Question

the slope of the tangent to the curve x=a sin t., y=a(cos t + log (tan t/2)) at the point t is​

Answer 1

Answer:

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Answer 2

SOLUTION

TO DETERMINE

The slope of the tangent to the curve

$\displaystyle\sf{x = a \sin t \: \: , \: y = a \bigg \{ \cos t + \log \bigg( \tan \frac{t}{2} \bigg) \: \bigg \} }$

EVALUATION

Here the given equation of the curve is

$\displaystyle\sf{x = a \sin t \: \: , \: y = a \bigg \{ \cos t + \log \bigg( \tan \frac{t}{2} \bigg) \: \bigg \} }$

Now

$\displaystyle\sf{x = a \sin t \: }$

Differentiating both sides with respect to t we get

$\displaystyle\sf{ \frac{dx}{dt} = a \cos t }$

Now

$\displaystyle\sf{ y = a \bigg \{ \cos t + \log \bigg( \tan \frac{t}{2} \bigg) \: \bigg \} }$

Differentiating both sides with respect to t we get

$\displaystyle\sf{ \frac{dy}{dt} = a \bigg \{ - \sin t + \frac{ \frac{1}{2} { \sec}^{2} \frac{t}{2} }{\tan \frac{t}{2}} \bigg \} }$

$\displaystyle\sf{ \implies \: \frac{dy}{dt} = a \bigg \{ - \sin t + \frac{ 1}{2 { \cos}^{2} \frac{t}{2} \tan \frac{t}{2}} \bigg \} }$

$\displaystyle\sf{ \implies \: \frac{dy}{dt} = a \bigg \{ - \sin t + \frac{ 1}{2 { \cos}^{} \frac{t}{2} \sin \frac{t}{2}} \bigg \} }$

$\displaystyle\sf{ \implies \: \frac{dy}{dt} = a \bigg \{ - \sin t + \frac{ 1}{ \sin t } \bigg \} }$

$\displaystyle\sf{ \implies \: \frac{dy}{dt} = a \bigg \{ \frac{ 1 - { \sin}^{2}t }{ \sin t } \bigg \} }$

$\displaystyle\sf{ \implies \: \frac{dy}{dt} = a \bigg \{ \frac{ { \cos}^{2}t }{ \sin t } \bigg \} }$

$\displaystyle\sf{ \implies \: \frac{dy}{dt} = a \bigg \{ \frac{ { \cos}^{2}t }{ \sin t } \bigg \} }$

$\displaystyle\sf{ \implies \: \frac{dy}{dt} = a \cos t \cot t}$

Hence the required slope of the tangent to the curve is

$\displaystyle\sf{ = \frac{dy}{dx} }$

$\displaystyle\sf{ = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}}$

$\displaystyle\sf{ = \frac{a \cos t \cot t}{a \cos t } }$

$\displaystyle\sf{ =\cot t }$

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