JEE MAIN - Mathematics (2019 - 9th January Evening Slot - No. 7)
If x $$=$$ 3 tan t and y $$=$$ 3 sec t, then the value of $${{{d^2}y} \over {d{x^2}}}$$ at t $$ = {\pi \over 4},$$ is :
$${1 \over {3\sqrt 2 }}$$
$${1 \over {6\sqrt 2 }}$$
$${3 \over {2\sqrt 2 }}$$
$${1 \over 6}$$
Explanation
x = 3 tan t and y = 3 sec t
So that $${{dx} \over {dt}}$$ = 3sec2t and $${{dy} \over {dt}}$$ = 3 sec t tan t
$${{dy} \over {dx}}$$ = $${{dy/dt} \over {dx/dt}}$$ = sin t
$${{{d^2}y} \over {d{x^2}}}$$ = (cos t)$$.{{dt} \over {dx}}$$
$${{{d^2}y} \over {d{x^2}}} = \left( {\cos t} \right).{1 \over {3{{\sec }^2}t}}$$
$${{{d^2}y} \over {d{x^2}}}$$ = $${1 \over 3}$$(cos3 t)
$${\left( {{{{d^2}y} \over {d{x^2}}}} \right)_{t = \pi /4}}$$ = $${1 \over 3} \times {\left( {{1 \over {\sqrt 2 }}} \right)^3} = {1 \over {6\sqrt 2 }}$$
So that $${{dx} \over {dt}}$$ = 3sec2t and $${{dy} \over {dt}}$$ = 3 sec t tan t
$${{dy} \over {dx}}$$ = $${{dy/dt} \over {dx/dt}}$$ = sin t
$${{{d^2}y} \over {d{x^2}}}$$ = (cos t)$$.{{dt} \over {dx}}$$
$${{{d^2}y} \over {d{x^2}}} = \left( {\cos t} \right).{1 \over {3{{\sec }^2}t}}$$
$${{{d^2}y} \over {d{x^2}}}$$ = $${1 \over 3}$$(cos3 t)
$${\left( {{{{d^2}y} \over {d{x^2}}}} \right)_{t = \pi /4}}$$ = $${1 \over 3} \times {\left( {{1 \over {\sqrt 2 }}} \right)^3} = {1 \over {6\sqrt 2 }}$$
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