \sin 2\theta \cos 2\theta + \sin 2\theta \cos 2\theta) = 2b(\sin 4\theta - \sin 2\theta) = 4b \cos 3\theta \sin \theta$.
Therefore,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{4b \cos 3\theta \sin \theta}{4a \cos 3\theta \cos \theta} = \frac{b}{a} \tan \theta$.

(B) Given $x = \frac{3at}{1 + t^3}$ and $y = \frac{3at^2}{1 + t^3}$.
Observe that $y = tx$.
Differentiating both sides with respect to $x$,we get:
$\frac{dy}{dx} = t + x \frac{dt}{dx}$ ... $(i)$
Now,differentiate $x$ with respect to $t$ using the quotient rule:
$\frac{dx}{dt} = 3a \left[ \frac{(1 + t^3)(1) - t(3t^2)}{(1 + t^3)^2} \right] = \frac{3a(1 - 2t^3)}{(1 + t^3)^2}$.
Thus,$\frac{dt}{dx} = \frac{(1 + t^3)^2}{3a(1 - 2t^3)}$.
Substituting this into $(i)$:
$\frac{dy}{dx} = t + \left( \frac{3at}{1 + t^3} \right) \left( \frac{(1 + t^3)^2}{3a(1 - 2t^3)} \right)$
$\frac{dy}{dx} = t + \frac{t(1 + t^3)}{1 - 2t^3} = \frac{t(1 - 2t^3) + t + t^4}{1 - 2t^3} = \frac{t - 2t^4 + t + t^4}{1 - 2t^3} = \frac{2t - t^4}{1 - 2t^3} = \frac{t(2 - t^3)}{1 - 2t^3}$.

(B) Given $x = t + \frac{1}{t}$ and $y = t - \frac{1}{t}$.
First,find the derivatives with respect to $t$:
$\frac{dx}{dt} = 1 - \frac{1}{t^2} = \frac{t^2 - 1}{t^2}$
$\frac{dy}{dt} = 1 + \frac{1}{t^2} = \frac{t^2 + 1}{t^2}$
Now,find $\frac{dy}{dx}$ using the chain rule:
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{(t^2 + 1)/t^2}{(t^2 - 1)/t^2} = \frac{t^2 + 1}{t^2 - 1}$
Next,differentiate $\frac{dy}{dx}$ with respect to $x$:
$\frac{d^2y}{dx^2} = \frac{d}{dt}\left( \frac{t^2 + 1}{t^2 - 1} \right) \cdot \frac{dt}{dx}$
Using the quotient rule for $\frac{d}{dt}\left( \frac{t^2 + 1}{t^2 - 1} \right)$:
$= \frac{(2t)(t^2 - 1) - (t^2 + 1)(2t)}{(t^2 - 1)^2} = \frac{2t^3 - 2t - 2t^3 - 2t}{(t^2 - 1)^2} = \frac{-4t}{(t^2 - 1)^2}$
Since $\frac{dt}{dx} = \frac{1}{dx/dt} = \frac{t^2}{t^2 - 1}$,we have:
$\frac{d^2y}{dx^2} = \left( \frac{-4t}{(t^2 - 1)^2} \right) \cdot \left( \frac{t^2}{t^2 - 1} \right) = \frac{-4t^3}{(t^2 - 1)^3} = -4t^3(t^2 - 1)^{-3}$.

(B) Given $x = t^2$ and $y = t^3$.
First,find the first derivative $\frac{dy}{dx}$ using the chain rule for parametric equations:
$\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = 3t^2$.
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3t^2}{2t} = \frac{3}{2}t$.
Now,differentiate $\frac{dy}{dx}$ with respect to $x$ to find $\frac{d^2y}{dx^2}$:
$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{3}{2}t\right) = \frac{d}{dt}\left(\frac{3}{2}t\right) \cdot \frac{dt}{dx}$.
Since $\frac{dt}{dx} = \frac{1}{dx/dt} = \frac{1}{2t}$,we have:
$\frac{d^2y}{dx^2} = \frac{3}{2} \cdot \frac{1}{2t} = \frac{3}{4t}$.

(C) Given $x = a \sin \theta$ and $y = b \cos \theta$.
First,find the derivatives with respect to $\theta$:
$\frac{dx}{d\theta} = a \cos \theta$ and $\frac{dy}{d\theta} = -b \sin \theta$.
Now,find $\frac{dy}{dx}$ using the chain rule:
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{-b \sin \theta}{a \cos \theta} = -\frac{b}{a} \tan \theta$.
Next,differentiate $\frac{dy}{dx}$ with respect to $x$:
$\frac{d^2y}{dx^2} = \frac{d}{dx} \left( -\frac{b}{a} \tan \theta \right) = -\frac{b}{a} \sec^2 \theta \cdot \frac{d\theta}{dx}$.
Since $\frac{dx}{d\theta} = a \cos \theta$,we have $\frac{d\theta}{dx} = \frac{1}{a \cos \theta}$.
Substituting this value:
$\frac{d^2y}{dx^2} = -\frac{b}{a} \sec^2 \theta \cdot \frac{1}{a \cos \theta} = -\frac{b}{a^2} \sec^3 \theta$.

(C) Given $y = t^{10} + 1$ and $x = t^8 + 1.$
First,we express $t$ in terms of $x$: $t^8 = x - 1 \Rightarrow t^2 = (x - 1)^{1/4}.$
Substitute $t^2$ into the expression for $y$: $y = (t^2)^5 + 1 = ((x - 1)^{1/4})^5 + 1 = (x - 1)^{5/4} + 1.$
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{5}{4}(x - 1)^{5/4 - 1} = \frac{5}{4}(x - 1)^{1/4}.$
Differentiating again with respect to $x$:
$\frac{d^2y}{dx^2} = \frac{5}{4} \times \frac{1}{4}(x - 1)^{1/4 - 1} = \frac{5}{16}(x - 1)^{-3/4} = \frac{5}{16(x - 1)^{3/4}}.$
Since $x - 1 = t^8$,we have $(x - 1)^{3/4} = (t^8)^{3/4} = t^6.$
Therefore,$\frac{d^2y}{dx^2} = \frac{5}{16t^6}.$

(D) Given $x = a \cos^3 \theta$ and $y = a \sin^3 \theta$.
First,find the derivatives with respect to $\theta$:
$\frac{dx}{d\theta} = 3a \cos^2 \theta (-\sin \theta) = -3a \cos^2 \theta \sin \theta$.
$\frac{dy}{d\theta} = 3a \sin^2 \theta (\cos \theta) = 3a \sin^2 \theta \cos \theta$.
Now,find $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3a \sin^2 \theta \cos \theta}{-3a \cos^2 \theta \sin \theta} = -\frac{\sin \theta}{\cos \theta} = -\tan \theta$.
Then,$\left( \frac{dy}{dx} \right)^2 = (-\tan \theta)^2 = \tan^2 \theta$.
Finally,$\sqrt{1 + \left( \frac{dy}{dx} \right)^2} = \sqrt{1 + \tan^2 \theta} = \sqrt{\sec^2 \theta} = |\sec \theta|$.

(A) Given the parametric equations:
$x = a(t + \sin t)$
$y = a(1 - \cos t)$
We need to find $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.
First,differentiate $y$ with respect to $t$:
$\frac{dy}{dt} = \frac{d}{dt}[a(1 - \cos t)] = a(0 - (-\sin t)) = a \sin t$
Next,differentiate $x$ with respect to $t$:
$\frac{dx}{dt} = \frac{d}{dt}[a(t + \sin t)] = a(1 + \cos t)$
Now,calculate $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{a \sin t}{a(1 + \cos t)} = \frac{\sin t}{1 + \cos t}$
Using trigonometric identities $\sin t = 2 \sin(t/2) \cos(t/2)$ and $1 + \cos t = 2 \cos^2(t/2)$:
$\frac{dy}{dx} = \frac{2 \sin(t/2) \cos(t/2)}{2 \cos^2(t/2)} = \frac{\sin(t/2)}{\cos(t/2)} = \tan(t/2)$
Therefore,the correct option is $A$.

(B) Given $x = \frac{2t}{1 + t^2}$ and $y = \frac{1 - t^2}{1 + t^2}$.
Substitute $t = \tan \theta$.
Then $x = \frac{2 \tan \theta}{1 + \tan^2 \theta} = \sin 2\theta$.
And $y = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \cos 2\theta$.
Now,find the derivatives with respect to $\theta$:
$\frac{dx}{d\theta} = 2 \cos 2\theta$.
$\frac{dy}{d\theta} = -2 \sin 2\theta$.
Using the chain rule,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{-2 \sin 2\theta}{2 \cos 2\theta} = -\tan 2\theta$.
Since $\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta} = \frac{2t}{1 - t^2}$,
$\frac{dy}{dx} = -\left( \frac{2t}{1 - t^2} \right) = \frac{2t}{t^2 - 1}$.

(D) Given $\cos x = \frac{1}{\sqrt{1 + t^2}}$. Let $t = \tan \theta$,then $\cos x = \frac{1}{\sqrt{1 + \tan^2 \theta}} = \frac{1}{\sec \theta} = \cos \theta$. Thus,$x = \theta = \tan^{-1} t$.
Given $\sin y = \frac{t}{\sqrt{1 + t^2}}$. Substituting $t = \tan \theta$,we get $\sin y = \frac{\tan \theta}{\sec \theta} = \sin \theta$. Thus,$y = \theta = \tan^{-1} t$.
Since $x = \tan^{-1} t$ and $y = \tan^{-1} t$,we have $x = y$.
Therefore,$\frac{dy}{dx} = \frac{d}{dx}(x) = 1$.

(B) Given parametric equations are $x = a(\cos \theta + \theta \sin \theta )$ and $y = a(\sin \theta - \theta \cos \theta )$.
First,differentiate $x$ with respect to $\theta$:
$\frac{dx}{d\theta} = a[-\sin \theta + (\sin \theta + \theta \cos \theta)] = a\theta \cos \theta$.
Next,differentiate $y$ with respect to $\theta$:
$\frac{dy}{d\theta} = a[\cos \theta - (\cos \theta - \theta \sin \theta)] = a\theta \sin \theta$.
Now,find $\frac{dy}{dx}$ using the chain rule:
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a\theta \sin \theta}{a\theta \cos \theta} = \tan \theta$.

(A) Given $x = a \cos^4 \theta$ and $y = a \sin^4 \theta$.
Differentiating $y$ with respect to $\theta$:
$\frac{dy}{d\theta} = 4a \sin^3 \theta \cos \theta$.
Differentiating $x$ with respect to $\theta$:
$\frac{dx}{d\theta} = 4a \cos^3 \theta (-\sin \theta) = -4a \cos^3 \theta \sin \theta$.
Now,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{4a \sin^3 \theta \cos \theta}{-4a \cos^3 \theta \sin \theta} = -\frac{\sin^2 \theta}{\cos^2 \theta} = -\tan^2 \theta$.
At $\theta = \frac{3\pi}{4}$,$\tan(\frac{3\pi}{4}) = -1$.
Therefore,$\frac{dy}{dx} = -(\tan(\frac{3\pi}{4}))^2 = -(-1)^2 = -1$.

(D) Given $x = \sin^{-1}(3t - 4t^3)$ and $y = \cos^{-1}(\sqrt{1 - t^2})$.
For $y = \cos^{-1}(\sqrt{1 - t^2})$,let $t = \sin \theta$,then $\sqrt{1 - t^2} = \cos \theta$.
So,$y = \cos^{-1}(\cos \theta) = \theta = \sin^{-1} t$.
For $x = \sin^{-1}(3t - 4t^3)$,let $t = \sin \theta$,then $x = \sin^{-1}(3 \sin \theta - 4 \sin^3 \theta) = \sin^{-1}(\sin 3 \theta) = 3 \theta = 3 \sin^{-1} t$.
Now,differentiate $x$ and $y$ with respect to $t$:
$\frac{dx}{dt} = 3 \cdot \frac{1}{\sqrt{1 - t^2}}$
$\frac{dy}{dt} = 1 \cdot \frac{1}{\sqrt{1 - t^2}}$
Therefore,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{1/\sqrt{1 - t^2}}{3/\sqrt{1 - t^2}} = \frac{1}{3}$.

(C) Given equations are $x = a(t - \frac{1}{t})$ $(i)$ and $y = a(t + \frac{1}{t})$ $(ii)$.
Squaring both equations,we get $x^2 = a^2(t^2 - 2 + \frac{1}{t^2})$ and $y^2 = a^2(t^2 + 2 + \frac{1}{t^2})$.
Subtracting $(i)^2$ from $(ii)^2$,we get $y^2 - x^2 = a^2(t^2 + 2 + \frac{1}{t^2} - (t^2 - 2 + \frac{1}{t^2})) = a^2(4) = 4a^2$.
Differentiating $y^2 - x^2 = 4a^2$ with respect to $x$,we get $2y \frac{dy}{dx} - 2x = 0$.
Therefore,$2y \frac{dy}{dx} = 2x$,which simplifies to $\frac{dy}{dx} = \frac{x}{y}$.

(C) Given $x = \sin t \cos 2t$ $(i)$ and $y = \cos t \sin 2t$ $(ii)$.
Differentiating $(i)$ with respect to $t$ using the product rule:
$\frac{dx}{dt} = \cos t \cos 2t - 2 \sin t \sin 2t$ $(iii)$
Differentiating $(ii)$ with respect to $t$ using the product rule:
$\frac{dy}{dt} = - \sin t \sin 2t + 2 \cos t \cos 2t$ $(iv)$
Now,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2 \cos t \cos 2t - \sin t \sin 2t}{\cos t \cos 2t - 2 \sin t \sin 2t}$.
At $t = \frac{\pi}{4}$,we have $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,$\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,$\cos 2(\frac{\pi}{4}) = \cos \frac{\pi}{2} = 0$,and $\sin 2(\frac{\pi}{4}) = \sin \frac{\pi}{2} = 1$.
Substituting these values:
$\frac{dy}{dx} = \frac{2(\frac{1}{\sqrt{2}})(0) - (\frac{1}{\sqrt{2}})(1)}{(\frac{1}{\sqrt{2}})(0) - 2(\frac{1}{\sqrt{2}})(1)} = \frac{-\frac{1}{\sqrt{2}}}{-\frac{2}{\sqrt{2}}} = \frac{1}{2}$.

(C) Let $y = {\cos ^{ - 1}}(\sqrt x )$ and $z = \sqrt {1 - x} $.
First,differentiate $y$ with respect to $x$:
$\frac{dy}{dx} = -\frac{1}{\sqrt{1 - (\sqrt{x})^2}} \times \frac{d}{dx}(\sqrt{x}) = -\frac{1}{\sqrt{1 - x}} \times \frac{1}{2\sqrt{x}}$.
Next,differentiate $z$ with respect to $x$:
$\frac{dz}{dx} = \frac{1}{2\sqrt{1 - x}} \times \frac{d}{dx}(1 - x) = \frac{1}{2\sqrt{1 - x}} \times (-1) = -\frac{1}{2\sqrt{1 - x}}$.
Now,find the differential coefficient of $y$ with respect to $z$:
$\frac{dy}{dz} = \frac{dy/dx}{dz/dx} = \frac{-\frac{1}{\sqrt{1 - x}} \times \frac{1}{2\sqrt{x}}}{-\frac{1}{2\sqrt{1 - x}}} = \frac{1}{\sqrt{x}}$.
Thus,the correct option is $C$.

(A) Let $y_1 = \sin^{-1}x$ and $y_2 = \cos^{-1}\sqrt{1-x^2}$.
For $y_2$,we know that $\cos^{-1}\sqrt{1-x^2} = \sin^{-1}x$ for $x \in [0, 1]$.
Thus,$y_2 = y_1$.
Now,differentiating both sides with respect to $x$:
$\frac{dy_1}{dx} = \frac{1}{\sqrt{1-x^2}}$
$\frac{dy_2}{dx} = \frac{d}{dx}(\sin^{-1}x) = \frac{1}{\sqrt{1-x^2}}$.
Therefore,the differential coefficient of $y_1$ with respect to $y_2$ is $\frac{dy_1}{dy_2} = \frac{dy_1/dx}{dy_2/dx} = \frac{1/\sqrt{1-x^2}}{1/\sqrt{1-x^2}} = 1$.

(B) Let $u = x^3$ and $v = x^2$.
We need to find the derivative of $u$ with respect to $v$,which is $\frac{du}{dv}$.
Using the chain rule,$\frac{du}{dv} = \frac{du/dx}{dv/dx}$.
First,differentiate $u$ with respect to $x$: $\frac{du}{dx} = \frac{d}{dx}(x^3) = 3x^2$.
Next,differentiate $v$ with respect to $x$: $\frac{dv}{dx} = \frac{d}{dx}(x^2) = 2x$.
Therefore,$\frac{du}{dv} = \frac{3x^2}{2x} = \frac{3}{2}x$.

(D) Let $y = \sin^2 x$ and $z = \cos^2 x$.
First,differentiate $y$ with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\sin^2 x) = 2 \sin x \cos x = \sin 2x$.
Next,differentiate $z$ with respect to $x$:
$\frac{dz}{dx} = \frac{d}{dx}(\cos^2 x) = 2 \cos x (-\sin x) = -2 \sin x \cos x = -\sin 2x$.
Now,find the derivative of $y$ with respect to $z$ using the chain rule:
$\frac{dy}{dz} = \frac{dy/dx}{dz/dx} = \frac{\sin 2x}{-\sin 2x} = -1$.
Thus,the correct option is $D$.

(A) Let $y = a \sin^3 t$ and $x = a \cos^3 t$.
First,find the derivatives with respect to $t$:
$\frac{dy}{dt} = 3a \sin^2 t \cos t$
$\frac{dx}{dt} = 3a \cos^2 t (-\sin t) = -3a \cos^2 t \sin t$
Now,find $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3a \sin^2 t \cos t}{-3a \cos^2 t \sin t} = -\tan t$
Next,differentiate $\frac{dy}{dx}$ with respect to $x$:
$\frac{d^2y}{dx^2} = \frac{d}{dx}(-\tan t) = -\sec^2 t \cdot \frac{dt}{dx}$
Since $\frac{dt}{dx} = \frac{1}{dx/dt} = \frac{1}{-3a \cos^2 t \sin t}$,we have:
$\frac{d^2y}{dx^2} = -\sec^2 t \cdot \frac{1}{-3a \cos^2 t \sin t} = \frac{\sec^2 t}{3a \cos^2 t \sin t} = \frac{1}{3a} \cdot \frac{\sec^4 t}{\sin t}$
At $t = \frac{\pi}{4}$,$\sec(\frac{\pi}{4}) = \sqrt{2}$ and $\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$:
$\left(\frac{d^2y}{dx^2}\right)_{t = \pi/4} = \frac{1}{3a} \cdot \frac{(\sqrt{2})^4}{1/\sqrt{2}} = \frac{1}{3a} \cdot \frac{4}{1/\sqrt{2}} = \frac{4\sqrt{2}}{3a}$

(C) Given $x = a \cos \theta + \frac{1}{2}b \cos 2\theta$ and $y = a \sin \theta + \frac{1}{2}b \sin 2\theta$.
First,find the derivatives with respect to $\theta$:
$\frac{dx}{d\theta} = -a \sin \theta - b \sin 2\theta$
$\frac{dy}{d\theta} = a \cos \theta + b \cos 2\theta$
Then,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a \cos \theta + b \cos 2\theta}{-a \sin \theta - b \sin 2\theta}$.
To find $\frac{d^2y}{dx^2}$,we use $\frac{d}{dx}(\frac{dy}{dx}) = \frac{d}{d\theta}(\frac{dy}{dx}) \cdot \frac{d\theta}{dx}$.
Setting $\frac{d^2y}{dx^2} = 0$ implies the numerator of $\frac{d}{d\theta}(\frac{dy}{dx})$ must be zero.
After simplification,this leads to the condition:
$a^2 + 2b^2 + 3ab(\cos 2\theta \cos \theta + \sin 2\theta \sin \theta) = 0$
Using the identity $\cos(A-B) = \cos A \cos B + \sin A \sin B$:
$a^2 + 2b^2 + 3ab \cos(2\theta - \theta) = 0$
$a^2 + 2b^2 + 3ab \cos \theta = 0$
Therefore,$\cos \theta = - \frac{a^2 + 2b^2}{3ab}$.

(D) The position coordinates are given by $x = a + bt - ct^2$ and $y = at + bt^2$.
First,we find the velocity components by differentiating with respect to $t$:
$v_x = \frac{dx}{dt} = b - 2ct$
$v_y = \frac{dy}{dt} = a + 2bt$
Next,we find the acceleration components by differentiating the velocity components with respect to $t$:
$a_x = \frac{dv_x}{dt} = \frac{d^2x}{dt^2} = -2c$
$a_y = \frac{dv_y}{dt} = \frac{d^2y}{dt^2} = 2b$
The resultant acceleration $a$ is given by the magnitude of the acceleration vector:
$a = \sqrt{a_x^2 + a_y^2}$
$a = \sqrt{(-2c)^2 + (2b)^2}$
$a = \sqrt{4c^2 + 4b^2}$
$a = 2\sqrt{b^2 + c^2}$

(B) Given the parametric equations of the curve are $x = t^2 + 3t - 8$ and $y = 2t^2 - 2t - 5$.
First,we find the value of $t$ at the point $(2, -1)$.
Substituting $x = 2$ into the equation for $x$: $2 = t^2 + 3t - 8 \implies t^2 + 3t - 10 = 0 \implies (t+5)(t-2) = 0$. Thus,$t = 2$ or $t = -5$.
Checking $t = 2$ in the equation for $y$: $y = 2(2)^2 - 2(2) - 5 = 8 - 4 - 5 = -1$. This matches the given point.
Now,calculate the derivatives with respect to $t$: $\frac{dx}{dt} = 2t + 3$ and $\frac{dy}{dt} = 4t - 2$.
The slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4t - 2}{2t + 3}$.
At $t = 2$,the slope is $\frac{4(2) - 2}{2(2) + 3} = \frac{8 - 2}{4 + 3} = \frac{6}{7}$.

(A) Given the parametric equations of the curve are $x = 3t^2 + 1$ and $y = t^3 - 1$.
First,we find the derivatives with respect to $t$:
$\frac{dx}{dt} = 6t$
$\frac{dy}{dt} = 3t^2$
Now,the slope of the tangent $\frac{dy}{dx}$ is given by:
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3t^2}{6t} = \frac{t}{2}$ (for $t \neq 0$).
At $x = 1$,we have $3t^2 + 1 = 1$,which implies $3t^2 = 0$,so $t = 0$.
As $t \to 0$,the slope $\frac{dy}{dx} = \frac{t}{2} \to 0$.
Thus,the slope of the tangent at $x = 1$ is $0$.

(C) Given the parametric equations $x = 2\cos^3\theta$ and $y = 3\sin^3\theta$.
At $\theta = \pi/4$,the coordinates are:
$x = 2(\cos(\pi/4))^3 = 2(1/\sqrt{2})^3 = 2/(2\sqrt{2}) = 1/\sqrt{2}$.
$y = 3(\sin(\pi/4))^3 = 3(1/\sqrt{2})^3 = 3/(2\sqrt{2})$.
Now,find the derivative $dy/dx = (dy/d\theta) / (dx/d\theta)$:
$dy/d\theta = 9\sin^2\theta \cos\theta$.
$dx/d\theta = -6\cos^2\theta \sin\theta$.
$dy/dx = (9\sin^2\theta \cos\theta) / (-6\cos^2\theta \sin\theta) = -3/2 \tan\theta$.
At $\theta = \pi/4$,$dy/dx = -3/2 \tan(\pi/4) = -3/2$.
The equation of the tangent is $y - y_1 = m(x - x_1)$:
$y - 3/(2\sqrt{2}) = -3/2(x - 1/\sqrt{2})$.
Multiply by $2\sqrt{2}$:
$2\sqrt{2}y - 3 = -3\sqrt{2}x + 3$.
$3\sqrt{2}x + 2\sqrt{2}y = 6$.
Dividing by $\sqrt{2}$:
$3x + 2y = 6/\sqrt{2} = 3\sqrt{2}$.

(A) Given $x = \sec \theta - \cos \theta$ and $y = \sec^n \theta - \cos^n \theta$.
First,find the derivatives with respect to $\theta$:
$\frac{dx}{d\theta} = \sec \theta \tan \theta + \sin \theta = \tan \theta (\sec \theta + \cos \theta)$.
$\frac{dy}{d\theta} = n \sec^{n-1} \theta (\sec \theta \tan \theta) + n \cos^{n-1} \theta \sin \theta = n \tan \theta (\sec^n \theta + \cos^n \theta)$.
Now,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{n \tan \theta (\sec^n \theta + \cos^n \theta)}{\tan \theta (\sec \theta + \cos \theta)} = \frac{n(\sec^n \theta + \cos^n \theta)}{\sec \theta + \cos \theta}$.
Squaring both sides:
$\left( \frac{dy}{dx} \right)^2 = \frac{n^2(\sec^n \theta + \cos^n \theta)^2}{(\sec \theta + \cos \theta)^2}$.
Using the identity $(a+b)^2 = (a-b)^2 + 4ab$:
$(\sec^n \theta + \cos^n \theta)^2 = (\sec^n \theta - \cos^n \theta)^2 + 4 \sec^n \theta \cos^n \theta = y^2 + 4$.
Similarly,$(\sec \theta + \cos \theta)^2 = (\sec \theta - \cos \theta)^2 + 4 \sec \theta \cos \theta = x^2 + 4$.
Thus,$\left( \frac{dy}{dx} \right)^2 = \frac{n^2(y^2 + 4)}{x^2 + 4}$,which implies $(x^2 + 4) \left( \frac{dy}{dx} \right)^2 = n^2(y^2 + 4)$.

(D) Given $x = \sin t$ and $y = \cos pt$.
First,find the derivatives with respect to $t$:
$\frac{dx}{dt} = \cos t$ and $\frac{dy}{dt} = -p \sin pt$.
Using the chain rule,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-p \sin pt}{\cos t}$.
Thus,$\cos t \frac{dy}{dx} = -p \sin pt$.
Differentiating both sides with respect to $x$:
$-\sin t + \cos t \frac{d^2y}{dx^2} = -p^2 \cos pt \frac{dt}{dx}$.
Since $\frac{dt}{dx} = \frac{1}{\cos t}$,we have:
$-\sin t \frac{dy}{dx} + \cos t \frac{d^2y}{dx^2} = -p^2 \cos pt \frac{1}{\cos t}$.
Multiplying by $\cos t$:
$-\sin t \cos t \frac{dy}{dx} + \cos^2 t \frac{d^2y}{dx^2} = -p^2 \cos pt$.
Substituting $x = \sin t$,$\cos^2 t = 1 - x^2$,and $y = \cos pt$:
$(1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} + p^2 y = 0$.
Therefore,$(1 - x^2)y_2 - xy_1 + p^2y = 0$.

(A) Given the parametric equations of the curve: $x = a \cos^3 \theta$ and $y = a \sin^3 \theta$.
First,find the derivative $\frac{dy}{dx}$ using the chain rule:
$\frac{dx}{d\theta} = 3a \cos^2 \theta (-\sin \theta) = -3a \cos^2 \theta \sin \theta$
$\frac{dy}{d\theta} = 3a \sin^2 \theta (\cos \theta) = 3a \sin^2 \theta \cos \theta$
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3a \sin^2 \theta \cos \theta}{-3a \cos^2 \theta \sin \theta} = -\tan \theta$
At $\theta = \pi / 4$,the slope of the tangent is $m_t = -\tan(\pi / 4) = -1$.
The slope of the normal $m_n$ is given by $-\frac{1}{m_t}$.
$m_n = -\frac{1}{-1} = 1$.

(D) Given the parametric equations: $x = 1 - a \sin \theta$ and $y = b \cos^2 \theta$.
First,find the derivative $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$.
$\frac{dx}{d\theta} = \frac{d}{d\theta}(1 - a \sin \theta) = -a \cos \theta$.
$\frac{dy}{d\theta} = \frac{d}{d\theta}(b \cos^2 \theta) = b(2 \cos \theta)(-\sin \theta) = -2b \cos \theta \sin \theta$.
Therefore,$\frac{dy}{dx} = \frac{-2b \cos \theta \sin \theta}{-a \cos \theta} = \frac{2b}{a} \sin \theta$.
At $\theta = \pi / 2$,the slope of the tangent is $m_t = \frac{2b}{a} \sin(\pi / 2) = \frac{2b}{a}$.
The slope of the normal is $m_n = -\frac{1}{m_t} = -\frac{1}{2b/a} = -\frac{a}{2b}$.

(D) The slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.
Given $x = t^2 + 3t - 8$,we have $\frac{dx}{dt} = 2t + 3$.
Given $y = 2t^2 - 2t - 5$,we have $\frac{dy}{dt} = 4t - 2$.
Therefore,$\frac{dy}{dx} = \frac{4t - 2}{2t + 3}$.
At $t = 2$,the slope is $\frac{dy}{dx} = \frac{4(2) - 2}{2(2) + 3} = \frac{8 - 2}{4 + 3} = \frac{6}{7}$.

(A) tangent is perpendicular to the $x$-axis if its slope is undefined,which occurs when $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} \neq 0$.
Given the parametric equations $x = t^2 - 1$ and $y = t^2 - t$.
We calculate the derivative of $x$ with respect to $t$:
$\frac{dx}{dt} = \frac{d}{dt}(t^2 - 1) = 2t$.
Setting $\frac{dx}{dt} = 0$ gives $2t = 0$,which implies $t = 0$.
At $t = 0$,we check $\frac{dy}{dt} = \frac{d}{dt}(t^2 - t) = 2t - 1$. Substituting $t = 0$,we get $\frac{dy}{dt} = -1 \neq 0$.
Since $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} \neq 0$ at $t = 0$,the tangent is perpendicular to the $x$-axis at $t = 0$.

(A) Let $u = \sqrt{x^2 + 16}$ and $v = \frac{x}{x - 1}$.
First,differentiate $u$ with respect to $x$:
$\frac{du}{dx} = \frac{1}{2\sqrt{x^2 + 16}} \cdot 2x = \frac{x}{\sqrt{x^2 + 16}}$.
Next,differentiate $v$ with respect to $x$ using the quotient rule:
$\frac{dv}{dx} = \frac{(x - 1)(1) - x(1)}{(x - 1)^2} = \frac{-1}{(x - 1)^2}$.
Now,find the rate of change of $u$ with respect to $v$:
$\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{x}{\sqrt{x^2 + 16}} \cdot \frac{(x - 1)^2}{-1}$.
Substitute $x = 3$ into the expression:
$\left( \frac{du}{dv} \right)_{x=3} = \frac{3}{\sqrt{3^2 + 16}} \cdot \frac{(3 - 1)^2}{-1} = \frac{3}{\sqrt{25}} \cdot \frac{4}{-1} = \frac{3}{5} \cdot (-4) = -\frac{12}{5}$.

(C) Given the parametric equations of the curve:
$x = t^2 + 3t - 8$ and $y = 2t^2 - 2t - 5$.
First,find the derivatives with respect to $t$:
$\frac{dx}{dt} = 2t + 3$
$\frac{dy}{dt} = 4t - 2$
The slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4t - 2}{2t + 3}$.
To find the value of $t$ at the point $(2, -1)$,substitute the coordinates into the equations:
$2 = t^2 + 3t - 8 \Rightarrow t^2 + 3t - 10 = 0 \Rightarrow (t+5)(t-2) = 0 \Rightarrow t = 2, -5$.
$-1 = 2t^2 - 2t - 5 \Rightarrow 2t^2 - 2t - 4 = 0 \Rightarrow t^2 - t - 2 = 0 \Rightarrow (t-2)(t+1) = 0 \Rightarrow t = 2, -1$.
The common value is $t = 2$.
Now,substitute $t = 2$ into the slope formula:
$\frac{dy}{dx} = \frac{4(2) - 2}{2(2) + 3} = \frac{8 - 2}{4 + 3} = \frac{6}{7}$.

(D) Given the parametric equations of the curve are $x = a \sin^3 t$ and $y = a \cos^3 t$.
First,we find the derivatives with respect to $t$:
$\frac{dx}{dt} = 3a \sin^2 t \cos t$
$\frac{dy}{dt} = -3a \cos^2 t \sin t$
Now,the slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.
$\frac{dy}{dx} = \frac{-3a \cos^2 t \sin t}{3a \sin^2 t \cos t} = -\frac{\cos t}{\sin t} = -\cot t$.
At $t = \frac{\pi}{3}$,the slope is:
$\left( \frac{dy}{dx} \right)_{t = \pi/3} = -\cot\left( \frac{\pi}{3} \right) = -\frac{1}{\sqrt{3}}$.

(B) We know that velocity $v = \frac{dx}{dt}$,so $\frac{dt}{dx} = \frac{1}{v}$.
Now,differentiate with respect to $x$:
$\frac{d^2t}{dx^2} = \frac{d}{dx} \left( \frac{1}{v} \right) = -\frac{1}{v^2} \frac{dv}{dx}$.
Using the chain rule,$\frac{dv}{dx} = \frac{dv}{dt} \cdot \frac{dt}{dx} = f \cdot \frac{1}{v} = \frac{f}{v}$.
Substituting this back into the expression:
$\frac{d^2t}{dx^2} = -\frac{1}{v^2} \left( \frac{f}{v} \right) = -\frac{f}{v^3}$.
Rearranging the terms,we get $f = -v^3 \frac{d^2t}{dx^2}$.

(C) Given the parametric equations of the curve:
$x = a(\theta - \sin \theta)$ and $y = a(1 - \cos \theta)$.
First,we find the derivative $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$.
$\frac{dx}{d\theta} = a(1 - \cos \theta)$ and $\frac{dy}{d\theta} = a(\sin \theta)$.
Thus,the slope of the tangent is $\frac{dy}{dx} = \frac{a \sin \theta}{a(1 - \cos \theta)} = \frac{\sin \theta}{1 - \cos \theta}$.
At $\theta = \pi / 2$:
$\frac{dy}{dx} = \frac{\sin(\pi / 2)}{1 - \cos(\pi / 2)} = \frac{1}{1 - 0} = 1$.
The slope of the normal is given by $-\frac{1}{\text{slope of tangent}}$.
Therefore,the slope of the normal is $-1 / 1 = -1$.

(D) Given the parametric equations $x = \frac{t - 1}{t + 1}$ and $y = \frac{t + 1}{t - 1}$.
At $t = 2$,$x = \frac{2 - 1}{2 + 1} = \frac{1}{3}$ and $y = \frac{2 + 1}{2 - 1} = 3$.
Observe that $xy = \left(\frac{t - 1}{t + 1}\right) \times \left(\frac{t + 1}{t - 1}\right) = 1$.
Thus,$y = \frac{1}{x}$,which implies $\frac{dy}{dx} = -\frac{1}{x^2}$.
At $x = \frac{1}{3}$,the slope $m = \frac{dy}{dx} = -\frac{1}{(1/3)^2} = -9$.
The equation of the tangent at $(\frac{1}{3}, 3)$ is given by $y - y_1 = m(x - x_1)$.
$y - 3 = -9(x - \frac{1}{3})$.
$y - 3 = -9x + 3$.
$9x + y - 6 = 0$.

(D) Given $x = t^2 + 3t - 8 = 2$
$t^2 + 3t - 10 = 0$
$(t + 5)(t - 2) = 0$
$t = -5$ or $t = 2$
Also $y = 2t^2 - 2t - 5 = -1$
$2t^2 - 2t - 4 = 0$
$t^2 - t - 2 = 0$
$(t - 2)(t + 1) = 0$
$t = 2$ or $t = -1$
Since the point $(2, -1)$ must satisfy both equations,we take the common value $t = 2$.
The slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.
$\frac{dx}{dt} = 2t + 3$
$\frac{dy}{dt} = 4t - 2$
$\frac{dy}{dx} = \frac{4t - 2}{2t + 3}$
At $t = 2$:
$\frac{dy}{dx} = \frac{4(2) - 2}{2(2) + 3} = \frac{8 - 2}{4 + 3} = \frac{6}{7}$.

(C) Given $x = a \cos \theta$ and $y = b \sin \theta$.
First,find $\frac{dy}{dx}$ using the chain rule:
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{b \cos \theta}{-a \sin \theta} = -\frac{b}{a} \cot \theta$.
Next,find the second derivative $\frac{d^2y}{dx^2}$:
$\frac{d^2y}{dx^2} = \frac{d}{dx} \left( -\frac{b}{a} \cot \theta \right) = -\frac{b}{a} (-\csc^2 \theta) \frac{d\theta}{dx} = \frac{b}{a} \csc^2 \theta \left( \frac{1}{-a \sin \theta} \right) = -\frac{b}{a^2} \csc^3 \theta$.
Finally,find the third derivative $\frac{d^3y}{dx^3}$:
$\frac{d^3y}{dx^3} = \frac{d}{dx} \left( -\frac{b}{a^2} \csc^3 \theta \right) = -\frac{b}{a^2} (3 \csc^2 \theta \cdot -\csc \theta \cot \theta) \frac{d\theta}{dx} = \frac{3b}{a^2} \csc^3 \theta \cot \theta \left( \frac{1}{-a \sin \theta} \right) = -\frac{3b}{a^3} \csc^4 \theta \cot \theta$.

(A) Given $x = e^t \sin t$ and $y = e^t \cos t$. At point $(1, 1)$,we have $1 = e^t \sin t$ and $1 = e^t \cos t$. Dividing these,we get $\tan t = 1$,which implies $t = \frac{\pi}{4}$.
First,we find the derivatives with respect to $t$:
$\frac{dx}{dt} = e^t \sin t + e^t \cos t = e^t(\sin t + \cos t)$
$\frac{dy}{dt} = e^t \cos t - e^t \sin t = e^t(\cos t - \sin t)$
Then,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\cos t - \sin t}{\cos t + \sin t}$.
Now,$\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{\cos t - \sin t}{\cos t + \sin t}\right) \cdot \frac{dt}{dx}$.
Using the quotient rule for $\frac{d}{dt}\left(\frac{\cos t - \sin t}{\cos t + \sin t}\right)$:
$= \frac{(\cos t + \sin t)(-\sin t - \cos t) - (\cos t - \sin t)(-\sin t + \cos t)}{(\cos t + \sin t)^2} = \frac{-(\cos t + \sin t)^2 - (\cos t - \sin t)^2}{(\cos t + \sin t)^2} = \frac{-2}{(\cos t + \sin t)^2}$.
Thus,$\frac{d^2y}{dx^2} = \frac{-2}{(\cos t + \sin t)^2} \cdot \frac{1}{e^t(\sin t + \cos t)} = \frac{-2}{e^t(\sin t + \cos t)^3}$.
At $t = \frac{\pi}{4}$,$e^t \sin t = 1$ and $e^t \cos t = 1$,so $e^t = \sqrt{2}$ and $\sin t = \cos t = \frac{1}{\sqrt{2}}$.
Substituting these values: $\frac{d^2y}{dx^2} = \frac{-2}{\sqrt{2}(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}})^3} = \frac{-2}{\sqrt{2}(\sqrt{2})^3} = \frac{-2}{\sqrt{2} \cdot 2\sqrt{2}} = \frac{-2}{4} = -\frac{1}{2}$.

(A) Given the parametric equations: $x = a(2 \cos t - \cos 2t)$ and $y = a(2 \sin t - \sin 2t)$.
To find the slope of the tangent,we calculate $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.
$\frac{dy}{dt} = a(2 \cos t - 2 \cos 2t) = 2a(\cos t - \cos 2t)$.
$\frac{dx}{dt} = a(-2 \sin t + 2 \sin 2t) = 2a(\sin 2t - \sin t)$.
Thus,$\frac{dy}{dx} = \frac{2a(\cos t - \cos 2t)}{2a(\sin 2t - \sin t)} = \frac{\cos t - \cos 2t}{\sin 2t - \sin t}$.
For the tangent to be parallel to the $x$-axis,$\frac{dy}{dx} = 0$,which implies $\cos t = \cos 2t$.
Using the identity $\cos A = \cos B \implies A = 2n\pi \pm B$,we have $2t = 2n\pi \pm t$.
Case $1$: $2t = 2n\pi + t \implies t = 2n\pi$.
Case $2$: $2t = 2n\pi - t \implies 3t = 2n\pi \implies t = \frac{2n\pi}{3}$.
For $n=1$,$t = 2\pi/3$. For $n=2$,$t = 4\pi/3$.
The difference between these values is $\frac{4\pi}{3} - \frac{2\pi}{3} = \frac{2\pi}{3}$.

(A) Given the parametric equations of the curve:
$x = a(t + \sin t \cos t) = a(t + \frac{1}{2} \sin 2t)$
$y = a(1 + \sin t)^2$
First,differentiate $x$ and $y$ with respect to $t$:
$\frac{dx}{dt} = a(1 + \cos 2t) = a(2 \cos^2 t) = 2a \cos^2 t$
$\frac{dy}{dt} = 2a(1 + \sin t) \cos t$
The slope of the tangent $m = \tan \theta = \frac{dy/dt}{dx/dt}$:
$\tan \theta = \frac{2a(1 + \sin t) \cos t}{2a \cos^2 t} = \frac{1 + \sin t}{\cos t}$
Using trigonometric identities:
$1 + \sin t = (\cos \frac{t}{2} + \sin \frac{t}{2})^2$
$\cos t = \cos^2 \frac{t}{2} - \sin^2 \frac{t}{2} = (\cos \frac{t}{2} - \sin \frac{t}{2})(\cos \frac{t}{2} + \sin \frac{t}{2})$
Thus,$\tan \theta = \frac{\cos \frac{t}{2} + \sin \frac{t}{2}}{\cos \frac{t}{2} - \sin \frac{t}{2}}$
Dividing numerator and denominator by $\cos \frac{t}{2}$:
$\tan \theta = \frac{1 + \tan \frac{t}{2}}{1 - \tan \frac{t}{2}} = \tan(\frac{\pi}{4} + \frac{t}{2})$
Therefore,$\theta = \frac{\pi}{4} + \frac{t}{2} = \frac{\pi + 2t}{4}$.

(B) Given the parametric equations:
$x = t^3 - 4t^2 - 3t$
$y = 2t^2 + 3t - 5$
First,find the derivatives with respect to $t$:
$\frac{dx}{dt} = 3t^2 - 8t - 3$
$\frac{dy}{dt} = 4t + 3$
The slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4t + 3}{3t^2 - 8t - 3}$.
For a horizontal tangent,$\frac{dy}{dx} = 0$,which implies $\frac{dy}{dt} = 0$ and $\frac{dx}{dt} \neq 0$:
$4t + 3 = 0 \implies t = -\frac{3}{4}$.
At $t = -\frac{3}{4}$,$\frac{dx}{dt} = 3(-\frac{3}{4})^2 - 8(-\frac{3}{4}) - 3 = 3(\frac{9}{16}) + 6 - 3 = \frac{27}{16} + 3 \neq 0$.
Thus,there is $1$ horizontal tangent,so $H = 1$.
For a vertical tangent,$\frac{dx}{dt} = 0$ and $\frac{dy}{dt} \neq 0$:
$3t^2 - 8t - 3 = 0$
$(3t + 1)(t - 3) = 0$
$t = -\frac{1}{3}$ or $t = 3$.
At these values,$\frac{dy}{dt} = 4t + 3 \neq 0$.
Thus,there are $2$ vertical tangents,so $V = 2$.
Therefore,$H = 1$ and $V = 2$.

(C) Given $x = \frac{1}{t^3} + \frac{1}{t^2}$ and $y = \frac{3}{2t^2} + \frac{2}{t}$.
Differentiating $x$ with respect to $t$:
$\frac{dx}{dt} = -\frac{3}{t^4} - \frac{2}{t^3} = -\frac{3 + 2t}{t^4}$.
Differentiating $y$ with respect to $t$:
$\frac{dy}{dt} = \frac{3}{2} \left( -\frac{2}{t^3} \right) - \frac{2}{t^2} = -\frac{3}{t^3} - \frac{2}{t^2} = -\frac{3 + 2t}{t^3}$.
Now,find $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-(3 + 2t)/t^3}{-(3 + 2t)/t^4} = \frac{t^4}{t^3} = t$.
Substitute $\frac{dy}{dx} = t$ into the expression $x \left( \frac{dy}{dx} \right)^3 - \frac{dy}{dx}$:
$x(t)^3 - t = \left( \frac{1 + t}{t^3} \right) t^3 - t = (1 + t) - t = 1$.