Ana 25 Nov 2015, 09:46
Please help me solve a rate of change problem about a conical tank wit vertex down. i dont know the equation i have to use
I read the examples on the page, but none of them were like my one.
X
Please help me solve a rate of change problem about a conical tank wit vertex down. i dont know the equation i have to use
Relevant page <a href="/applications-differentiation/4-related-rates.php">4. Related Rates</a> What I've done so far I read the examples on the page, but none of them were like my one.
Newton 26 Nov 2015, 03:43
Hello Ana
You will need to provide more details before I can help you.
You may get some inspiration from this page, which is also talking about a conical tank:
4. The Graph of a Function
(go to the bottom of the page.)
Regards
X
Hello Ana You will need to provide more details before I can help you. You may get some inspiration from this page, which is also talking about a conical tank: <a href="/functions-and-graphs/4-graph-of-function.php">4. The Graph of a Function</a> (go to the bottom of the page.) Regards
Ana 27 Nov 2015, 05:09
The question says the tank has diameter 3 m at the top, and is 5 m high. What is the rate of change of the height of the water at time t?
X
The question says the tank has diameter 3 m at the top, and is 5 m high. What is the rate of change of the height of the water at time t?
Murray 28 Nov 2015, 08:34
You need to show us your attempts at the problem.
Did you look at that page I suggested?
Hint 1: What is the formula for the volume of a cone?
Hint 2: What is the rate of flow of the water?
X
You need to show us your attempts at the problem. Did you look at that page I suggested? <b>Hint 1: </b> What is the formula for the volume of a cone? <b>Hint 2: </b> What is the rate of flow of the water?
Ana 29 Nov 2015, 07:43
I did look at that example, but it says r = h and my question does not have that. So I got stuck.
Anyway, volume of a cone is pir2h/3
The question doesn't say what the water flow rate is.
X
I did look at that example, but it says r = h and my question does not have that. So I got stuck. Anyway, volume of a cone is pir2h/3 The question doesn't say what the water flow rate is.
Murray 30 Nov 2015, 11:08
I encourage you to use the math input system, so we can read your answer more easily.
You just need to put the following in between back ticks,
V = (pi r^2 h)/2
So it looks like this:
`V=(pi r^2 h)/3`
You're right - the question doesn't give us a number for the flow rate, but we can just give it a letter, say `f` and assume it is constant at liters per minute, say.
Hint 3: How long will it take the water to flow out?
Hint 4: How do you find the rate of change of the volume?
X
I encourage you to use the math input system, so we can read your answer more easily. You just need to put the following in between back ticks, <code>V = (pi r^2 h)/2</code> So it looks like this: `V=(pi r^2 h)/3` You're right - the question doesn't give us a number for the flow rate, but we can just give it a letter, say `f` and assume it is constant at liters per minute, say. <b>Hint 3: </b> How long will it take the water to flow out? <b>Hint 4: </b> How do you find the rate of change of the volume?
Ana 30 Nov 2015, 20:21
If it flows at `f` liters per minute, then it will take `V/f` minutes to empty out.
Do we use `(dV)/(dt)` to find rate of change of volume?
Then what?
X
If it flows at `f` liters per minute, then it will take `V/f` minutes to empty out. Do we use `(dV)/(dt)` to find rate of change of volume? Then what?
Murray 01 Dec 2015, 20:40
Yes, `(dV)/(dt)` is correct.
On the right hand side of our volume equation, what is constant and what varies as the water empties out?
X
Yes, `(dV)/(dt)` is correct. On the right hand side of our volume equation, what is constant and what varies as the water empties out?
Ana 02 Dec 2015, 06:19
`pi` is constant, but `r` and `h` vary as time goes on.
I use product rule:
so `(dV)/(dt) = pi( r^2 (dh)/(dt) + (h) 2r (dr)/(dt))`
But I'm lost again.
X
`pi` is constant, but `r` and `h` vary as time goes on. I use product rule: so `(dV)/(dt) = pi( r^2 (dh)/(dt) + (h) 2r (dr)/(dt))` But I'm lost again.
Murray 03 Dec 2015, 01:15
What you have is correct, but it's more complicated than it needs to be.
What's the relationship between `r` and `h`?
Can you simplify `V` now?
X
What you have is correct, but it's more complicated than it needs to be. What's the relationship between `r` and `h`? Can you simplify `V` now?
Ana 03 Dec 2015, 19:33
Radius is 1.5 m and height is 5 m, so
`r/h = 1.5/5 = 0.3`
So `r = 0.3h`
So now
`V = (pi r^2 h)/3 ` `= (pi (0.3h)^2 h)/3 ` `= 0.03 pi h^3`
I see it now
`(dV)/(dt) = 3(0.03 pi h^2)(dh)/(dt)` `=0.09 pi h^2(dh)/(dt)`
Where do we use `f`?
X
Radius is 1.5 m and height is 5 m, so `r/h = 1.5/5 = 0.3` So `r = 0.3h` So now `V = (pi r^2 h)/3 ` `= (pi (0.3h)^2 h)/3 ` `= 0.03 pi h^3` I see it now `(dV)/(dt) = 3(0.03 pi h^2)(dh)/(dt)` `=0.09 pi h^2(dh)/(dt)` Where do we use `f`?
Murray 04 Dec 2015, 16:46
Well, `f` is just the rate of flow, so it's equal to the change in volume of the water. So we have
`f=0.09 pi h^2(dh)/(dt)`
Can you get the expression of the rate of change of the height now?
X
Well, `f` is just the rate of flow, so it's equal to the change in volume of the water. So we have `f=0.09 pi h^2(dh)/(dt)` Can you get the expression of the rate of change of the height now?
X
So is it this? `(dh)/(dt) = f/(0.09pih^2)`
Murray 06 Dec 2015, 06:06
Yes, you are correct. What are the units?
X
Yes, you are correct. What are the units?
X
The units will be m/min
X
Correct.
Serious34 17 May 2022, 02:18
Wow, I have the same problem with my class. Van Nuys Concrete Contractors - Cardea Concrete
X
Wow, I have the same problem with my class. <a href="https://cardeaconcrete.com/van-nuys-concrete-contractors/">Van Nuys Concrete Contractors - Cardea Concrete</a>
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