Water is pumped into a cylindrical tank, standing vertically, at a decreasing rate given at time minutes r(t) = 130 - 10t ft^3/min for 0≤t≤5.The tank has radius 6 ft and is empty when t=0. Find the depth of the gasoline in the tank at t=3.
dv/dt = 130 - 10 t
so
v = 130 t - 5 t^2 + constant
since v = 0 when t = 0
v = 130 t - 5 t^2
so at t = 3
v = 130(3) - 5(9) = pi r^2 h = pi (36) h
solve for h
rate of height =rateofvolume/area=(130-10t)/(PI*36)
depth= INT rate*dt= (130t-5t^2)/(PI*36) over limits
=(130*3-5*9)/(PI*36) =3.05 ft check all that.
To find the depth of the water in the cylindrical tank at t = 3 minutes, we need to integrate the rate of change of the water level with respect to time from 0 to 3.
Given that the rate of change of the water level is given by r(t) = 130 - 10t ft^3/min for 0 ≤ t ≤ 5, we can integrate this function with respect to t over the interval [0, 3] to find the total change in the water level up to t = 3.
Let's calculate the integral:
∫[0,3] (130 - 10t) dt
Integrating this function with respect to t, we get:
= [130t - 5t^2/2] evaluated from 0 to 3
Substituting the values into the integral:
= [130(3) - 5(3)^2/2] - [130(0) - 5(0)^2/2]
Simplifying the equation:
= [390 - 45/2] - [0 - 0]
= [390 - 45/2] ft^3
Now, to find the depth of the water in the tank, we need to divide the volume by the cross-sectional area.
The cross-sectional area of the tank is given by A = πr^2, where r is the radius of the tank.
Given that the radius is 6 ft, we can plug in the values:
A = π(6)^2
A = π(36)
A = 36π ft^2
The depth of the water in the tank at t = 3 minutes can be found by dividing the volume by the cross-sectional area:
Depth = (390 - 45/2) / (36π)
Depth ≈ 12.04 ft
Therefore, the depth of the water in the tank at t = 3 minutes is approximately 12.04 ft.
To find the depth of the water in the tank at a specific time, we need to integrate the rate function with respect to time and then use the formula for the volume of a cylinder.
Given:
Rate function: r(t) = 130 - 10t ft^3/min
Radius of tank: r = 6 ft
1. Integrate the rate function to obtain the volume function of the tank:
Integrating r(t) with respect to t gives us the volume function V(t):
V(t) = ∫ (130 - 10t) dt
To find V(t), we integrate (130 - 10t) with respect to t:
∫ (130 - 10t) dt = 130t - 5t^2 + C
C is the constant of integration, which we can determine using the initial condition that the tank is empty when t = 0:
V(0) = 130(0) - 5(0)^2 + C = 0
C = 0
Therefore, the volume function is:
V(t) = 130t - 5t^2
2. Find the depth of water in the tank at t = 3:
The volume of a cylinder is given by the formula V = πr^2h, where r is the radius and h is the height (or depth) of the cylinder.
To find the height of the water in the tank at t = 3, we substitute V(t) into the volume formula and solve for h:
V(t) = πr^2h
130t - 5t^2 = π(6^2)h
130(3) - 5(3)^2 = π(6^2)h
390 - 45 = 36πh
345 = 36πh
Now, solve for h:
h = 345 / (36π)
h ≈ 3.022 ft
Therefore, the depth of the water in the tank at t = 3 minutes is approximately 3.022 feet.