Question

A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.5 ft/s, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 6 ft from the wall?

Answer 1

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

tan ∅ = y/x

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