Solution:
d y / d t = 4 c m / m i n d x / d t = − 3 c m / m i n x = 5 c m y = 2 c m dy/dt=4\ cm/min
\\ dx/dt=-3\ cm/min
\\x=5\ cm
\\y=2\ cm d y / d t = 4 c m / min d x / d t = − 3 c m / min x = 5 c m y = 2 c m
sec θ = x 2 + y 2 x ⇒ sec 2 θ = x 2 + y 2 x 2 . . . ( i ) \sec\theta=\dfrac {\sqrt{x^2+y^2}}{x}
\\\Rightarrow \sec^2\theta=\dfrac {{x^2+y^2}}{{x^2}} \ \ ...(i) sec θ = x x 2 + y 2 ⇒ sec 2 θ = x 2 x 2 + y 2 ... ( i )
Also,
tan θ = A C A B ⇒ tan θ = y x \tan \theta=\dfrac{AC}{AB}
\\\Rightarrow\tan \theta=\dfrac yx tan θ = A B A C ⇒ tan θ = x y
On differentiating both sides w.r.t t t t ,
sec 2 θ d θ d t = x ( d y / d t ) − y ( d x / d t ) x 2 ⇒ x 2 + y 2 x 2 d θ d t = x ( d y / d t ) − y ( d x / d t ) x 2 ⇒ ( 5 2 + 2 2 5 2 ) d θ d t = 5 ( 4 ) − 2 ( − 3 ) 5 2 \sec^2\theta \dfrac{d\theta}{dt}=\dfrac{x(dy/dt)-y(dx/dt)}{x^2}
\\\Rightarrow\dfrac {x^2+y^2}{{x^2}} \dfrac{d\theta}{dt}=\dfrac{x(dy/dt)-y(dx/dt)}{x^2}
\\ \Rightarrow (\dfrac{5^2+2^2}{5^2}) \dfrac{d\theta}{dt}=\dfrac{5(4)-2(-3)}{5^2} sec 2 θ d t d θ = x 2 x ( d y / d t ) − y ( d x / d t ) ⇒ x 2 x 2 + y 2 d t d θ = x 2 x ( d y / d t ) − y ( d x / d t ) ⇒ ( 5 2 5 2 + 2 2 ) d t d θ = 5 2 5 ( 4 ) − 2 ( − 3 )
⇒ ( 29 25 ) d θ d t = 26 25 ⇒ d θ d t = 26 29 r a d i a n s / m i n \\ \Rightarrow (\dfrac{29}{25}) \dfrac{d\theta}{dt}=\dfrac{26}{25}
\\ \Rightarrow \dfrac{d\theta}{dt}=\dfrac{26}{29} \ radians/min ⇒ ( 25 29 ) d t d θ = 25 26 ⇒ d t d θ = 29 26 r a d ian s / min
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