Answer to Question #248141 in Calculus for jhonh

Question #248141

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. ?


1
Expert's answer
2021-10-12T02:05:29-0400

Solution:

dy/dt=4 cm/mindx/dt=3 cm/minx=5 cmy=2 cmdy/dt=4\ cm/min \\ dx/dt=-3\ cm/min \\x=5\ cm \\y=2\ cm

secθ=x2+y2xsec2θ=x2+y2x2  ...(i)\sec\theta=\dfrac {\sqrt{x^2+y^2}}{x} \\\Rightarrow \sec^2\theta=\dfrac {{x^2+y^2}}{{x^2}} \ \ ...(i)

Also,

tanθ=ACABtanθ=yx\tan \theta=\dfrac{AC}{AB} \\\Rightarrow\tan \theta=\dfrac yx

On differentiating both sides w.r.t tt,

sec2θdθdt=x(dy/dt)y(dx/dt)x2x2+y2x2dθdt=x(dy/dt)y(dx/dt)x2(52+2252)dθdt=5(4)2(3)52\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}

(2925)dθdt=2625dθdt=2629 radians/min\\ \Rightarrow (\dfrac{29}{25}) \dfrac{d\theta}{dt}=\dfrac{26}{25} \\ \Rightarrow \dfrac{d\theta}{dt}=\dfrac{26}{29} \ radians/min


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