Question

A machine starts dumping sand at the rate of 20 m3/min, forming a pile in the shape of a cone. The height of the pile is always twice the length of the base diameter.After 5 minutes, how fast is the area of the base increasing?
When after 5 min the height is 11.516, after 5 min the height is increasing at .7678m/min, after 5 min the base radius is increasing at .19196 m/min

Answer 1

Hello, S. W.!
Nothing like doing a little calculus bright and early on a fine Sunday morning, eh?

Have you done this type of problem before? If so, what's the topic?

Looking at your problem statement, the following to-do list comes to mind:

1) Write the equations of the volume of a cone and that of the area of the base.
2) Write a formula for the height of the pile in terms of the base radius (not diameter). Base it upon "The height of the pile is always twice the length of the base diameter."
3) Re-write the formula for the volume in terms of the base radius only.
4) Determine the proper label for the quantity described here: "A machine starts dumping sand at the rate of 20 m3/min."
5) Determine the proper label for "how fast is the area of the base increasing"

There's more to this process, but my list should get you started.

Answer 2

Hello, I found that After 5 minutes, how fast is the area of the base increasing?
When after 5 min the height is 11.516, after 5 min the height is increasing at .7678m/min, after 5 min the base radius is increasing at .19196 m/min, so I already workd out the volume formula v=1/3pir^2h to solve these

Answer 3

It is rate of change

Answer 4

Have you any remaining questions or concerns about this problem? If so, please be specific about what you still need to know to solve it completely.

Answer 5

Yes I need to figure out how to find "after 5 minutes, how fast is the area of the base increasing"

Answer 6

If you were able to figure out the first part of this problem, the rest should be fairly straightforward, and should draw upon the work you've already done. At this point I'd ask you to explain what you've done so far and where you're unsure of what to do next. While you don't absolutely have to answer my questions (above), doing so might be useful.

Answer 7

Re: "It is rate of change:" I'd also call this a "related rates" problem.

Answer 8

This is what I have done so far 1/ What is the height of a cone volume 100m^3 with a base diameter 1/2 the height?
h = 4r
V = 1/3 pi r^2*h
100 = 1/3 pi r^2*4r
r^3 = 100/(4/3 pi)
r = cuberoot(100/(4/3 pi))
4r = 4*cuberoot(100/(4/3 pi))
h = 11.5176 m

2/ V = 1/3 pi (1/4 h)^2*h = 1/3 pi 1/16 h^3
dV/dt = pi 1/16 h^2 dh/dt
20 = pi 1/16 (11.5176)^2 dh/dt
dh/dt = 20/(pi 1/16 (11.5176)^2) = .7678 m/min

3/ As h = 4r
Then dr/dt = 1/4 dh/dt
dr/dt = .7678/4 = .1920 m/min

Answer 9

Thank you for sharing this considerable amount of work. Having done this much work, where do you need help?

Answer 10

I don't see a formula for the area of the base in what you've shared with me. Care to write that formula?

Care to differentiate that formula with respect to t?

Answer 11

Ok so in respect to h if the radius is 1/4*h the area of the base would be a=pi(1/4*h)^2?

Answer 12

Why in respect to t?

Answer 13

Let me ask you this: Why did you find the derivatives with respect to t here? " Then dr/dt = 1/4 dh/dt "

Answer 14

I was looking for a time rate

Answer 15

soo.. area might be pi 1.4^2 dh/dt

Answer 16

Finding the derivative of the formula for the area of a circle with respect to t gives you precisely that: the time rate of change of the area with respect to t.

Answer 17

1. write the formula for the area of the base in terms of the radius, r.
2. differentiate this formula with respect to t.
3. plug in the known values. Identify any values that are not yet known; find them.

Answer 18

1. it would be pi(1/4h)^2 because radius is in respect h right?

Answer 19

Well r is cubrt(100/(4/3pi) so I guess the area in respect to r is pi[100/(4/3pi)]^2?

Answer 20

That looks reasonable. However, the question is not what the area is, but rather how fast the area is changing with respect to time. Agree with that?

If A=Pi*r^2,
differentiate both sides with respect to t, please.\[\frac{ d }{ dt }(A=\pi*r^2)\]

Answer 21

Although instead of doing that couldn't I just substitute the rate of change of the height .7678 so the problem would look like da/dt = pi (1/4 * .7678)^2?

Answer 22

If I understand your math problem correctly, the part we're focusing on now is "After 5 minutes, how fast is the area of the base increasing? "

If you see the phrase, "how fast is the area of the base increasing?" you know, or should know, that we need to evaluate a derivative, and that this derivative is (dA/dt).

If \(A=\pi*r^2,\) then \(\frac{ dA }{ dt }=2\pi*r\frac{ dr }{ dt}\).

Answer 23

If you already know r and (dr/dt), all you have to do is to substitute those two quantities into the above equationf or dA/dt.

Answer 24

After 5 minutes, what is the radius, r? Have you already calculated that?
After 5 minutes, what is the rate of change of the radius with respect to t? Have you already calculated that?

Answer 25

I have "After 5 minutes, what is the rate of change of the radius"

Answer 26

and I have r = cuberoot(100/(4/3 pi))

Answer 27

Although if I know that the height is 11.52 and teh radius in respect to h is 1/4h. Than the starting radius would be 2.8

Answer 28

Are you saying that "at t=5 minutes, r = cuberoot(100/(4/3 pi)) ? If so, good.

To find \(\frac{ dA }{ dt },\) you need \(\frac{ dr }{ dt }\) at the specific time t=5 minutes. Do you have that? If not, please find it.

Answer 29

It's very important that you keep the two different kinds of formulas separate:
1) formulas that relate your variables (except for t) to each other, such as \(A=\pi*r^2\), and 2) formulas that relate rates of change.

Answer 30

Stop for a moment and ask y ourself, "At this moment, my goal is to find the ( ? ), and to do that, I need this information: ( ? )."

Answer 31

my goal is to find how fast the area of the base is increasing

Answer 32

Right, and to do that, "I need ... "

Answer 33

for that I need to find r and dr/dt

Answer 34

dr/dt after 5 minutes is .19196

Answer 35

"and I have to do that for the specific time, t=5 minutes" (assuming that either or both depends upon the time).l

Answer 36

Your "dr/dt after 5 minutes is .19196" sounds great (although I am not actually checking your numeric results).

Answer 37

"and my radius, r, after 5 minutes, is ... "

Answer 38

if the rate of change is .19196 and the original radius is 2.8 than .19196*5=.9598. Then 2.8 +0.9598 = 3.7598?

Answer 39

The radius of the base of the conical pile after 5 minutes will be greater than the original radius, since the machine continues to dump sand onto the pile. Do you have a formula for the radius as a function of time, t? If you do, write out that formula:\(r(t) = 2.8 + ???\). Then substitute t=5 (minutes).

Once again, I'm not checking your work, but rather asking questions that will help you check your results yourself.

Answer 40

ummm do you meant something like...dV /dt = (4/3) 3r^2 dr/dt

Answer 41

I think you've already tackled that.
If I'm reading your math problem correctly, your focus right now is to find how fast
the area of the base is changing. Is (dV/dt) relevant to that? Why or why not?

Answer 42

hmmm not dV is for the volume I am looking for dr for the radius

Answer 43

Let's do a quick review here.
Our goal is to determine r and dr/dt at t=5 minutes.
\[V _{cone}=\frac{ 1 }{ 3 }\pi*r^2*h\]
and h=4r (you already have this result).

Answer 44

Are you saying do 2.8 +(1/4) dh/dt because dr/dt =(1/4) dh/dt

Answer 45

Thus, \[V _{cone}=\frac{ 4 }{ 3 }\pi*r^3\]

Answer 46

Are you completely comfortable with this discussion, so far? If not, where do you need clarification?

Answer 47

well I was until we got to solving for r

Answer 48

It didn't seem like another derivative was needed to solve

Answer 49

OK: So, we have a formula for V in terms of r, and you want to solve for r?

\[V=\frac{ 4 }{ 3 }\pi*r^3\rightarrow 3V=4\pi*r^3, OK?\]

Answer 50

(All I did was to multiply b oth sides of this equation by 3.)

Answer 51

I'm waiting for y our "OK."

Answer 52

that isn't finished is it? you must go on to solve for r in that equation?

Answer 53

Right, you must.

Answer 54

Ok

Answer 55

If 3V=4*Pi*r^3, divide both sides of this equation by (4*Pi). What do you get?

Answer 56

3v/4pi= r^2

Answer 57

That's fine, except that it'd be r^3, not r^2, right?

Answer 58

right so it would be cubrt(3c/4pi)= cubrt(r^3)

Answer 59

You're on the right track. I'd hope that you'd label this quantity:\[r=\sqrt[3]{\frac{ 3V }{ 4\pi}}\]

Answer 60

Here you have the formula for r that you wanted.
To calculate r after 5 minutes, you'll need to know the volume of the cone. Simply substitute that volume into the formula for r, above, and you'll then have the radius of the base of the cone.

Answer 61

Your final goal is to determine how fast the area of the base is changing after 5 minutes.
Since\[A=\pi*r^2,\]

Answer 62

Wait but for after five minutes (20m^3/min*5= 100m^3/min) I had r = cuberoot(100/(4/3 pi))

Answer 63

\[\frac{ dA }{dt }=2\pi*r \frac{ dr }{ dt }\]

Answer 64

You already have a formula for r (see above). I believe we also found a formula for (dr/dt). If so, we're all set. Just substitute your r and your (dr/dt) into the above formula for (dA/dt) and you're done (assuming that you've calculated r and (dr/dt) for time t=5 min.).

Answer 65

\[\large h = 4r\] \[\large V = \pi r^2 \frac h3\]
so \[\large V = \pi r^2 \frac {4r}3\] \[\large V = \pi r^3 \frac 43\]
when t = 5mins:
\[\large V = \frac {20m^3}{min} \times 5mins\] \[\large V = 100m^3\]
therefore r = ?
\[\large V = \pi r^3 \frac 43\] \[\large 100 = \pi r^3 \frac 43\]
\[\large r^3 = \frac {300}{4 \pi}\] \[\large r = \sqrt[3]{75/\pi}\]
\[\large h = 4r\] \[\large h = 4 \sqrt[3]{75/\pi}\]

base area change as a function of volume change
volume is increasing at 20m^3/min
if \[\large V = \pi r^3 \frac 43\] then \[\Large r = \sqrt[3]{ \frac {3V}{4 \pi}}\]

if volume at 5mins = 100, and volume increases at 20m^3/min
\[\large V = 20t\] therefore \[\Large r = \sqrt[3]{ \frac {3\times 20t}{4 \pi}}\]
\[\Large r = \sqrt[3]{ \frac {60t}{4 \pi}}\]
\[\Large r = \sqrt[3]{ \frac {15t}{\pi}}\]

\[\Large \frac{dr}{dt} = r' = \frac {(\frac 5\pi)^{1/3}}{3^{2/3} \times \sqrt[3]{t^2}}? \]
i think...?

Answer 66

@Jack1: Thank you for contributing. I'll leave it to you to compare what you have done, on one side, and what S.W. and I, on the other side, have done. if you find discrepancies, we could discuss and resolve them.

Note that S.W. needs to find the actual (numeric) rate of change of the area of the base with respect to time after 5 minutes.

Answer 67

xx confusald

Answer 68

@S.W.: I'd suggest you read over our entire conversation and determine whether there's anything left that you're not sure about. I'll give you some time to do that. If you feel confused, it's up to you to explain where and why you feel confused, so that Jack1 or I could address that.

Answer 69

of course

Answer 70

I may not answer right away, if and when you reply again, but eventually I'll see your work and respond to your questions/concerns. In the meantime, you might want to move on to another math problem.

Answer 71

This is the only one

Answer 72

k @S.W. so ur equations and mine match up for the volume and height
at t = 5 mins V = 100, h = 11.518, and r = 2.879

Answer 73

I feel like I had already had r and dr/dt... actually I did... derp. r=2.8 after 5 min and dr/dt= .19196 after five min

Answer 74

Now it seems as though I just need to substitute. 2pi * 2.9* 0.19196= 3.38

Answer 75

Great to see you two working together. By continuing to compare your results and to resolve any differences, you'll most likely come up with the correct solution. Good going!

Answer 76

it seems like dA/dt= 2pir dr/dt would be the circumference

Answer 77

V always = 20 t, where t is the time in minutes
therefore \[\Large r = \sqrt[3]{ \frac {15t}{\pi}}\]
so 15/pi approximately = 4.7746, therefore
\[\Large r = \sqrt[3]{ 4.7746t}\]
\[\Large r = 1.6839 t^{1/3}\]
\[\large dr/dt = \frac {0.5613}{t^{2/3}}\]

rate of change for r = {0.5613} / {t^{2/3}

Area of base = pi x r^2
\[Area= pi \times r^2\]
\[\large Area= pi \times (1.6839 t^{1/3})^2\]
\[\large Area= pi \times 2.83552 t^{2/3}\]
therefore base area as a function of time =
\[\large Area= 8.908 t^{2/3}\]

therefore if r is doubled, area is quadrupled, as area is a function of r^2, not r
so area of base is increasing at

Answer 78

Ok, but I don't need to know what the area of the base is I need to know how fast it is increasing

Answer 79

\[\Large dA/dt (8.908 t^(2/3)) = \frac {5.93867}{t^{1/3}}\]

Answer 80

so that's how fast it's increasing

Answer 81

ummm some how this should be in m/min

Answer 82

units of this are meters / min, simply plug in your required value for time
if u want when t = 5 mins
5.93867/t^1/3
5.93867/5^1/3
=...?

Answer 83

Hmmm oh I see. Are you sure this is correct?

Answer 84

not sure, checkin now, one sec

Answer 85

my bad, the units are in m^2/minute, not meters/minute
otherwise...yep, i'm sure, above is ROC of cones' base area, so when t = 5, instantaneous ROC is 3.4729 m^2/min

Answer 86

how did you get 3.4729 m^2/min?

Answer 87

dA/dt = 5.93867/t^1/3... so if t = 5
= 5.93867/5^1/3
= 3.4729 m^2/min

Answer 88

Oh ok I calculated wrong . This answer is very similar that that of dA/dt=2 pi* 2.8*.19192 do you think they correlate?

Answer 89

unsure, where/how did u derive that formula from?

Answer 90

Actually they are the same answer. @mathmale gave it to me

Answer 91

cools... but importantly do u understand how we got here?

Answer 92

hahah yes I've been there for the past 3 hours, but my way was not so complex. I am sure your ways were the true mathematical ways to solve

Answer 93

all good dude, so long as u can arrive at the right answer in all situations use ur way if it's easier

Answer 94

yeah right that wouldn't fly over so well with the teacher, he's straight by the book, it's just that eventually I get confused by the details

Answer 95

Thanks for the help man!

Answer 96

all good dude, u got there hey, we just confirmed u were right, s'all
anytime, slaters man!

Answer 97

A one line solution using Mathematica 9 for the calculations is attached.