Interval with the leingth of 3

[rizza]

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)

 

[Muzza]

I'm afraid I don't understand your question. f illustrates an interval? Maybe you could post the question in Danish instead.

 

[HallsofIvy]

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= $$\sqrt{y'^2+1}dx$$ (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= $$\sqrt{4x^2-8x+ 5}= \sqrt{4(x- 1)^2+1}$$ is 25/4

To find the arc-length, we need find $$\integral_a^{a+3}\sqrt{4(x-1)^2+1}dx$$.
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

 

[HallsofIvy]

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.

 

[humanino]

I guess if he can undestand post #3, it is probably this one he should solve.