Question #b4d3e

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Question #b4d3e

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Answer 1 · 2014-08-17T18:11:44

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

The tangential acceleration is:

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

Here's the second part:

The normal acceleration is:

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

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

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

We just take the derivatives component-wise:

$r'(t)=(e^t,(2t)^(-1/2),-e^(-t))$
$r''(t)=(e^t,-(1/2)(2t)^(-3/2)(2),e^(-t))$
$=(e^t,-(2t)^(-3/2),e^(-t))$

We need to find the unit tangent vector:

$T(t)=(r'(t))/(|r'(t)|)$

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

$|r'(t)|=sqrt((e^t)^2+((2t)^(-1/2))^2+(e^(-t))^2)$
$=sqrt(e^(2t)+1/(2t)+e^(-2t))$

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

$a_T=(r'(t)*r''(t))/(|r'(t)|)$
$=((e^t*e^t,(2t)^(-1/2)*-(2t)^(-3/2),-e^(-t)*e^(-t)))/(|r'(t)|)$
$=((e^(2t)-(2t)^(-2)-e^(-2t)))/(|r'(t)|)$

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

$a_T T(t)=(r'(t)*r''(t))/(|r'(t)|) (r'(t))/(|r'(t)|)$
$=((e^(2t),-(2t)^(-2),-e^(-2t)))/sqrt(e^(2t)+1/(2t)+e^(-2t)) (e^t,(2t)^(-1/2),-e^(-t))/sqrt(e^(2t)+1/(2t)+e^(-2t))$
$=((e^(2t)-(2t)^(-2)-e^(-2t)))/(e^(2t)+1/(2t)+e^(-2t)) (e^t,(2t)^(-1/2),-e^(-t))$

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 $a_N N(t)$

$kappa=1/(|r'(t)|) |(dT)/(dt)|$
$a_N=kappa |r'(t)|^2=(|r'(t) xx r''(t)|)/(|r'(t)|)$

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

$a_N=(|(e^t,(2t)^(-1/2),-e^(-t)) xx (e^t,-(1/2)(2t)^(-3/2)(2),e^(-t))|)/(|r'(t)|)$
$=(|(1/(sqrt(2t)e^t)-1/(2tsqrt(2t)e^t),-(1-(-1)), -(e^t)/(2tsqrt(2t))-(e^t)/sqrt(2t))|)/(|r'(t)|)$
$=(|(1/(sqrt(2t)e^t)(1-1/(2t)),-2, -(e^t)/sqrt(2t)(1/(2t)+1))|)/(sqrt(e^(2t)+1/(2t)+e^(-2t)))$

To compute $N(t)$ :

$N(t)=1/kappa (dT)/(ds)$

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

$a_N N(t)=a-a_T T(t)$
$=r''(t)-a_T T(t)$

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