If x=3≤ft[\;t-≤ft(/2)] and y=6≤ft[ t+≤ft(/2)] then \;dy/dx= [AP EAPCET 18 May 2024 Shift 1]

Correct Answer is : 2cos⁡2t1+sin⁡ tcos⁡t\;2² t/1+\;t t1+sintcost2cos2t

Correct Answer is : 2cos⁡2t1+sin⁡ tcos⁡t\;2² t/1+\;t t1+sintcost2cos2t

Test Index

Differentiation

Section: Mathematics

Question : 34 of 34

Marks: +1, -0

x=3\left[\sin\;t-\log\left(\cot\frac{t}{2}\right)\right]

y=6\left[\cos t+\log\left(\tan\frac{t}{2}\right)\right]

\;\frac{dy}{dx}=

Solution:

To determine

\;\frac{dy}{dx}

, we begin with the expressions given for

x

and

y

x=3\left[\sin\;t-\log\left(\cot\frac{t}{2}\right)\right]

Differentiating

x

with respect to

t

\;\frac{dx}{dt}\;=3\left[\cos t-\frac{1}{\cot\frac{t}{2}}\cdot\left(-\csc^2\frac{t}{2}\right)\cdot\frac{1}{2}\right]

\;=3\left[\cos t+\frac{1}{\sin\;t}\right]

\;=3\left[\frac{\cos t\sin\;t+1}{\sin\;t}\right]

Next, differentiating

y

with respect to

t

y=\;6\left[\cos t+\log\left(\tan\frac{t}{2}\right)\right]

\;\frac{dy}{dt}\;=6\left[-\sin\;t+\frac{1}{\tan\frac{t}{2}}\cdot\sec^2\frac{t}{2}\cdot\frac{1}{2}\right]

\;=6\left[-\sin\;t+\frac{1}{\sin\;t}\right]

\;=6\left[\frac{-\sin^2 t+1}{\sin\;t}\right]

Now, to find

\;\frac{dy}{dx}

\;\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{6\left[\frac{-\sin^2 t+1}{\sin\;t}\right]}{3\left[\frac{\cos t\sin\;t+1}{\sin\;t}\right]}

The

\sin\;t

terms cancel out:

=\;\frac{6(-\sin^2 t+1)}{3(\cos t\sin\;t+1)}

Simplifying further:

=\;\frac{2(-\sin^2 t+1)}{\cos t\sin\;t+1}

Rewriting the numerator,

1-\sin^2 t=\cos^2 t

=\;\frac{2\cos^2 t}{\cos t\sin\;t+1}

Therefore, the expression for

\;\frac{dy}{dx}

is:

\;\frac{dy}{dx}=\;\frac{2\cos^2 t}{1+\sin\;t\cos t}

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