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If x and y are connected parametrically by the equations given

Question

If x and y are connected parametrically by the equations given, without eliminating the parameter, Find $\frac{d y}{d x}$ .
$x = \frac{sin ^{3} t}{cos 2 t}, y = \frac{cos ^{3} t}{cos 2 t}$

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If x and y are connected parametrically by the equations given, without eliminating the parameter, Find

\frac{d y}{d x}

x = \frac{sin ^{3} t}{cos 2 t}, y = \frac{cos ^{3} t}{cos 2 t}

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Text solutionVerified

Solution:

y = \frac{cos ^{3} t}{cos 2 t}

\frac{d y}{d t} = \frac{1}{cos 2 t} (cos 2 t (3 cos^{2} t) (- sin t) - cos^{3} t (\frac{1}{2} (\frac{1}{cos 2 t} (- sin 2 t) (2)))

\frac{d y}{d t} = \frac{1}{cos 2 t} (- cos 2 t (3 cos^{2} t) (sin t) + \frac{cos ^{3} t sin 2 t}{cos 2 t})

\frac{d y}{d t} = \frac{1}{( cos 2 t ) ( cos 2 t )} (cos^{3} t sin 2 t - 3 cos^{2} t sin t cos 2 t) \to (1)

x = \frac{sin ^{3} t}{cos 2 t}

\frac{d x}{d t} = \frac{1}{cos 2 t} (cos 2 t (3 sin^{2} t) (cos t) - sin^{3} t (\frac{1}{2} (\frac{1}{cos 2 t} (- sin 2 t) (2)))

\frac{d x}{d t} = \frac{1}{cos 2 t} (cos 2 t (3 sin^{2} t) (cos t) + \frac{sin ^{3} t sin 2 t}{cos 2 t})

\frac{d x}{d t} = \frac{1}{( cos 2 t ) ( cos 2 t )} (3 sin^{2} t cos t cos 2 t + sin^{3} t sin 2 t) \to (2)

Dividing (1) by (2),

\frac{d y}{d x} = \frac{cos ^{3} t sin 2 t - 3 cos ^{2} t sin t cos 2 t}{3 sin ^{2} t cos t cos 2 t + sin ^{3} t sin 2 t}

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Question Text	If x and y are connected parametrically by the equations given, without eliminating the parameter, Find $\frac{d y}{d x}$ . $x = \frac{sin ^{3} t}{cos 2 t}, y = \frac{cos ^{3} t}{cos 2 t}$
Answer Type	Text solution:1
Upvotes	151