Let θθ (in radians) be an acute angle in a right triangle and let xx and yy, respectively, be the lengths of the sides adjacent to and opposite θθ. Suppose also that xx and yy vary with time. At a certain instant x=8x=8 units and is increasing at 77 unit/s, while y=8y=8 and is decreasing at 1414 units/s. How fast is θθ changing at that instant?

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Answer:θ is decreasing at the rate of [tex]\frac{21}{16}[/tex] units/secor [tex]\frac{d}{dt}[/tex](θ) = [tex]\frac{-21}{16}[/tex]Step-by-step explanation:Given :Length of side opposite to angle θ is yLength of side adjacent to angle θ is xθ is part of a right angle triangleAt this instant,x =  8 , [tex]\frac{dx}{dt}[/tex] = 7 ( [tex]\frac{dx}{dt}[/tex] denotes the rate of change of x with respect to time)y = 8 , [tex]\frac{dy}{dt}[/tex] = -14( The negative sign denotes the decreasing rate of change )Here because it is a right angle triangle,tanθ = [tex]\frac{y}{x}[/tex]-------------------------------------------------------------------1At this instant,tanθ = [tex]\frac{8}{8}[/tex] = 1Therefore θ = π/4We differentiate equation (1) with respect to time in order to obtain the rate of change of θ or [tex]\frac{d}{dt}[/tex](θ)[tex]\frac{d}{dt}[/tex] (tanθ) = [tex]\frac{d}{dt}[/tex] (y/x) ( Applying chain rule of differentiation for R.H.S as y*1/x)[tex]sec^{2}[/tex]θ[tex]\frac{d}{dt}[/tex](θ) = [tex]\frac{1}{x}[/tex][tex]\frac{dy}{dt}[/tex] - [tex]\frac{y}{x*x}[/tex][tex]\frac{dx}{dt}[/tex]-----------------------2Substituting the values of x , y , [tex]\frac{dx}{dt}[/tex] , [tex]\frac{dy}{dt}[/tex] , θ at that instant in equation (2)2[tex]\frac{d}{dt}[/tex](θ) = [tex]\frac{1}{8}[/tex]*(-14)- [tex]\frac{8}{8*8}[/tex]*7[tex]\frac{d}{dt}[/tex](θ) = [tex]\frac{-21}{16}[/tex]Therefore θ is decreasing at the rate of [tex]\frac{21}{16}[/tex] units/secor  [tex]\frac{d}{dt}[/tex](θ) = [tex]\frac{-21}{16}[/tex]
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general 1 month ago 7690