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Differential Calculus

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Question : 67 of 69

Marks: +1, -0

if

x=a t^{3} ; y=3 b t^{2} ;

\; \frac{dy}{dx}=

$\; \frac{3x}{2y}$
$\; \frac{2y}{3x}$
$\; \frac{3y}{2x}$
$\; \frac{2x}{3y}$

Solution:

Concept:

Parametric differentiation:

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}

.

Explanation:

Given

x = a t^3

and

y = 3b t^2

.

Compute

\frac{dx}{dt} = 3a t^2

and

\frac{dy}{dt} = 6b t

.

Then

\frac{dy}{dx} = \frac{6b t}{3a t^2} = \frac{2b}{a t}

.

Express

t

in terms of

x

and

y

: from

y = 3b t^2

we get

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

, and from

x = a t^3 = a t \cdot t^2 = a t \cdot \frac{y}{3b}

, so

t = \frac{3b x}{a y}

.

Substitute:

\frac{dy}{dx} = \frac{2b}{a \cdot \frac{3b x}{a y}} = \frac{2b \cdot a y}{a \cdot 3b x} = \frac{2y}{3x}

.

Answer:

\frac{dy}{dx} = \frac{2y}{3x}

, which corresponds to option B.

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