Hi everyone!
As an additional post related to calculator techniques, I will be discussing about the calculator technique related to calculus: Volumetric Flow Rate.
So, what is Volumetric Flow Rate?
The volumetric flow rate, (also known as volume flow rate, rate of fluid flow or volume velocity) is the volume of fluid which passes per unit time; usually represented by the symbol Q (sometimes V̇). It is expressed sometimes m³/s (cubic meter per second) and also in ft³/sec (cubic feet per second). Source
Let us now solve some problems: Using Manual Solution and Calculator Technique
Problem and Manual Solution - Source
A water tank in the shape of a right circular cone has a height of 10 feet. The top rim of the tank is a circle with a radius of 4 feet. If water is being pumped into the tank at the rate of 2 cubic feet per minute, what is the rate of change of the water depth, in feet per minute, when the depth is 5 feet?
Now, let us try to solve this first using the Manual Solution
4/10 = r/h ; rearranging the equation
r = (2/5)h
Volume of the water inside the tank;
V = (1/3)πr²h ; substitute r = (2/5)h to the equation
V = (1/3)π[(2/5)h]²h
V = (4/75)πh³
Take the derivative of the equation with respect to time;
dV/dt = (12/75)πh² dh/dt
dV/dt = (4/25)πh² dh/dt ; substitute dV/dt = 2 ft³/min
2 = (4/25)πh² dh/dt ; when the depth is 5 ft (h = 5 ft)
2 = (4/25)π(5)² dh/dt
2 = (4/25)π(25) dh/dt
2 = 4π dh/dt; rearranging the equation
dh/dt = 1/2π ft/min = 0.1592 ft/min
The rate of how fast is the water rising at water depth of 5 ft is 1/2π ft/min or 0.1592 ft/min.
Now, let's try the Calculator Technique
Press in sequence;
[Mode] [3:STAT] [3:_+cX²]
Since the cross-sectional shape is a circle, we will use A = πr².
*Since on a front view, a cone has a shape of triangle; it is linear. By ratio and proportion of similar triangles, @ h = 5 ft;
R/H = r/h
4/10 = r/5 ; rearranging the equation
r = 2 ft
[AC] 2÷5ŷ = 1/2π ft/min = 0.1592 ft/min
ŷ --> [Shift] [1] [5:Reg] [6: ŷ]
We arrived with same answer as with the manual solution, the difference is that the length and the time you'd save using this calculator technique.
Let's try another problem.
Problem and Manual Solution - Source
Water is being drained from a cone-shaped reservoir 10 ft. in diameter and 10 ft. deep at a constant rate of 3 ft3/min. How fast is the water level falling when the depth of the water is 6 ft?
Let us try to solve this first using the Manual Solution
Similar Triangles
R/W = r/w
5/10=r/w ; rearranging the equation
r = w/2
Volume of the water inside the reservoir;
V = (1/3)πr²w ; substitute r = w/2 to the equation
V = (1/3)π[w/2]²w
V = (1/12)πw³
Take the derivative of the equation with respect to time;
dV/dt = (3/12)πw² dw/dt
dV/dt = (1/4)πw² dw/dt; substitute dV/dt = -3 ft³/min [ “negative (-)“ since it is draining]
-3 = (1/4)πw² dw/dt
-3/[(1/4)(πw²)] = dw/dt
-12/(πw²)= dw/dt ; rearranging the equation
dw/dt = -12/(πw²) ; when the depth is 6 ft (w = 6 ft)
dw/dt = -12/[π(6)²]
dw/dt = -12/36π
dw/dt = -(1/3)π ft/min = -0.1061 ft/min
The rate of how fast is the water draining at water depth of 6 ft is -(1/3)π ft/min = -0.1061 ft/min.
Now, let us use the Calculator Technique
Press in sequence;
[Mode] [3:STAT] [3:_+cX²]
Since the cross-sectional shape is a circle, we will use A = πr².
Ration and proportion of similar triangles, @ w = 5 ft;
R/W = r/w
5/10 = r/5 ; rearranging the equation
r = 2.5 ft
[AC] -3÷6ŷ = -(1/3)π ft/min = -0.1061 ft/min
In case you forgot, ŷ --> [Shift] [1] [5:Reg] [6: ŷ]
Still, we got the same and exact answer using both the manual and calculator technique. We can clearly see that manual solution is quite lengthy compared to the calcu tech. I believe that this calcu tech would really save your time, especially in answering time-limited tests like the licensure examinations.
I hope this post helped you.
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Yours truly,
