Need help with derivatives.

2007-11-26 18:18:06

#1

Liberal-Sl@yer: Certified BF2S Asshole; +131|6936|The edge of sanity

I cant figure out this problem so maybe you guys can help.

So bases of the problem is this

At one minute past midnight, April 14th, the river level was 110.0 feet. Estimate the height 24 hours later and 48 hours later. Explain how you made your estimates.

Your given the background information that April 14th 12:01am = 0 for t(or time)

and that dh/dt=.2 feet per hour (as the water rises at 2 feet per hour at t=o)

h=110 at t=0

so i figured that i would take the formula h=rt or height= rate x time

so i should be able to take the derivative of the h=rt equation, but one problem i cant get dr/dt and i need it for the equation.

So right now my equation is

(dh/dt) x h = (dr/dt) x r x t

2007-11-26 18:20:09

#2

AussieReaper: ( ͡° ͜ʖ ͡°); +5,761|6633|what

how can you have time equal zero?

2007-11-26 18:22:22

#3

Hurricane: Banned; +1,153|7110|Washington, DC

wut?

If it rises .2 feet per hour, couldn't you multiply that by 24 and 48, then add those results to 110.0?

2007-11-26 18:22:38

#4

White-Fusion: Fuck; +616|7031|Scotland

The answer is B: Firetruck

2007-11-26 18:24:50

#5

nukchebi0: Пушкин, наше всё; +387|6804|New Haven, CT

Can't you just cheat and multiply .2 by 24 for the first, and then by another 2 for the second?

Edit: Doh.

Edit^2: What are you guys studying in calculus currently? It would make it much easier to figure out how to solve.

Edit^3: I don't see why this is calculus, other than in you know the rate of change, which is shown by a derivative.

Last edited by nukchebi0 (2007-11-26 18:27:57)

2007-11-26 18:26:13

#6

Liberal-Sl@yer: Certified BF2S Asshole; +131|6936|The edge of sanity

no its a rate of change question i cant do the cheat shit

2007-11-26 18:28:30

#7

White-Fusion: Fuck; +616|7031|Scotland

White-Fusion wrote:
The answer is B: Firetruck

2007-11-26 18:30:36

#8

nukchebi0: Пушкин, наше всё; +387|6804|New Haven, CT

I figured it out.

You know the rate of change as .2.

It is the derivative of the h function.

Therefore, h=.2x+b.

Since h(0)=110, b must be 110.

With function for height as h=.2x+110, substitute 24 and 48 to find heights.

Make sure you show that x=number of hours river is rising.

The dr/dt you were trying to find is a known constant, hence why you couldn't find it.

Last edited by nukchebi0 (2007-11-26 18:31:58)

2007-11-26 18:33:49

#9

Liberal-Sl@yer: Certified BF2S Asshole; +131|6936|The edge of sanity

nukchebi0 wrote:
I figured it out.

You know the rate of change as .2.

It is the derivative of the h function.

Therefore, h=.2x+b.

Since h(0)=110, b must be 110.

With function for height as h=.2x+110, substitute 24 and 48 to find heights.

Make sure you show that x=number of hours river is rising.

The dr/dt you were trying to find is a known constant, hence why you couldn't find it.

damn that seems to simple

2007-11-26 18:34:43

#10

White-Fusion: Fuck; +616|7031|Scotland

Liberal-Sl@yer wrote:
nukchebi0 wrote:
I figured it out.

You know the rate of change as .2.

It is the derivative of the h function.

Therefore, h=.2x+b.

Since h(0)=110, b must be 110.

With function for height as h=.2x+110, substitute 24 and 48 to find heights.

Make sure you show that x=number of hours river is rising.

The dr/dt you were trying to find is a known constant, hence why you couldn't find it.
damn that seems to simple

God damnit! I was so close aswell, damn that dr/dt!

2007-11-26 18:35:55

#11

nukchebi0: Пушкин, наше всё; +387|6804|New Haven, CT

Liberal-Sl@yer wrote:
nukchebi0 wrote:
I figured it out.

You know the rate of change as .2.

It is the derivative of the h function.

Therefore, h=.2x+b.

Since h(0)=110, b must be 110.

With function for height as h=.2x+110, substitute 24 and 48 to find heights.

Make sure you show that x=number of hours river is rising.

The dr/dt you were trying to find is a known constant, hence why you couldn't find it.
damn that seems to simple

It isn't. You were almost there.

2007-11-26 18:37:22

#12

Liberal-Sl@yer: Certified BF2S Asshole; +131|6936|The edge of sanity

nukchebi0 wrote:
Liberal-Sl@yer wrote:
nukchebi0 wrote:
I figured it out.

You know the rate of change as .2.

It is the derivative of the h function.

Therefore, h=.2x+b.

Since h(0)=110, b must be 110.

With function for height as h=.2x+110, substitute 24 and 48 to find heights.

Make sure you show that x=number of hours river is rising.

The dr/dt you were trying to find is a known constant, hence why you couldn't find it.
damn that seems to simple
It isn't. You were almost there.

wait lol i left out something the peak height of the river was 114.4. damn it i wish we had a fucking graph for this shit.

2007-11-26 18:38:20

#13

nukchebi0: Пушкин, наше всё; +387|6804|New Haven, CT

Liberal-Sl@yer wrote:
nukchebi0 wrote:
Liberal-Sl@yer wrote:
damn that seems to simple
It isn't. You were almost there.
wait lol i left out something the peak height of the river was 114.4. damn it i wish we had a fucking graph for this shit.

That makes it slightly harder.

2007-11-26 18:38:27

#14

SharkyMcshark: I'll take two; +132|7265|Perth, Western Australia

Is it SHM? Or Rectilinear?

2007-11-26 18:40:09

#15

nukchebi0: Пушкин, наше всё; +387|6804|New Haven, CT

As I said earlier, what are you studying currently?

2007-11-26 18:40:44

#16

White-Fusion: Fuck; +616|7031|Scotland

SharkyMcshark wrote:
Is it SHM? Or Rectilinear?

SHM

2007-11-26 18:40:56

#17

Liberal-Sl@yer: Certified BF2S Asshole; +131|6936|The edge of sanity

nukchebi0 wrote:
Liberal-Sl@yer wrote:
nukchebi0 wrote:

It isn't. You were almost there.
wait lol i left out something the peak height of the river was 114.4. damn it i wish we had a fucking graph for this shit.
That makes it slightly harder.

so that would mean that there cant be a constant rate of change.

2007-11-26 18:41:39

#18

nukchebi0: Пушкин, наше всё; +387|6804|New Haven, CT

Yes.

2007-11-26 18:41:54

#19

Ender2309: has joined the GOP; +470|7051|USA

wow we did this like three months ago in my AP calculus class and i already forgot it.

2007-11-26 18:43:46

#20

SharkyMcshark: I'll take two; +132|7265|Perth, Western Australia

Ender2309 wrote:
wow we did this like three months ago in my AP calculus class and i already forgot it.

LOL yeah me too... my Calc exam was like... two and a half weeks ago and Im clueless already!

2007-11-26 18:45:02

#21

Liberal-Sl@yer: Certified BF2S Asshole; +131|6936|The edge of sanity

nukchebi0 wrote:
As I said earlier, what are you studying currently?

Like i said applications of derivatives.

2007-11-26 18:45:37

#22

nukchebi0: Пушкин, наше всё; +387|6804|New Haven, CT

Doh.

I never had to do something like this.

Can you write up the problem in full, how it is worded in the book?

Last edited by nukchebi0 (2007-11-26 18:46:34)

2007-11-26 18:50:31

#23

SharkyMcshark: I'll take two; +132|7265|Perth, Western Australia

Ok I've got this....

H = 4.4 sin(0.2t) + 110 ----> at t=0 gives 110, and the max value attainable is 114.4

therefore

rate of change of height = 0.2 x (4.4 cos(0.2t)) -----> gives a rate of change of 0.2 at t=0

therefore

h(24) = 4.4 sin(4.8) + 110

= 105.617 (3 dp)

h(48) = 4.4 sin (9.6) + 110

= 109.232 (3 dp)

That would be how I would approach this question (I got an A for calculus too... a year mark of 81%)
Can I has biscuit?

EDIT: Got this one wrong, see next page (one miscalculation screwed the entire thing over, fixed it though!)

Last edited by SharkyMcshark (2007-11-26 18:59:59)

2007-11-26 18:50:35

#24

Liberal-Sl@yer: Certified BF2S Asshole; +131|6936|The edge of sanity

I talk to my friend and apparently i didnt get the second paper with the graph on it. Yes, it does have a derivative graph with the rates of change on it. My question is with the rates of change on the graph can i take all of the rates of change and do an average of the rates of change over a 24 hour and 48 hour interval to get my dr/dt in the situation?

EDIT: BTW my freind is e-mailing graph

Last edited by Liberal-Sl@yer (2007-11-26 18:50:58)

2007-11-26 18:51:53

#25

nukchebi0: Пушкин, наше всё; +387|6804|New Haven, CT

SharkyMcshark wrote:
Ok I've got this....

H = 4.4 sin(0.2t) + 110 ----> at t=0 gives 110, and the max value attainable is 114.4

therefore

rate of change of height = 0.2 x (4.4 cos(0.2t)) -----> gives a rate of change of 0.2 at t=0

therefore

h(24) = 4.4 sin(4.8) + 110

= 105.617 (3 dp)

h(48) = 4.4 sin (9.6) + 110

= 109.232 (3 dp)

That would be how I would approach this question (I got an A for calculus too... a year mark of 81%)
Can I has biscuit?

Haxs.

Liberal, will you post graph?

Last edited by nukchebi0 (2007-11-26 18:53:04)

BF2S Forums Need help with derivatives.

White-Fusion wrote:

nukchebi0 wrote:

Liberal-Sl@yer wrote:

nukchebi0 wrote:

Liberal-Sl@yer wrote:

nukchebi0 wrote:

nukchebi0 wrote:

Liberal-Sl@yer wrote:

nukchebi0 wrote:

Liberal-Sl@yer wrote:

nukchebi0 wrote:

Liberal-Sl@yer wrote:

SharkyMcshark wrote:

nukchebi0 wrote:

Liberal-Sl@yer wrote:

nukchebi0 wrote:

Ender2309 wrote:

nukchebi0 wrote:

SharkyMcshark wrote:

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