Hi, can anyone please help me solving these problems?
any help will be greatly appreciated
1-If a tank holds 1000L of water, which takes an hour to drain from the bottom in the tank, then the volume V of water remaining in the tank after t minutes is V=1000(1-t/60)^2 0<=t<=60. Find the rate at which water is flowing out of the tank (the instantaneous rate of change of V with respect to t- given by dV/dT) after 10 minutes.
2-A waterskier skis over the ramp at a speed of 12m/s.How fast is she rising as she leaves the ramp? * triangle measures opposite 1 m adjacent 5m.*
3-A man starts walking north at a speed of 1.5 m/s and a woman starts at the same time from the same place but walking west at 2m/s. At what rate is the distance between the man and the woman increasing one minute later?
4-At 1pm, ship A is 80km south of shipB.Ship A is sailing north at 30kph and ship B is sailling east at 40kph.How fast is the distance between the man and the woman increasing one minute later?
1-Sand is being dumped from a coveyor belt at 1.2 m^3/min and forms a pile in the shape of a cone whose base diameter and height are always equal . How fast is the height of the pile growing when the pile is 3m high?
2-The position of a particle moving in a straight line is given by the equation s(t)=2t^3-21t+60t, when t=0 and is measured in seconds and s is measured in meters.
*What is the velocity after 3s? 6?
*When is the particle at rest?
*When is the particle moving in a positive direction?
*Find the total distance traveled by the particle during the first 6s.
Related rates are quite difficult. I'm still mastering them. I would recommend googling "related rates" without the quotes. There is a great website called "Pauls Online Notes: Calculus I - Related Rates" that gives detailed analyses for many different problems. There are three basic steps that you should follow for all problems of this type:
(1) Write the general equation for the problem, like volume or area.
(2) Implicitly differentiate the equation with respect to time. Do not substitute in any variables except consonants. That's very important.
(3) Plug in what you know, and solve for what you need.
If you want to work on those problems, I'll be happy to critique your thinking.
For your other problem, this information will be helpful (and should be memorized):
s(t) position function
s'(t) = v(t) velocity function
s''(t) = v'(t) = a(t) acceleration function
For the velocity, differentiate the position function and plug in your values for t.
The velocity will be 0 when the particle is at rest.
Just plug t=6 into the position function s(t) to find the distance traveled.
If you want to start from there and work on that, I'll be happy to help you.
WOW
Do you actually expect somebody to just do all those homework questions for you??
I will set the second #1 question for you.
let the radius of the base be r m, then the height will be 2r m.
given dV/dt = 1.2 m^3/min
find dr/dt when h = 3, then r = 3/2
V = (1/3)pi r^2 h
=(1/3)pi r^2(2r)
V = (2/3)pi r^3
dV/dt = 2pi r^2 dr/dt
sub in dV/dt, the value of r, and the solve for dr/dt.
let me know what you got.
Ok for the last (1 and 2) I was able to do this
1)
V = (1/3) π h³
dv/ dt = 1/3 π 3h² dh/dt
dv/ dt = π h² dh/ dt
dv/dt = rate sand is being dumped = 1.2m³ / minute
1.2 = π 3² dh/dt
dh / dt = 1.2 / 9π = .0425 m/ minute
2)
velocity = ds/ dt = 6t² - 42t + 60
v(3) = 6(3²) - 42(3) + 60 = 54 - 126 +60 = -12m/s
v(6) = 216 -252 +60 = 24 m/s
particle at rest means v = 0
6t² - 42t + 60 = 0
t² - 7t + 10 = 0 (divided both sides by 6)
(t-5)(t-2) = 0
t = 5 seconds and t = 2 seconds
particle moving in positive direction means velocity is positive
velocity positive from t = negative infinity to t = 2 (open), and
velocity positive from t = 5 to t = positive infinity (open)
s(0) = 0, s(6) = 36
the displacement is 36
Michael and Reiny can you check if its right?
You had
V = (1/3) π h³ , I don't know how you got that
here is the correct equation:
V =(1/3)pi(r^2)(h)
but 2r = h, given, so r = h/2
then V = (1/3)pi(h^2/4)h
V = (1/12)pi(h^3)
dV/dt = (1/4)pi(h^2)dh/dt
1.2 = (1/4(pi)(9)dh/dt
dh/dt = .1698 m/min
Ok thanks Reiny, I was able to solve (1.1),(2.2),(4)
However I cant think of a way to solve (2) and (3), could you help me out?
For (2) I don't know how to develope the formula to find the derivative since both sides are given and so is the rate. My only assumption is that the skis go upward so I have to find that rate of change.
<<3-A man starts walking north at a speed of 1.5 m/s and a woman starts at the same time from the same place but walking west at 2m/s. At what rate is the distance between the man and the woman increasing one minute later? >>
Let the man's coordinates be (x1,y1) and the woman coordiates be (x2,y2). Let them be at (0,0) at t=0.
x1 = 0 y1 = 1.5 t
x2 = -2t y2 = 0
Separation: sqrt [(x2-x1)^2 + (y2-y1)^2]
= sqrt (4t^2 + 2.25 t^2) = 2.5 t
d(separation)/dt = 2.5 m/s
The rate of separation change is 2.5 m/s at all times.