# Second derivatives when pouring juice into a cup

1. Oct 31, 2017

### mindy

Its a question about volume increase (in units cm^3) and height increase (cm) when pouring juice into a cup. Its stated that the volume of the juice in the cup increases at a constant rate, so I know the volume derivatives are zero. But the shape of the cup is inconsistent and there is alot of variance in it, so the height increase is not constant. I know the first derivative of the height increase is positive... my professor hinted that the second derivative of height increase would be negative... how is that so?

Last edited by a moderator: Oct 31, 2017
2. Oct 31, 2017

The derivatives of interest are $\frac{dh}{dt}$ and $\frac{d^2 h }{dt^2}$. You are given $\frac{dV}{dt}=constant$. Suggestion is to write $V=V(h)$, and compare derivatives $\frac{dV}{dt}$ and $\frac{d^2 V}{dt^2}$ using the chain rule. Also $\frac{d^2 V}{dt^2}=0$. $\\$ Additional hint: What can you say about $\frac{d^2 V}{dh^2}$ if the volume is written as $V(h)=\int\limits_{0}^{h} A(h') \, dh'$, and $\frac{d A(h)}{dh} >0$ ?

Last edited: Oct 31, 2017
3. Oct 31, 2017

### Staff: Mentor

The first derivative measures the rate of change, so it's not zero for the volume. It's a positive constant. The second derivative of the volume is zero, because it measures the rate to which the rate of change changes. The same is true for the height. As it doesn't change constantly, it's first derivative actually changes with time. And as this change isn't constant, the first derivative of this change isn't zero, which is the second derivative of the height function. Whether this is positive or negative depends on whether the rate of change is sublinear or hyperlinear.

4. Oct 31, 2017

### mindy

What does the A stand for in your last equation?

5. Oct 31, 2017

### Staff: Mentor

The surface area at height $h$.

6. Oct 31, 2017

I had a slight error. I corrected it. (I originally wrote $V(h)=A(h) h$. The correct expression is $V(h)=\int\limits_{0}^{h} A(h') \, dh' \,$).

7. Oct 31, 2017

### mindy

i know that the height increase and volume increase first derivatives are positive. thats helpful
the units of the first volume derivative is cm^3/s (seconds)
second derivative is cm^3/s^2
height first derivative is cm/s
second derivative of height is cm/s^2

8. Oct 31, 2017

### mindy

not sure why the photo of the problem didn't go through last time. here it is

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9. Oct 31, 2017

### mindy

I guess I was confused because I thought the derivative of a constant was always zero

10. Oct 31, 2017

The trickiest part of this for someone just learning calculus is applying the chain rule a couple of times. Suggestion: $dV/dt =(dV/dh)(dh/dt)=constant$. Finding $d^2V/dt^2 =0=...$ using the chain rule takes a little more work, but it should give you the result you need. $\\$ Additional hint: By the chain rule, $d^2 V/dt^2=(d(dV/dh)/dt)(dh/dt)+(dV/dh)(d^2 h/dt^2)=0$. The first term there needs to be evaluated with the chain rule one more time..

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11. Oct 31, 2017

### FactChecker

You will get answers that are more specific to your questions if you show us the Figure 7 of the cup and the rest of the problem statement.

12. Oct 31, 2017

### mindy

thanks everyone
thanks charles =)

13. Oct 31, 2017

See my edited post of post 10. That is very close to the result you need.

14. Oct 31, 2017

### mindy

here it is, for those curious

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15. Oct 31, 2017

### mindy

nevermind. my phones too blurry. i give up

16. Oct 31, 2017

I was able to read it. They want to know the sign of the terms $(dh/dt)$ and $(d^2 h/dt^2)$, etc. If you look at my hint in post 10, you need to compute $\frac{d (\frac{dV}{dh})}{dt }$ using the chain rule. Once you have that, it will allow you to compare $d^2 h/dt^2$ with $d^2 V/dh^2$. (See also my hint for $d^2V/dh^2$ in the last part of post 2).

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17. Oct 31, 2017

### FactChecker

First you should consider how the shape of the cup effects the rate of height increase and the derivative of that. I think that your statement in the original post about the first derivative of height increase being positive is not always true as the cup fills up. [CORRECTION: Of course it is always positive. The second dirivative is not always positive.]
Consider the derivatives where the cup diameter is expanding, where it is contracting, the sudden change in the middle, and the constant curvature of the top and bottom halves.

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18. Oct 31, 2017

@FactChecker input can be quantified by computing $d^2 V/dh^2 =dA(h)/dh$ as my hint in the last part of post 2 is suggesting. Using that, along with the equation in post 10, and the problem is really very straightforward.

19. Oct 31, 2017

### FactChecker

Yes, I know. I can do it easily. But I think that the original poster is less familiar with the application and meaning of derivatives in real problems and I'm not sure that the problem was asking for actual formulas as opposed to a more qualitative characterization.

20. Oct 31, 2017

### mindy

you're right factchecker, I am new to all this. I took a 6 year break from pre calc to this class (calc 1) and prior to that, it'd been like 6 years since I took algebra or trig. so I am very behind. I own calculus for dummies, and a complete idiots guide to calculus, and I watch alot of kahn academy and youtube videos. but i am struggling and am very behind academically

21. Oct 31, 2017

### mindy

"Today at 10:46 AM#17
FactChecker
3,085 / 887
Gold Member
First you should consider how the shape of the cup effects the rate of height increase and the derivative of that. I think that your statement in the original post about the first derivative of height increase being positive is not always true as the cup fills up.
Consider the derivatives where the cup diameter is expanding, where it is contracting, the sudden change in the middle, and the constant curvature of the top and bottom halves."

hey factchecker. even if the height derivative fluctuates at the shape of the glass (I know itll slow down drastically at the wide parts, and increase in height quicker at skinnier parts of the cup) wouldn't it still be a positive derivative? i guess i'm trying to imagine this on a graph. I've attached the graph. lets say that at the skinny parts of the cup, it increase 10cm/s. and then at the widest parts of the cup, it decreases to an increase of 4cm/s. isn't it still positive because it's still an increase? or am i just not understanding

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22. Oct 31, 2017

### mindy

i forgot to include my other increases. i'm not sure if my image came up clearly. so if its 10cm/s at the skinniest part, and decreases down to 4cm/s at the widest part.... the first second would be 10cm increase, the second second would be 8cm increase, the third second would be 6 cm increase, the fourth second 4cm increase, and then progressively back up to 10cm increase (I'm just using these numbers as an example based loosely upon the shape)... isn't it all still increases? even if the increases fluctuate? or would the derivative not be a positive nor a negative nor a zero, would it be a fluctuating derivative (if that makes sense?) would it bounce between being a positive and a negative?

23. Oct 31, 2017

In this complicated orange juice cup, if you worked with the formulas that you can derive from calculus, you could treat every different part of the cup very quickly. Otherwise, it is tricky trying to do a qualitative explanation for the different parts of the cup. $\\$ The formulas will tell you that if $dA/dh >0$ , that $d^2 h/dt^2 <0$, once you compute the various derivatives, (from the chain rule), and equate them to each other. Alternatively, you have parts of the cup that have $dA/dh<0$, and for those parts, $d^2 h/dt^2>0$.And if the cup has a part that is like a cylinder, then $dA/dh=0$, so that $d^2 h/dt^2=0$.

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24. Oct 31, 2017

### FactChecker

Note: I corrected my earlier post. The rate of increase (first derivative) is always positive as you originally stated. The second derivative is not always positive.

25. Nov 1, 2017

### mindy

thankyou guys. handing this in today =)