# Find the length of the curve x=e^(t)-t, y=4e^(t/2) ,-8≤x≤3

mlehuzzah | Student, Graduate | (Level 1) Associate Educator

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The graph of `x=e^t-t` is always positive (in fact, it is always greater than 1)

To see this, `(dx)/(dt) = e^t - 1`

If `t<0`, then `(dx)/(dt)` is negative, so decreasing

If `t>0`, then `(dx)/(dt)` is positive, so increasing

If `t=0`, then `(dx)/(dt) = 0` and we have a local minimum

At `t=0`, `x=1`, so that is the minimum for `x`, so `x` can't ever equal -8.

So, I'm guessing you mean `-8 <= t <= 3`

If so, the arc length formula in parametric equations is:

`int _(-8)^3 sqrt(((dx)/(dt))^2 + ((dy)/(dt))^2) dt `

`(dx)/(dt) = e^t - 1`

`( (dx)/(dt) )^2 = e^(2t) - 2e^t + 1

`(dy)/(dt) = 4 e^(t/2)* (1/2) = 2e^(t/2)`

`( (dy)/(dt) )^2 = 4e^t`

`((dx)/(dt))^2 + ((dy)/(dt))^2 = e^(2t) + 2e^t + 1`

`=(e^t+1)^2`

So, arc length =

`int _(-8)^3 sqrt(((dx)/(dt))^2 + ((dy)/(dt))^2) dt `

`= int _(-8)^3 sqrt( (e^t+1)^2 ) dt`

`= int _(-8)^3 (e^t+1) dt `

`= [e^t+t]_(-8)^3`

`= e^3 + 3 - e^(-8)+8`

`=11 + e^3 + (1)/(e^8)`