Interval with the leingth of 3

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In summary, the conversation discusses finding an interval with a length of 3 where the function f(x)=x^2-2x+2 illustrates on an interval with a length of 25/4. Two possible methods for finding this interval are mentioned, one involving arclength and the other involving a change in y. The conversation also mentions the use of substitutions and trigonometric methods to solve the problem.
  • #1
rizza
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f(x)=x^2-2x+2
Draw the function (I can do that) and deside an interval with the leingth of 3, there by f illustrates on an interval in the leingh of 25/4. (how??)

(hope you can read it, I am from Denmark, so i'll try)
 
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  • #2
I'm afraid I don't understand your question. f illustrates an interval? Maybe you could post the question in Danish instead.
 
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  • #3
You seem to be asking for an interval [a, a+3] (so that the interval has length 3) such that the length of the arc y= x2- 2x+ 2 is 25/4.

Since y= x2- 2x+ 2, y'= 2x- 2 and ds= [tex]\sqrt{y'^2+1}dx[/tex] (if you don't know what that means, either you need to go back and review "arclength" or I've completely misunderstood your question!).

This is ds= [tex]\sqrt{4x^2-8x+ 5}= \sqrt{4(x- 1)^2+1}[/tex] is 25/4

To find the arc-length, we need find [tex]\integral_a^{a+3}\sqrt{4(x-1)^2+1}dx[/tex].
Do that integral (I suggest the substitution u= 2(x-1) followed by a trig substitution) and then determine what a must be so that the arc-length is
 
  • #4
Another, simpler, possiblity is that you just want an interval from x to x+3 such that the change in y is 25/3. Since y= x2- 2x+ 2= (x-1)2+1, the change in y between x and x+3 is ((x+2)2+ 1)- ((x-1)2+ 1)= x2+ 4x+ 5- x2- 2x+ 1= 2x+ 6= 25/4. Then 2x= 25/4- 6= 1/4 and x= 1/8.
 
  • #5
I guess if he can undestand post #3, it is probably this one he should solve.
 

1. What is an interval with the length of 3?

An interval with the length of 3 refers to a sequence of numbers that are evenly spaced apart by a difference of 3. This can be represented as (3, 6, 9, 12...) or (1, 4, 7, 10...).

2. How do you calculate the length of an interval with the length of 3?

To calculate the length of an interval with the length of 3, you simply subtract the first number from the last number. For example, if the interval is (3, 6, 9, 12), the length would be 12 - 3 = 9.

3. What is the significance of an interval with the length of 3 in mathematics?

In mathematics, an interval with the length of 3 is often used in number patterns and sequences. It is also important in determining the slope of a line, as it represents the difference in y-values for every 3 unit increase in x-value.

4. Can an interval with the length of 3 be negative?

Yes, an interval with the length of 3 can be negative. This would be represented as (-3, -6, -9, -12...). The only requirement is that the numbers are evenly spaced apart by a difference of 3.

5. How can you use an interval with the length of 3 in real-life situations?

An interval with the length of 3 can be applied in various real-life situations, such as calculating monthly salary increases, measuring distance traveled in a fixed amount of time, or tracking changes in stock prices over time.

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