Question

please help me here : A ladder 20 feet long is leaning against the wall of a house. The base of the ladder is pulled away
from the wall at a rate of 2 feet per second. Find the rate at which the angle between the ladder and the wall
of the house is changing when the base of the ladder is 7 feet from the wall.

Answer 1

whoa...

Answer 2

so equation might look something like :
angle (theta) = arcsin (initial distance out + 2ft/sec / 20)

rewritten may look something like:
y = sin^-1 [(7 +2x)/20]
\[y = \sin^{-1} [\frac {7 +2x}{20}]\]

so rate of change = instantaneous derivative
so y' = ??? @tryingharder ???

Answer 3

when base of ladder is 7ft from wall, angle theta = about 20.49 degrees (calculated from SOH CAH TOA rules)

it get's a little advanced here, but stay with me
we'll use the chain rule to calculate the derivative of (sin^(-1)(1/20 (7+2 x)))

u = 1/20 (2 x+7)
now:
d/dx(sin^(-1)(1/20 (2 x+7))) = ( dsin^(-1)(u))/( du) ( du)/( dx)

so 2 parts:
( d)/( du)(sin^(-1)(u)) = 1/sqrt(1-u^2)

(d/dx(1/20 (7+2 x)))/sqrt(1-1/400 (7+2 x)^2)

(d/dx(x))/(10 sqrt(1-1/400 (7+2 x)^2))

= 1/(10 sqrt(1-1/400 (7+2 x)^2))

expanded out:

so your rate of change of the angle (lets still call it theta) is y', which equals:
\[y' = \frac {2}{\sqrt(-4 x^2-28 x+351)}\]

where x is your time in seconds since the ladder was 7 feet from the wall

Answer 4

thanks :) this is really a big help :)

Answer 5

all good, good luck ;)