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AP DSC School Assistant Math Model Test 3

Section: Subject Specific

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Question : 113 of 160

Marks: +1, -0

If

x=a\left(t+\frac{1}{t}\right)

y=a\left(t-\frac{1}{t}\right)

\frac{d x}{d y}

is:

$\frac{x}{y}$
$\frac{y}{x}$
$\frac{1}{x}$
1

Solution:

Concept:

Differentiation of parametric functions: If

x

and

y

are given in terms of a parameter

t

, then

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

.

Explanation:

Given

x = a\left(t + \frac{1}{t}\right)

and

y = a\left(t - \frac{1}{t}\right)

.

1. Differentiate

x

with respect to

t

:

\frac{dx}{dt} = a\left(1 - \frac{1}{t^2}\right)

because

\frac{d}{dt}(t) = 1

and

\frac{d}{dt}(t^{-1}) = -t^{-2}

.

2. Differentiate

y

with respect to

t

:

\frac{dy}{dt} = a\left(1 + \frac{1}{t^2}\right)

.

3. Using the parametric formula:

\frac{dx}{dy} = \frac{dx/dt}{dy/dt} = \frac{a\left(1 - \frac{1}{t^2}\right)}{a\left(1 + \frac{1}{t^2}\right)}

.

4. Cancel

a

and multiply numerator and denominator by

t

:

\frac{dx}{dy} = \frac{t - \frac{1}{t}}{t + \frac{1}{t}} = \frac{a\left(t - \frac{1}{t}\right)}{a\left(t + \frac{1}{t}\right)} = \frac{y}{x}

.

Answer:

Option B:

\frac{y}{x}

is correct.

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