Another simple question that I don't get

  • Thread starter r6mikey
  • Start date
In summary: I always get lost trying to figure out the problems and end up struggling. It's really discouraging when I can't figure something out and I'm not the only one, it seems like a lot of people have this same issue. In summary, the author is struggling to solve a homework problem and is confused.
  • #1
r6mikey
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[SOLVED] Another simple question that I don't get :(

Homework Statement


Find the Length of the indicated curve

y=2x-3 x[1-3]


Homework Equations



Arc Length=[tex]\int[/tex][tex]\sqrt{1+[f'(x)]^{2}dx}[/tex]

The Attempt at a Solution



ok so y'=2 of course,

so it becomes a problem of [tex]\int from 1 to 3 for \sqrt{5}[/tex]

using substitution, u=5 ,and this is where i get a little confused...

du=0dx which would make dx=du/0

the answer becomes 2[tex]\sqrt{5}[/tex]

I'm sure I'm making it harder than it is, I always do... :(
 
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  • #2
and thinking about it, i should probably not be using the formula for arc length for this problem, even though it is in the arc length chapter...since this is a linear function
 
  • #3
using the arc length formula the only thing i can think of is that when i get

[tex]\sqrt{5}[/tex]dx and dx ends up being 0, it becomes neglible so i only have [tex]\sqrt{5}[/tex] to evaluate from 1 to 3... but then my question would be, I know subtracting 3[tex]\sqrt{5}[/tex] from 1[tex]\sqrt{5}[/tex] gives me the answer 2[tex]\sqrt{5}[/tex] but there is no X in front of the [tex]\sqrt{5}[/tex]


:(
 
  • #4
In the formula for arc length, the dx should be OUTSIDE of the radicand. Does this help you?

Hint: it should!

What is [tex]\int_1^3\sqrt{5}*dx[/tex]
 
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  • #5
yea sorry, forgot to put the dx on the outside, but still confused... using substitution and writing the problem like this...

u[tex]^{1/2}[/tex] where u=5, du=0dx so dx=du/0 which is ?

continuing, if it is right u[tex]^{1/2}[/tex]du would become u[tex]^{3/2}[/tex] times 2/3 so I'm doing something wrong with the substituion...I'm just confusing meself :(

I can do all the problems except this one b/c when I take the derivative of u it becomes nothing... so that is where I am getting lost
 
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  • #6
What are you doing? Why the need for u sub? Again I ask you what is [tex]\int c*dx[/tex]

where c is any constant...any constant like, oh let's say [itex]\sqrt 5[/itex]
 
  • #7
so you're saying it would just become (integral)c=x so x(sqrt 5)


I'm not seeing the obvious, for some reason i 'm thinking i need substitution...wtf
 
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  • #8
r6mikey said:
[tex]\intc[/tex]=x so you're saying it would just become x[tex]\sqrt{5}[/tex]

Of course! sqrt5 is just a number, just some constant. Pull it through the integral and proceed as normal. Evaluate what you have from x=3--->x=1

---->(3-1)*[itex]\sqrt 5[/itex]
[tex]\int_{x_0}^x cdx=c\int_{x_o}^x dx=c[x]_{x_o}^x[/tex]
What is there to substitute? What are you going to replace with u?
 
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  • #9
you with me?
 
  • #10
I believe so, i just think i got to step away from these books for awhile:yuck:
 
  • #11
We all do! I feel the same way most of the time!
 

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