Question

Let θ

Answer 1

Answer:

θ is decreasing at the rate of $\frac{21}{16}$ units/sec

or $\frac{d}{dt}$(θ) = $\frac{-21}{16}$

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 , $\frac{dx}{dt}$ = 7

( $\frac{dx}{dt}$ denotes the rate of change of x with respect to time)

y = 8 , $\frac{dy}{dt}$ = -14

( The negative sign denotes the decreasing rate of change )

Here because it is a right angle triangle,

tanθ = $\frac{y}{x}$-------------------------------------------------------------------1

At this instant,

tanθ = $\frac{8}{8}$ = 1

Therefore θ = π/4

We differentiate equation (1) with respect to time in order to obtain the rate of change of θ or $\frac{d}{dt}$(θ)

$\frac{d}{dt}$ (tanθ) = $\frac{d}{dt}$ (y/x)

( Applying chain rule of differentiation for R.H.S as y*1/x)

$sec^{2}$θ$\frac{d}{dt}$(θ) = $\frac{1}{x}$$\frac{dy}{dt}$ - $\frac{y}{x*x}$$\frac{dx}{dt}$-----------------------2

Substituting the values of x , y , $\frac{dx}{dt}$ , $\frac{dy}{dt}$ , θ at that instant in equation (2)

2$\frac{d}{dt}$(θ) = $\frac{1}{8}$*(-14)- $\frac{8}{8*8}$*7

$\frac{d}{dt}$(θ) = $\frac{-21}{16}$

Therefore θ is decreasing at the rate of $\frac{21}{16}$ units/sec

or  $\frac{d}{dt}$(θ) = $\frac{-21}{16}$