Question

What is the arclength of `(t,t-1)` on `t in [-7,1]`?

Answer 1

`8sqrt(2)`

Explanation:

The Arc Length for a Parametric Curve is given by

`L=int_alpha^beta sqrt((dx/dt)^2+(dy/dt)^2) \ dt`

So in this problem we have

`x=t => dx/dt=1`
`y= t-1 => dy/dt = 1`

So the Arc Length is;

`L=int_-7^1 sqrt((1)^2+(1)^2) \ dt`
`\ \ \= sqrt(2) \ int_-7^1 \ dt`
`\ \ \= sqrt(2) \ [t color(white)int]_-7^1 \ dt`
`\ \ \= sqrt(2) \ {(1)-(-7)}`
`\ \ \= 8sqrt(2)`

Additionally, If we look at the actual graph of the parametric curve:

we can see that the equations represent a straight line, so in fact we can easily calculate the are length (coloured blue) from a triangle using Pythagoras:

`L=sqrt((1-(-7))^2 + (0-8)^2)`
`\ \ \=sqrt(8^2+8^2)`
`\ \ \=sqrt(64+64)`
`\ \ \=sqrt(128)`
`\ \ \=8sqrt(2)`, as above

So, Maths Works!