A ladder of 8 m is leaning against a wall begins to slide. If its upper end slides down the wall at a rate of 0.25 m/sec, at what rate is the angle between the ladder and the ground changing when the foot of the ladder is 5 m from the wall?
I would like oobleck to check my answer for this question that I asked yesterday and confirm my answer.
My answer for this question is: -1/20 m/s
Also, I have been given this function: y = 2cos(x) - 1 domain: [0,2pi]
I have to determine:
a) intervals of increase & decrease
b) local max and mins
for my answers:
a) increase: [0, pi)
decrease: (pi, 2pi]
b) local min: (pi, -3)
if these answers are correct based on the given function and domain, please let me know
I agree with your answer
As to the 2nd question, it was answered yesterday by both Damon and I.
I suggest you give yourself a nickname instead of anonymous, that way
previous posts by you can be quickly found.
Most of your questions can be answered by looking at the graph of
y = 2cos(x) - 1
https://www.wolframalpha.com/input/?i=y+%3D+2cos%28x%29+-+1+for++0+%3C+x+%3C+2%CF%80
I don't get -1/20
How did you arrive at -1/20 ?
Note that when x = 5, y = √39
for the 2nd problem,
y' = -2sinx
increasing when y'>0, so you got these backwards
min is correct
oobleck I followed the method you showed me yesterday. I think I asked this question at 5:29 pm and you answered me at 6:01 pm using the sec^2 theta method
I would have posted the URL but I am not allowed to
@Reiny I just asked again because I wasn't sure about the set notation method I am supposed to use. And also, my prof insists that we don't use the graph because this is calculus so we have to solve by derivatives and other class taught methods. Also, you said that my intervals of concavity were wrong and so I am confused if what I did was right for that and the inflection points.
I think I misled you how did you get a value for dx/dt?
Note that since we only know dy/dt, my answer, while correct, could not have been used very easily. SO, expressing everything in terms of y, I should have written
tanθ = y/√(64-y^2)
sec^2θ dθ/dt = 64/(64-y^2)^(3/2) dy/dt
I'm surprised that Reiny agreed with you.
So. what did you have for dx/dt?
for dx/dt i have √39/20
I said this (Pls understand that I am on a computer so my work may be not clear as I would want it to be)
tanθ = y/x
sec^2θ dθ/dt = x(dy/dt) - y(dx/dt)/(x^2)
dθ/dt = x(dy/dt) - y(dx/dt)/ [(x)^2 * sec^2θ]
then I filled in what I know and found that dθ/dt = -1/20 m/s
Let the distance of the top of the ladder up the wall be y m
given dy/dt = -.25 m/s
let θ be the angle at the base
so sin θ = y/8
8sinθ = y
8cosθ dθ/dt = dy/dt = -1/4
when x = 5
8(5/8) dθ/dt = -1/4
5 dθ/dt = -1/4
dθ/dt = -1/20