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Differential Calculus

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Question : 65 of 69

Marks: +1, -0

If

x=\frac{2t}{1+t^{2}}, y=\frac{1-t^{2}}{1+t^{2}}

\frac{dy}{dx}+\frac{x}{y}

is

1
2
0
$4t^{2}$

Solution:

Concept:

The parametric equations represent the unit circle

x^2 + y^2 = 1

, since

x = \sin \theta

and

y = \cos \theta

with

t = \tan(\theta/2)

.

Explanation:

From

x = \frac{2t}{1+t^2}

and

y = \frac{1-t^2}{1+t^2}

, we compute

x^2 + y^2 = 1

.

Thus,

x

and

y

satisfy

x^2 + y^2 = 1

.

Differentiating implicitly with respect to

x

:

2x + 2y \frac{dy}{dx} = 0

.

This gives

\frac{dy}{dx} = -\frac{x}{y}

.

Therefore,

\frac{dy}{dx} + \frac{x}{y} = -\frac{x}{y} + \frac{x}{y} = 0

.

Answer:

0

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