## Snail problem... find the equation algebrically..

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### Snail problem... find the equation algebrically..

A snail decides to climb a 12 meter well.

Every day, he climbs 3 meters in the daytime and slides 2 meters at night.

After how many days will he finally arrive at the other side of the well?
helpmeplz!
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### Re: Snail problem... find the equation algebrically..

helpmeplz! wrote:After how many days will he finally arrive at the other side of the well?

That depends on whether I salt it or not.

What do you have so far?

CRGreathouse
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### Re: Snail problem... find the equation algebrically..

helpmeplz! wrote:A snail decides to climb a 12 meter well.

Every day, he climbs 3 meters in the daytime and slides 2 meters at night.

After how many days will he finally arrive at the other side of the well?

After 9 days the snail will be at the 9 meter mark. On the 10th day it will reach the top - hopefully it won't slide back any more.
mathman
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Suppose that the rate at which the snail traveled decreased at a non-constant rate depending upon how long he had been traveling, and also depending upon the amount of salt being poured on him?
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Infinity
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### Re: Snail problem... find the equation algebrically..

mathman wrote:
helpmeplz! wrote:A snail decides to climb a 12 meter well.

Every day, he climbs 3 meters in the daytime and slides 2 meters at night.

After how many days will he finally arrive at the other side of the well?

After 9 days the snail will be at the 9 meter mark. On the 10th day it will reach the top - hopefully it won't slide back any more.

Yeah I get it but how do you do it algebrically? I need an equation.
helpmeplz!
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Infinity wrote:Suppose that the rate at which the snail traveled decreased at a non-constant rate depending upon how long he had been traveling, and also depending upon the amount of salt being poured on him?

Ooh, differential equations. I like.

CRGreathouse
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### Re: Snail problem... find the equation algebrically..

helpmeplz! wrote:
mathman wrote:
helpmeplz! wrote:A snail decides to climb a 12 meter well.

Every day, he climbs 3 meters in the daytime and slides 2 meters at night.

After how many days will he finally arrive at the other side of the well?

After 9 days the snail will be at the 9 meter mark. On the 10th day it will reach the top - hopefully it won't slide back any more.

Yeah I get it but how do you do it algebrically? I need an equation.

H(n)=3+(n-1), where H is height after n days. Since you want H(n)=10, I think you can do the rest.
mathman
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11 day
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helpmeplz! wrote:After how many days will he finally arrive at the other side of the well?

I don't understand. The snail is simply climbing upwards, so why does it ever reach the other side of the well? The standard (but tricky) problem of this type specifies a wall rather than a well, and you have to calculate how long it takes the snail to get to the point directly on the other side of the wall from the point where the snail started (so you have to include the time needed for the descent of the other side).
skipjack
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skipjack wrote:
helpmeplz! wrote:After how many days will he finally arrive at the other side of the well?

I don't understand. The snail is simply climbing upwards, so why does it ever reach the other side of the well? The standard (but tricky) problem of this type specifies a wall rather than a well, and you have to calculate how long it takes the snail to get to the point directly on the other side of the wall from the point where the snail started (so you have to include the time needed for the descent of the other side).

lol so the distance from it started point to finish point is 24 meters =) how ever I believe the question maken by just for 12 meters.
"As far as the laws of mathematics refer to reality, they are not certain; and as far as they are certain, they do not refer to reality." A.Einstein

Enochiche
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### Re: Snail problem... find the equation algebrically..

mathman wrote:
helpmeplz! wrote:
mathman wrote:
helpmeplz! wrote:A snail decides to climb a 12 meter well.

Every day, he climbs 3 meters in the daytime and slides 2 meters at night.

After how many days will he finally arrive at the other side of the well?

After 9 days the snail will be at the 9 meter mark. On the 10th day it will reach the top - hopefully it won't slide back any more.

Yeah I get it but how do you do it algebrically? I need an equation.

H(n)=3+(n-1), where H is height after n days. Since you want H(n)=10, I think you can do the rest.

Re u sure this equation is giving the general solution? So if we think we got 12meters to take and if we got [one in a day] 3 steps up & 2 steps down( totally 1 step up) in 9 days we take 9 meters and 10th day first of all we ll take 3 steps up and we re 12 meters up already. Now c it at the solution;

u say H(n)=10 so then 3+(n-1)=10 and we've got n=8 here :/ I thnk thats not true.
"As far as the laws of mathematics refer to reality, they are not certain; and as far as they are certain, they do not refer to reality." A.Einstein

Enochiche
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I think "10" was just "12" mistyped.

Here's the trickier problem. Forget algebra; just try to solve it correctly.

A snail decides to climb a 20 foot wall (whose width may be ignored) and then climb down the other side of the wall.

He can climb up 3 feet in the daytime, but slides down 2 feet at night. Each day consists of 12 hours of daytime followed by 12 hours of night-time, so the snail is 1 foot up the wall at the end of the first 24-hour day.

After how long will he finally arrive at the foot of the other side of the wall?

(I've used feet to distinguish this problem from the previous one.)

Boring answers are wrong; you will know when you have the right answer!
Last edited by skipjack on Wed Oct 03, 2007 4:37 pm, edited 5 times in total.
skipjack
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skipjack wrote:I think "10" was just "12" mistyped.

Here's the trickier problem. Forget algebra; just try to solve it correctly.

A snail decides to climb a 20 foot wall (whose width may be ignored) and then climb down the other side of the wall.

Every day, he climbs 3 feet in the daytime and slides 2 feet at night.

After how long will he finally arrive at the foot of the other side of the wall?

(I've used feet to distinguish this problem from the previous one.)

Boring answers are wrong; you will know when you have the right answer!

lol it can never arrive at the foot of the other side of the wall 'coz when it ll reach at the top of the wall it ll be 17th day and on the 18th day "climbin 3 feet in the daytime and slides 2 feet at the night" means ll be different(opposite)for it. thats why it can never reach to the foot point of the otherside of the wall.
"As far as the laws of mathematics refer to reality, they are not certain; and as far as they are certain, they do not refer to reality." A.Einstein

Enochiche
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Enochiche wrote:. . . it. . .'ll be different(opposite)for it.

That doesn't make sense. The wall's thickness can be ignored, so the snail can certainly descend on the other side of the wall after reaching the top. However, I changed the wording slightly to remove any suggestion that the snail can't switch from ascending to descending. You should, of course, assume the snail doesn't make any unnecessary detours!
skipjack
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