Try This Calculus Problem
Water drips into the cup, whose shape is shown in the image, at a steady rate.
What can we say about the rate of change of the height of water level?
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
5 solutions
I agree. The area of a cross section increases and the rate of change of the of the water volume decreases since the water inflow is constant. Hence the rate of change of the water height decreases.
but that should have been given that we should assume the glass to brim to be incremental with height.
Log in to reply
Water drips into the cup above at a steady rate
It can be seen that the radius of the glass increases with height.
Log in to reply
One should never assume by "looks" of a mathematical question. One can infer the glass is sloped outward, but a simple visual does not suffice the burden of proof requisite of such an assumption.
Sorry, but the information remains absent from the problem.
But if the area of the cross section increases, shouldn't the rate if change also increase? Because rate of change is slope which is rise/run? So you'd have to do greater riser/run?? Sorry, still kind of confusing.
Log in to reply
@M J – Does it take more time to fill up a small beaker or a large beaker (by cross sectional area) if we use the same tap (water flow)? Because Volume = Height * Cross sectional area, we get that Height = Volume / Cross sectional area. So, given the same rate of flow of volume, what happens to the height as the cross sectional area increases?
Nice one and great idea
I am pretty sure the rate of change is constant. Negative, but constant. It is like you where saying there is a car with linear increasing. What is the rate change of the speed. It is the acceleration so it's the derivative of the speed. And this a constant.
Log in to reply
yes it would be if glass was instead a beaker with a constant height.
after highest water level it goes out or as the the top diameter is not to scale the the rate of change of height can also be said to be zero
The picture shows a sloped edge cup, and while thinking about the problem it's easy to envision a cylindrical cup. The answer differs based on the shape of the cup.
Because the sides of the drop have vertical tangents at the water level, the volume of water arriving per time is instantaneously constant at its maximum. Although it was zero a moment before (between drops), and will be zero again a moment later (between drops). So based on the assumptions of increasing cup interior area with height (inside wall slopes outward as the single line representing the cup does) and drop sides at water level are vertical as shown, then the water level increase rate is decreasing as you say. The problem would be more interesting if the drop sides were shown rising outward at a small angle to water level, so the correct answer was increasing, despite the glass widening. Key point: constant drip implies intermittent water arrival with continuously varying rate of water volume arriving throughout the arrival of a drop.
The shape of the cup should have been given... Aren't we not to assume that diagrams are to scale?
The question is about the interval of time where the function is like polynomial or exponential or something but if you really insist it can be an infinitely tall cup and an incorporeal tap that raises so it always stays the right height above the water dripping infinitely...
I think the answer is wrong. Obviously the rate of rise in water level decreases, but the rate of change could equally be considered to be increasing in a negative direction. In other words, as the level rises the rate at which it rises diminishes and so the rate of change increases. (First or second derivative?)
Log in to reply
Henry Adam please refine
funny thing, I just realized I gave up and looked at the solution for this problem but it is so simple that's weird I totally understand the answer but apparently I didn't originally...
if the cup is not a cylinder then it SHOULD be mentioned in the statement!!! it's not right when one has to make that sort of guesses based on some schematic drawing!! if the solution is based on the fact that the shape of that cup wasn't cylindrical then this particular puzzle is rubbish!
Log in to reply
The problem states "Water drips into the cup below ". How would you rephrase it to make that clearer?
Log in to reply
simply: you mention that the cup shape is like it's shown on the drawing! or "consider cup shape to be..."... smth like that...
Log in to reply
@Nik Gibson – Thanks. I've elaborated that "below" refers to "whose shape is shown in the image below".
Log in to reply
@Calvin Lin – well it didn't... i don't think i'm the only "silly bugger" to have considered that bloody cup to be a cylinder ..:)) no offence!
But isnt the rate of change in hight constant since the rate of change in radius of a cross-section, hence the cross-sectional area, is constant?
Log in to reply
Note that if the rate of change in radius of a cross section is a non-zero constant, then this means that the cross-sectional area is strictly increasing, or strictly decreasing (In particular, not constant). So the second part of your statement isn't true.
It is true that if the cross sectional area is a constant, then the rate of change in height is a constant. However, you're vocalising a very common misconception when the rate of change in radius of the cross section is a non-zero constant.
Even though we have V = A × h , it is not true that d t d V = d t d A × d t d h (which I'm presuming you're using to get that the rates have to be a constant).
Insetad, we have to apply the chain rule to obtain d t d V = A × d t d h + d t d A × h . In this case, we know that the rate of change of volume is a constant. The rate of change of area is a positive constant, hence to keep the RHS to be a constant, the second term will have to be negative . Thus, we can conclude that the rate of change of the height is decreasing.
Log in to reply
I think I misunderstood the answer opportunities. I thought “The rate of change is decreasing” meant that the rate change in speed (dV/dt), that is the acceleration, and not the speed itself. However, thank you for your quick answer 👍
Log in to reply
@Mads Carlsen – Ah. The question was about the "rate of change of the height of the water level", so we're asking for the "speed".
You raise a good question about the "rate of change of the rate of change of the height of the water level". What can we say about it?
@Mads Carlsen but it isnt??
I didn't know I was supposed to take the graphic that seriously, goddamnit.
A single drop of water causes the same Δh and h increases as t increases, therefore Δh/h decreases.
No.. single drop of water would cause constant volume increase. Now since area is increasing delta_h would decrease for constant volume.
I assume the glass has the same diameter all the way up, so each level fills at the same rate.
It does not say how much water was in the cup to begin with. You can't assume it was empty. Over time the system will reach a steady state, either of water evaporating at an equal rate to it dripping into the cup, or of the water overflowing a full cup. Either way, my opinion is that the rate of change is zero.
when did you see the rate to be determined as "dh/h"? the rate has nothing to do with "h" itself!! by the way in physics the word "rate" is mostly used when it is about d(smth)/dt (when t is time)!
Rate decreasing as surface area increases
Well, if you notice, the area of cross section of the container increases from bottom to top. Each drop increases the height of the water level by the same amount. Thus, it is quite intuitive to say that when the water level increases more drops of water is required to raise the level by equal amount. Hence, the rate of change of water level is decreasing.
You could also view this is as a graphical problem: as you get closer and closer to a maximum, your rate of change of the rate of change (aka double derivative) decreases.
it's the tallness of the cup so: a×x=2π but its still not equaled.
In the cup, the cross-sectional area increases as the height increases. Thus, as the height of water level increases, it will take more water to fill up the next unit, and hence the rate of change will be decreasing .