# What is the arclength of f(t) = (t-sqrt(t^2+2),t+te^(t-2)) on t in [-1,1]?

Arc length $s = 3.2352212144206 \text{ }$units

#### Explanation:

The formula to obtain the arc length $s$

$s = {\int}_{{t}_{1}}^{{t}_{2}} \sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2}} \cdot \mathrm{dt}$

From the given $f \left(t\right) = \left(t - \sqrt{{t}^{2} + 2} , t + t \cdot {e}^{t - 2}\right)$

We have $x = t - \sqrt{{t}^{2} + 2} \text{ }$and $y = t + t \cdot {e}^{t - 2}$

We need to obtain the derivatives $\frac{\mathrm{dx}}{\mathrm{dt}}$ and $\frac{\mathrm{dy}}{\mathrm{dt}}$

$\frac{\mathrm{dx}}{\mathrm{dt}} = 1 - \frac{t}{\sqrt{{t}^{2} + 2}} \text{ }$ and $\frac{\mathrm{dy}}{\mathrm{dt}} = 1 + t \cdot {e}^{t - 2} + {e}^{t - 2}$

We solve now the arclength $s$

$s = {\int}_{{t}_{1}}^{{t}_{2}} \sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2}} \cdot \mathrm{dt}$

$s = {\int}_{- 1}^{1} \sqrt{{\left(1 - \frac{t}{\sqrt{{t}^{2} + 2}}\right)}^{2} + {\left(1 + t \cdot {e}^{t - 2} + {e}^{t - 2}\right)}^{2}} \cdot \mathrm{dt}$

The integral is complicated and requires the use of calculator or Simpson's Rule formula

$s = {\int}_{- 1}^{1} \sqrt{{\left(1 - \frac{t}{\sqrt{{t}^{2} + 2}}\right)}^{2} + {\left(1 + t \cdot {e}^{t - 2} + {e}^{t - 2}\right)}^{2}} \cdot \mathrm{dt}$

$s = 3.2352212144206 \text{ }$units

God bless....I hope the explanation is useful.