Calculus Related Rate Problems help!?

Question

Can anyone help me solving this please!!!
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.

Answer 1

1)
Volume of Cone = V = π r² h / 3

since the radius and height are equal h = r , and

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, but I not sure this is what you want.
Demiurge below sounds right about this part.

Answer 2

1) So you have the cone's volume:
(1/3)(pi)(r^2)h = V

You know your dV/dt = 1.2 m^3/min
But you also know that the base diameter is always equal to height so:
d= h
but also d = 2r

Substitute the ratios in, take the derivative, I think you know what to do next.

2) Well you have your position function, and taking the derivative of position yields the velocity function. Just plug in the appropriate times and find velocity.

For the particle at rest, set velocity equal to 0 and solve for t. These t values will get when the object is changing direction (which will help in the third part) and when it is not moving at all.

Use a number line test, knowing that you found when velocities in the previous part. Just plug in any number between the 0 velocities to find out which one is moving in a negative and positive direction.

The total distance during first 6 seconds: From the previous part, you know when the velocities are negative and positive. So, go back to the position function and find the difference in distance between the two consecutive intervals and also don't forget to add the ( l change in distance between the interval that had the negative velocity l); that was absolute value by the way.

Answer 3

The formula for the volume of a cone is:
V = pi * r^2 * h / 3
The diameter d = 2r = h
So r = h/2
Substitute this in the volume formula.
Take the derivative with respect to t of the volume formula.
Solve for dh/dt.

The velocity function v(t) = ds/dt.
Find the derivative. plug in 3s and 6s into v(t).
Solve v(t) = 0 to get the time (or times) that the particle at rest.
The particle is moving in a positive direction when v(t) > 0.

The total distance traveled will be the sum of the distance traveled to the right plus the distance traveled to the left. You need to split The time interval up into time when it is moving left and right. The distance traveled during each time interval is the absolute value of the change in s(t) over the time interval.

Good luck!

Answer 4

1)

V = π r² h / 3 = π h^3 / 12

dV/dt = 1.2 m^2/min = π h² / 4 dh/dt

dh/dt = 4 dV/dt / (π h²) = 4*1.2 / (π 3²) m/min = 0.1698 m/min

2)

s(t)=2t^3-21t^2+60t

v(t) = 6 t² - 42 t + 60

v(3) = -12 m/s
v(6) = 24 m/s

v(t) = 0 = 6 t² - 42 t + 60 = 6(t-2)(t-5) => t=2s or t=5s

v(t) >0 => 6 t² - 21 > 0 => t<2 s or t>5s

s(0) = 0, s(6)=36 => it travels 36 m

Answer 5

[a] floor area = circumference * top = 2? r * h top = 10cm + 0.1cm * t h = 10 + 0.1t A = 10? * (10 + 0.1t) = 100? + ? t dA/dt = fee of substitute of area with relation to time dA/dt = (100? + ? t)' = ? cm^2/sec [b] quantity = area of base * top = ? r^2 * (10 + 0.1t) V = 25? (10 + 0.1t) V = 250? + 2.5? t dV/dt = 2.5? cm^3/sec desire this permits!