Question

Let θ

θ
(in radians) be an acute angle in a right triangle and let x
x
and y
y
, respectively, be the lengths of the sides adjacent to and opposite θ
θ
. Suppose also that x
x
and y
y
vary with time.
At a certain instant x=8
x
=
8
units and is increasing at 7
7
unit/s, while y=8
y
=
8
and is decreasing at 14
1
4
units/s.
How fast is θ
θ
changing at that instant?

Answer 1

θ is decreasing at the rate of
`(21)/(16)` units/sec

or
`(d)/(dt)`(θ) =
`(-21)/(16)`

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 ,
`(dx)/(dt)` = 7

(
`(dx)/(dt)` denotes the rate of change of x with respect to time)

y = 8 ,
`(dy)/(dt)` = -14

( The negative sign denotes the decreasing rate of change )

Here because it is a right angle triangle,

tanθ =
`(y)/(x)`-------------------------------------------------------------------1

At this instant,

tanθ =
`(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
`(d)/(dt)`(θ)

`(d)/(dt)` (tanθ) =
`(d)/(dt)` (y/x)

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

`sec^(2)`θ
`(d)/(dt)`(θ) =
`(1)/(x)`
`(dy)/(dt)` -
`(y)/(x*x)`
`(dx)/(dt)`-----------------------2

Substituting the values of x , y ,
`(dx)/(dt)` ,
`(dy)/(dt)` , θ at that instant in equation (2)

2
`(d)/(dt)`(θ) =
`(1)/(8)`*(-14)-
`(8)/(8*8)`*7

`(d)/(dt)`(θ) =
`(-21)/(16)`

Therefore θ is decreasing at the rate of
`(21)/(16)` units/sec

or
`(d)/(dt)`(θ) =
`(-21)/(16)`