Curve in space - "wrong" answer

[Anne-Cathrine]

Hi there

I have some trouble with this task:

"A curve in space is given by:

x = cos(2t)
y = sin(2t)
z = 2ln(t)

​

where the parameter t "passes through" the positive real numbers. Mark the correct exspression for the arc lenght from t = 1 to t = 2"

I have attached a picture showing how I solved it, but the answer in the assignment is not the one I get. I get the correct answer, but I can't figure out how it is the same as the exspression in the multible choice assignment.

I have written my calculations and my answer in blue and the answer in the assignment in red. (I don't know why it's upside-down?? )

Thank you in advance! :grin:

 

[Otis]

Hi Anne:

They first factored the radicand, and then they simplified the resulting radical.

4 + 4/t^2 factors as (4)(1 + 1/t^2)

The square root of 4 is 2, leaving 1 + 1/t^2 inside the radical sign.

Also, using the definition of negative exponents, they rewrote 1/t^2 as t^(-2).

Questions?

 

[Anne-Cathrine]

Hi Otis

Thank you so much for your reply!

I'm pretty sure I understand what you are saying, but I have a few questions too check if I've got it right

I've written the steps factoring the radicand and simplified the radical, as you explained:

And I've tried to figure out why t^-2 =1/t^2:

If you could tell me if these calculations are correct it would be so great!

Hi Anne:

They first factored the radicand, and then they simplified the resulting radical.

4 + 4/t^2 factors as (4)(1 + 1/t^2)

The square root of 4 is 2, leaving 1 + 1/t^2 inside the radical sign.

Also, using the definition of negative exponents, they rewrote 1/t^2 as t^(-2).

Questions?

 

[HallsofIvy]

I am puzzled by this. Finding the arc length, the problem you first stated, is a fairly deep Calculus problem. But the fact that \(\displaystyle t^{-n}= \frac{1}{t^n}\), for any n, is basic algebra.