Truncation: The value of Young’s Modulus of Elasticity (E) for a specific steel alloy

[JimCrown]

Truncation: The value of Young’s Modulus of Elasticity (E) for a specific steel alloy

Hello, first time poster. I am preparing myself on maths as I am going back to study as a mature student. I am doing questions in my spare time. This one confuses me slightly but I have the general grasp of it.

The value of Young’s Modulus of Elasticity (E)fora specific steel alloy is 214 854 226 000 N m^-2


Pick a suitable place to truncate this number andgive the truncation error.

I am pretty sure that I would truncate this as 21485 as you don't round up. My problem is now do I leave it at hat or would I write it out as 21485 x 10^5 and would I write the truncation error as 4226000 x 10^5 or 4226000





Take the new value and give to;

• Standardform to 3 significant figures (Which I know would be 2.15 x 10^11)
  
  
• engineeringnotation to 1 decimal place (Which I know would be 0.2 x 10^12)
  
  The first part stumps me as I do not know how to write out the answer and if I got the truncation error correct. Also, would I have to put NM in my answers. Sorry if this is so obvious
  
  

 

[ksdhart2]

Well, "pick a suitable place" is a very vague instruction and allows for a lot of individual discretion, so your choice is probably adequate. As for what the truncation error is, that depends entirely on the precise definition your book/class/instructor uses. The definition I'm used to seeing is "Truncation error is the difference between a truncated value and the actual value." Under that definition, the truncation error would be: 214854226000 - 21485 x 10⁵.

As for the other parts, the scientific notation part looks correct as well, but I have an issue with the engineering notation. Perhaps you learned differently in your class, but I learned engineering notation as scientific notation, but operating on powers of 1000 instead of powers of 10. i.e.

1678 = 1.678 * 1000¹ = 1.678 * 10³
646493127 = 646.493127 * 1000² = 646.493127 * 10⁶

So, for engineering notation I'd write 214.9 * 10⁹. However, if you were taught differently, of course defer to what your instructor is expecting.

 

[JimCrown]

Well, "pick a suitable place" is a very vague instruction and allows for a lot of individual discretion, so your choice is probably adequate. As for what the truncation error is, that depends entirely on the precise definition your book/class/instructor uses. The definition I'm used to seeing is "Truncation error is the difference between a truncated value and the actual value." Under that definition, the truncation error would be: 214854226000 - 21485 x 10⁵.

As for the other parts, the scientific notation part looks correct as well, but I have an issue with the engineering notation. Perhaps you learned differently in your class, but I learned engineering notation as scientific notation, but operating on powers of 1000 instead of powers of 10. i.e.

1678 = 1.678 * 1000¹ = 1.678 * 10³
646493127 = 646.493127 * 1000² = 646.493127 * 10⁶

So, for engineering notation I'd write 214.9 * 10⁹. However, if you were taught differently, of course defer to what your instructor is expecting.

Thank you I see where I went wrong. I was also working on this question and feel like I have the right answer.

A new build estate requires a concrete path to be laid around the edge of each back garden. The first group of houses to be built have a garden 8m wide and 10m long. The path has constant width and is laid around the edge of the garden. If the area of the path is 100m2, calculate, by deriving a quadratic equation the width of the path (w) Find a quadratic expression to calculate the width of the path hence find the width using the known values. Create a spreadsheet to check your calculations and find an expression that could be used for gardens of different areas.

Working out:

8 x 10 = 80m^2
8+2x multiplied by 10+2x minus 80
(8+2x)(10+2x) - 80 = 100m^2
80 + 16x = 20x + 4x^2 - 80 = 100m^2
36x + 4x^2 = 100m^2

x^2 + 9x - 25 = 0

-b +/- √b^2 - 4ac
______
2a

-9 +/- √81-4(-25)
______
-50

-9 +/- √81+100
______
-50

-9 +/- √81-4x(x-25)

-9 +/- √181
______
2
x= -9 1
____ + ____ √181 = 2.22681202 metres for the width of the path