Question
A 24 foot ladder is leaning against a building. Let x be the distance between the bottom of the ladder and the building and let theta be the angle between the ladder and the ground.
a)Express theta as a function of x.
cos theta = x/24
theta = arccos(x/24) correct/incorrect?
b)When the angle is 40 degrees (2 Pi/9 radians) how big is x?
40 degrees = arccos(x/24)
cos(40 deg.)= cos(arccos)(x/24)
x= 18.385 feet coorect/incorrect?
c)Suppose the bottom of the ladder is sliding away from the wall at a rate of 2 ft/sec. When x=6, how fast is theta(in radians/sec.) changing?
I really don't know about this last part... help/hints would be greatly appreciated! Seems sort of like a physics problem.
a)Express theta as a function of x.
cos theta = x/24
theta = arccos(x/24) correct/incorrect?
b)When the angle is 40 degrees (2 Pi/9 radians) how big is x?
40 degrees = arccos(x/24)
cos(40 deg.)= cos(arccos)(x/24)
x= 18.385 feet coorect/incorrect?
c)Suppose the bottom of the ladder is sliding away from the wall at a rate of 2 ft/sec. When x=6, how fast is theta(in radians/sec.) changing?
I really don't know about this last part... help/hints would be greatly appreciated! Seems sort of like a physics problem.
Answers
Reiny
your a) and b) are correct
c) write your opening equation as
x = 24cosØ
differentiate with respect to t, (t in seconds)
dx/dt = -24sinØ dØ/dt (#1)
we are given dx/dt = 2 when x = 6
when x=6 we can get the height h of the triangle by Pythagoras h^2 + 6 ^2 = 24^2
h = √540 and sinØ = √540/24
so from #1
dØ/dt = (dx/dt) / (-24sinØ)
= 2/(-24(√540/24)
= -2/√540 radians/sec or appr. -0.086 radians/sec
You titled your subject as "Calculus".
This last part belongs to a part of Calculus called,
"Rates of Change" or "Related Rates"
Several chapters of your text should be devoted to it, depending on the depth of the course.
c) write your opening equation as
x = 24cosØ
differentiate with respect to t, (t in seconds)
dx/dt = -24sinØ dØ/dt (#1)
we are given dx/dt = 2 when x = 6
when x=6 we can get the height h of the triangle by Pythagoras h^2 + 6 ^2 = 24^2
h = √540 and sinØ = √540/24
so from #1
dØ/dt = (dx/dt) / (-24sinØ)
= 2/(-24(√540/24)
= -2/√540 radians/sec or appr. -0.086 radians/sec
You titled your subject as "Calculus".
This last part belongs to a part of Calculus called,
"Rates of Change" or "Related Rates"
Several chapters of your text should be devoted to it, depending on the depth of the course.