Find the linear density of the wire when x=4m.

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SUMMARY

The linear density of a wire, defined by the mass function x(1+sqrt(x)) kilograms, can be calculated by differentiating the mass with respect to length. When evaluating at x=4 meters, the correct approach involves finding the derivative, resulting in a linear density of 4 kg/m. The initial assumption of simply dividing mass by length was incorrect, as it does not account for the continuous nature of the mass distribution along the wire.

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Homework Statement



The mass of part of a wire is x(1+sqrt(x)) kilograms, where x
is measured in meters from one end of the wire. Find the
linear density of the wire when x=4m.



2. The attempt at a solution

Actually what i did was put in x = 4 in the mass equation given and divide by x since linear density = mass/length. And i get the answer to be 3.

But i don't know whether that's right or not.
Any help appreciated.
 
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The legend said:

Homework Statement



The mass of part of a wire is x(1+sqrt(x)) kilograms, where x
is measured in meters from one end of the wire. Find the
linear density of the wire when x=4m.



2. The attempt at a solution

Actually what i did was put in x = 4 in the mass equation given and divide by x since linear density = mass/length. And i get the answer to be 3.

But i don't know whether that's right or not.
Any help appreciated.
The way I interpret the question is the expression x(1+sqrt(x)), gives you the mass at a point x, i.e. that at x=1 the mass is 2, at x=2 the mass is 2+2sqrt(2), etc. If that is the case, then your method wouldn't be correct. Instead, to get the total mass of the wire, what you need to do as sum over the contribution of each point on the wire. Can you think of any method that will do that for you?
 


i would say it is integration that will help, but i haven't been taught that yet.
 


The legend said:
i would say it is integration that will help, but i haven't been taught that yet.
Perhaps I am misinterpreting the question then.
 
Hi Hoot! Hi The legend! :smile:

(have a square-root: √ :wink:)
The legend said:
The mass of part of a wire is x(1+sqrt(x)) kilograms, where x
is measured in meters from one end of the wire. Find the linear density of the wire when x=4m.

Isn't it differentiation?

I think they're saying that the total mass from 0 to x is x(1 + √x), and they want the density for a tiny bit round x = 4.
 


Hootenanny said:
Perhaps I am misinterpreting the question then.

I took this sum from Stewart's book for calculus (Early Transcendentals 6th edition) and this sum was found under the chapter on derivatives(differentiation)...just to say

so how do I approach it then?
 


The legend said:
I took this sum from Stewart's book for calculus (Early Transcendentals 6th edition) and this sum was found under the chapter on derivatives(differentiation)...just to say

so how do I approach it then?
If this is under differentiation, then I would suggest that tiny-tim has the correct interpretation. That is, the expression defines the mass of the segment of wire between 0 and x. In which case, the density is simply the rate of change of mass with respect to length.

P.S. Hi tim!
 
Hi Hoot! :wink:

How's it goin'? :smile:​
 


tiny-tim said:
Hi Hoot! Hi The legend! :smile:

(have a square-root: √ :wink:)Isn't it differentiation?

I think they're saying that the total mass from 0 to x is x(1 + √x), and they want the density for a tiny bit round x = 4.


Hello tiny-tim! :smile:

But the question says that it is for a part of the wire.

The "mass of part" of a wire is x(1+sqrt(x)) kilograms,

But i did try it taking your way.
So what i did was this:

i differentiated the given mass function w.r.t length which gives

<br /> \frac{dm}{dl} = 1 + \frac{3 \sqrt(x)}{2}<br />
this is the linear density at any tiny little place. So what i did is I put in x=4 in the equation and got the answer to be 4 kg/m3
 
  • #10


...that took some editing there^...(latex)

is it right?
 
Last edited:
  • #11
Yup! 1 + 3√4/2 = 4 looks fine. :smile:
 
  • #12


thanks a lot! :approve:
 

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