Water is Being Pumped Continuously Into a Tank at a Rate That is Inversely Proportional
Water is leaking out of an inverted conical tank at a rate of 11,000 cm3/min at the same time that water is being pumped into the tank at a constant rate. The tank has height 6 m and the diameter at the top is 4 m. If the water level is rising at a rate of 20 cm/min when the height of the water is 2 m, find the rate at which water is being pumped into the tank. (Round your answer to the nearest integer.)
Best Answer
+117964 The radius of the water surface is 1/3 the height of the water.
let the height be h and the radius be h/3 (in cm)
$$\\V=\frac{1}{3}\pi r^2 h\\\\
V=\frac{1}{3}\pi(\frac{h}{3})^2 h\\\\
V=\frac{\pi h^3}{27} \\\\
\frac{dV}{dh}=\frac{\pi h^2}{9}$$
$$\\When \;\;h=200\qquad radius=200/3\\\\
V=\frac{\pi\times 200^3}{27}=\frac{8\times 10^6 \pi}{27}\\\\
\frac{dV}{dh}=\frac{\pi h^2}{9}=\frac{\pi\times 200^2}{9}=\frac{4\times 10^4 \pi}{9}\\\\
\frac{dh}{dt}=20\\\\\\$$
$$\\\frac{dV}{dt}=\frac{dV}{dh}\times \frac{dh}{dt}\\\\
\frac{dV}{dt}=\frac{4\times 10^4 \pi}{9}\times 20\\\\
\frac{dV}{dt}=\frac{80\times 10^4 \pi}{9}\\\\\\
\frac{80\times 10^4 \pi}{9}=\frac{dV_{in}}{dt}-\frac{dV_{out}}{dt}\\\\
\frac{80\times 10^4 \pi}{9}=\frac{dV_{in}}{dt}-11000\\\\
\frac{dV_{in}}{dt}=\frac{80\times 10^4 \pi}{9}+11000\\\\
\frac{dV_{in}}{dt}=290,253\;\;cm^3/min\\\\$$
I think the water is being pumped into the tank at the rate of 290 litres per minute to the nearest litre. (edited)
---------------------------------------------------------
LancelotLink will probably have some great retort for my benefit LOL
----------------------------------------------------------
Okay LancelotLink has has some fun at my expense. He got Chimp Ayumu to do it. She is one smart Chimp.
Our answers are still different. Mine does look rather huge.
I'll have to look some more!
How annoying.
----------------------------------------------------------
It's all fixed now and I have a very good excuse!
Someone changed the metric system. Who did that!
Now that there are only 100cm in a metre instead of my original 1000cm in a metre, it works perfectly.
DOH!
+117964 The radius of the water surface is 1/3 the height of the water.
let the height be h and the radius be h/3 (in cm)
$$\\V=\frac{1}{3}\pi r^2 h\\\\
V=\frac{1}{3}\pi(\frac{h}{3})^2 h\\\\
V=\frac{\pi h^3}{27} \\\\
\frac{dV}{dh}=\frac{\pi h^2}{9}$$
$$\\When \;\;h=200\qquad radius=200/3\\\\
V=\frac{\pi\times 200^3}{27}=\frac{8\times 10^6 \pi}{27}\\\\
\frac{dV}{dh}=\frac{\pi h^2}{9}=\frac{\pi\times 200^2}{9}=\frac{4\times 10^4 \pi}{9}\\\\
\frac{dh}{dt}=20\\\\\\$$
$$\\\frac{dV}{dt}=\frac{dV}{dh}\times \frac{dh}{dt}\\\\
\frac{dV}{dt}=\frac{4\times 10^4 \pi}{9}\times 20\\\\
\frac{dV}{dt}=\frac{80\times 10^4 \pi}{9}\\\\\\
\frac{80\times 10^4 \pi}{9}=\frac{dV_{in}}{dt}-\frac{dV_{out}}{dt}\\\\
\frac{80\times 10^4 \pi}{9}=\frac{dV_{in}}{dt}-11000\\\\
\frac{dV_{in}}{dt}=\frac{80\times 10^4 \pi}{9}+11000\\\\
\frac{dV_{in}}{dt}=290,253\;\;cm^3/min\\\\$$
I think the water is being pumped into the tank at the rate of 290 litres per minute to the nearest litre. (edited)
---------------------------------------------------------
LancelotLink will probably have some great retort for my benefit LOL
----------------------------------------------------------
Okay LancelotLink has has some fun at my expense. He got Chimp Ayumu to do it. She is one smart Chimp.
Our answers are still different. Mine does look rather huge. I'll have to look some more!
How annoying.
----------------------------------------------------------
It's all fixed now and I have a very good excuse!
Someone changed the metric system. Who did that!
Now that there are only 100cm in a metre instead of my original 1000cm in a metre, it works perfectly.
DOH!
+132 Chimp Ayumu thinks you climbed up the wrong tree.:)
She says it is 290,253 cm3/min
Chimp Loki is not here to do the LaTex.
It's the weekend, so I only care if there is enough water for swimming!
We Chimps know what's really important in life.
+117964 Well Lancelotlink you have a whole team working with you.
Is that akin to cheating?
Maybe that is why chimp starts with ch ??
+132 Yes! That's true. Any member on the board can match wits with an average human, but our collective numbers are the only way we can compete with high-level humans.
You might say we are a sodality of a sedulity. :)
Lancelot_Link (A.P.E.)
+117964 Yes, well, you are smarter than Yogi. That's for sure. :)
(and he is smarter than the average bear)
+124650 Those chimps aren't monkeying around.......and they'd better not be.....
By my calculations, that tank holds pi/3*(200cm)2*(600cm) ≈ 25,132,741 cm3 of water
Which means that, if Melody's calculations are accurate......that the whole tank will be filled in about (25,132,741 / 27, 936, 268) min ≈ 54 seconds
Thus, taking into account the "Primate Float Factor" and "Imperial Minims," means that the chimps will be "washed" out of the tank after a "Total Recreation Time" of less than one minute.....!!!!!
Someone is NOT going to be a happy camper........
+33178 Here's my contribution:
.
+33178 A somewhat simpler way of tackling at this problem is as follows:
.
+117964 I'll repair my answer.
This time I will have 100cm in a metre instead of the 1000cm in a metre that I had before.
That might help :))
+124650 I understood your method, Alan......but, am I misreading this?? You say "r" at the "time of interest" = 1m ??? I thought it would be 2/3m based on similar triangles.....
+124650 Good job, Alan and Melody !!!
+117964 Thanks Chris
+33178 Chris wrote: "I understood your method, Alan......but, am I misreading this?? You say "r" at the "time of interest" = 1m "
Well spotted Chris - you are right: r = 2/3 m. The r = 1m is there because I inadvertently set the whole worksheet to round all numbers to the nearest integer, instead of just setting the final result to be rounded to the nearest integer.
The correct value is used in the calculation though.
.
9 Online Users
Source: https://web2.0calc.com/questions/water-is-leaking-out-of-an-inverted-conical-tank-at-a-rate-of-11-000-cm3-min-at-the-same-time-that-water-is-being-pumped-into-the-tank-at-a
0 Response to "Water is Being Pumped Continuously Into a Tank at a Rate That is Inversely Proportional"
Post a Comment