Answer:
d∅/dt = - 0.18101999999 rad/s
Explanation:
Length of the ladder = 10 ft
distance of the bottom of the ladder to the wall = x
where y = height of the wall
using SOHCAHTOA principle
tan ∅ = opposite/adjacent = y/x
use pythagora's theorem to get y
y² + x² = 10²
remember we are looking for how fast the angle is changing when x = 6ft
y² + 6² = 100
y² + 36 = 100
y² = 100 - 36
y² = 64
y = √64
y = 8
Rates of change
dx/dt = 1.5 ft/s
we are asked to find
d∅/dt when x = 6 ft
<em>tan ∅ = y/x</em>
find the derivative using the quotient rule
sec²∅ d∅/dt = (x dy/dt - y dx/dt) / x²
sec²∅ d∅/dt = (6 dy/dt - 8 dx/dt )/ 36
recall sec²∅ = (sec ∅)²
sec ∅ = 1/cos ∅
since cos ∅ = adjacent/ hypotenuse
sec ∅ = hypotenuse/adjacent
sec ∅ = 10/6
(sec ∅)² = 100/36
To get dy/dt
y² + x² = 10² dh/dt
find the derivative
2y dy/dt + 2x dx/dt = 100 (0)
note the length of ladder does not change
2 (8) dy/dt + 2 (6) 1.5 = 0
16 dy/dt + 18 = 0
dy/dt = -18/16
dy/dt = - 9/8
dy/dt = - 1.125
Therefore,
sec²∅ d∅/dt = (6 dy/dt - 8 dx/dt )/ 36
100/36 d∅/dt = (6 (-1.125) - 8 (1.5) )/36
100/36 d∅/dt = ( -6.75 - 12)/36
100/36 d∅/dt = - 18.75/36
100/36 d∅/dt = - 0.52083333333
2.77777777778 d∅/dt = -0.52083333333
divide both sides by 2.77777777778
d∅/dt = -0.52083333333/2.77777777778
d∅/dt = - 0.18101999999 rad/s
