# Question #b4d3e

##### 1 Answer
Aug 17, 2014

This is a long problem, so here is the first part:

The tangential acceleration is:

$\frac{{e}^{2 t} - \frac{1}{4 {t}^{2}} - {e}^{- 2 t}}{{e}^{2 t} + \frac{1}{2 t} + {e}^{- 2 t}} \left({e}^{t} , \frac{1}{\sqrt{2 t}} , - {e}^{- t}\right)$

Here's the second part:

The normal acceleration is:

$\left({e}^{t} \left(1 - \frac{{e}^{2 t} - \frac{1}{4 {t}^{2}} - {e}^{- 2 t}}{{e}^{2 t} + \frac{1}{2 t} + {e}^{- 2 t}}\right) , - \frac{1}{\sqrt{2 t}} \left(\frac{1}{2 t} + \frac{{e}^{2 t} - \frac{1}{4 {t}^{2}} - {e}^{- 2 t}}{{e}^{2 t} + \frac{1}{2 t} + {e}^{- 2 t}}\right) , {e}^{- t} \left(1 + \frac{{e}^{2 t} - \frac{1}{4 {t}^{2}} - {e}^{- 2 t}}{{e}^{2 t} + \frac{1}{2 t} + {e}^{- 2 t}}\right)\right)$

The round brackets should be angle brackets to indicate a vector.

$r \left(t\right) = \left({e}^{t} , \sqrt{2 t} , {e}^{- t}\right)$
$= \left({e}^{t} , {\left(2 t\right)}^{\frac{1}{2}} , {e}^{- t}\right)$

We just take the derivatives component-wise:

$r ' \left(t\right) = \left({e}^{t} , {\left(2 t\right)}^{- \frac{1}{2}} , - {e}^{- t}\right)$
$r ' ' \left(t\right) = \left({e}^{t} , - \left(\frac{1}{2}\right) {\left(2 t\right)}^{- \frac{3}{2}} \left(2\right) , {e}^{- t}\right)$
$= \left({e}^{t} , - {\left(2 t\right)}^{- \frac{3}{2}} , {e}^{- t}\right)$

We need to find the unit tangent vector:

$T \left(t\right) = \frac{r ' \left(t\right)}{| r ' \left(t\right) |}$

We already have $r ' \left(t\right)$, so let's compute the other:

$| r ' \left(t\right) | = \sqrt{{\left({e}^{t}\right)}^{2} + {\left({\left(2 t\right)}^{- \frac{1}{2}}\right)}^{2} + {\left({e}^{- t}\right)}^{2}}$
$= \sqrt{{e}^{2 t} + \frac{1}{2 t} + {e}^{- 2 t}}$

We need to find the acceleration component (this is the magnitude of the acceleration in the direction of the tangent vector):

${a}_{T} = \frac{r ' \left(t\right) \cdot r ' ' \left(t\right)}{| r ' \left(t\right) |}$
$= \frac{\left({e}^{t} \cdot {e}^{t} , {\left(2 t\right)}^{- \frac{1}{2}} \cdot - {\left(2 t\right)}^{- \frac{3}{2}} , - {e}^{- t} \cdot {e}^{- t}\right)}{| r ' \left(t\right) |}$
$= \frac{\left({e}^{2 t} - {\left(2 t\right)}^{- 2} - {e}^{- 2 t}\right)}{| r ' \left(t\right) |}$

To get the tangential acceleration, we multiply the magnitude by the unit tangent vector:

${a}_{T} T \left(t\right) = \frac{r ' \left(t\right) \cdot r ' ' \left(t\right)}{| r ' \left(t\right) |} \frac{r ' \left(t\right)}{| r ' \left(t\right) |}$
$= \frac{\left({e}^{2 t} , - {\left(2 t\right)}^{- 2} , - {e}^{- 2 t}\right)}{\sqrt{{e}^{2 t} + \frac{1}{2 t} + {e}^{- 2 t}}} \frac{{e}^{t} , {\left(2 t\right)}^{- \frac{1}{2}} , - {e}^{- t}}{\sqrt{{e}^{2 t} + \frac{1}{2 t} + {e}^{- 2 t}}}$
$= \frac{\left({e}^{2 t} - {\left(2 t\right)}^{- 2} - {e}^{- 2 t}\right)}{{e}^{2 t} + \frac{1}{2 t} + {e}^{- 2 t}} \left({e}^{t} , {\left(2 t\right)}^{- \frac{1}{2}} , - {e}^{- t}\right)$

If you are comfortable with projections, the tangential acceleration is the acceleration vector projected onto the unit tangent vector.

End of part 1

Part 2:
As you can see the normal acceleration is really messy to calculate. The formula for normal acceleration ${a}_{N} N \left(t\right)$

$\kappa = \frac{1}{| r ' \left(t\right) |} | \frac{\mathrm{dT}}{\mathrm{dt}} |$
${a}_{N} = \kappa | r ' \left(t\right) {|}^{2} = \frac{| r ' \left(t\right) \times r ' ' \left(t\right) |}{| r ' \left(t\right) |}$

Since $\kappa$ is usually difficult to compute, we'll use the second formula (we've computed $| r ' \left(t\right) |$, so we'll substitute later):

${a}_{N} = \frac{| \left({e}^{t} , {\left(2 t\right)}^{- \frac{1}{2}} , - {e}^{- t}\right) \times \left({e}^{t} , - \left(\frac{1}{2}\right) {\left(2 t\right)}^{- \frac{3}{2}} \left(2\right) , {e}^{- t}\right) |}{| r ' \left(t\right) |}$
$= \frac{| \left(\frac{1}{\sqrt{2 t} {e}^{t}} - \frac{1}{2 t \sqrt{2 t} {e}^{t}} , - \left(1 - \left(- 1\right)\right) , - \frac{{e}^{t}}{2 t \sqrt{2 t}} - \frac{{e}^{t}}{\sqrt{2 t}}\right) |}{| r ' \left(t\right) |}$
$= \frac{| \left(\frac{1}{\sqrt{2 t} {e}^{t}} \left(1 - \frac{1}{2 t}\right) , - 2 , - \frac{{e}^{t}}{\sqrt{2 t}} \left(\frac{1}{2 t} + 1\right)\right) |}{\sqrt{{e}^{2 t} + \frac{1}{2 t} + {e}^{- 2 t}}}$

To compute $N \left(t\right)$:

$N \left(t\right) = \frac{1}{\kappa} \frac{\mathrm{dT}}{\mathrm{ds}}$

Again, this is messy. If you need the normal acceleration, just compute this with:

${a}_{N} N \left(t\right) = a - {a}_{T} T \left(t\right)$
$= r ' ' \left(t\right) - {a}_{T} T \left(t\right)$