Answer to Question #248141 in Calculus for jhonh

Question

Consider the given right triangle with legs of length x cm. and y cm. and angle θ radians. If x is decreasing at the rate of 3 cm/min. and y is increasing at the rate of 4 cm/min., at what rate is angle θ changing when x=5 cm. and y=2 cm. ?

Answer 1

Solution:

$dy/dt=4\ cm/min \\ dx/dt=-3\ cm/min \\x=5\ cm \\y=2\ cm$

$\sec\theta=\dfrac {\sqrt{x^2+y^2}}{x} \\\Rightarrow \sec^2\theta=\dfrac {{x^2+y^2}}{{x^2}} \ \ ...(i)$

Also,

$\tan \theta=\dfrac{AC}{AB} \\\Rightarrow\tan \theta=\dfrac yx$

On differentiating both sides w.r.t $t$,

$\sec^2\theta \dfrac{d\theta}{dt}=\dfrac{x(dy/dt)-y(dx/dt)}{x^2} \\\Rightarrow\dfrac {x^2+y^2}{{x^2}} \dfrac{d\theta}{dt}=\dfrac{x(dy/dt)-y(dx/dt)}{x^2} \\ \Rightarrow (\dfrac{5^2+2^2}{5^2}) \dfrac{d\theta}{dt}=\dfrac{5(4)-2(-3)}{5^2}$

$\\ \Rightarrow (\dfrac{29}{25}) \dfrac{d\theta}{dt}=\dfrac{26}{25} \\ \Rightarrow \dfrac{d\theta}{dt}=\dfrac{26}{29} \ radians/min$