Respuesta :
Answer:
θ is decreasing at the rate of [tex]\frac{21}{16}[/tex] units/sec
or [tex]\frac{d}{dt}[/tex](θ) = [tex]\frac{-21}{16}[/tex]
Step-by-step explanation:
Given :
Length of side opposite to angle θ is y
Length of side adjacent to angle θ is x
θ is part of a right angle triangle
At 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]-------------------------------------------------------------------1
At this instant,
tanθ = [tex]\frac{8}{8}[/tex] = 1
Therefore θ = π/4
We 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]-----------------------2
Substituting 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/sec
or [tex]\frac{d}{dt}[/tex](θ) = [tex]\frac{-21}{16}[/tex]
