Question

Question #b4d3e

Answer 1

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

The tangential acceleration is:

`(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-(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(2t),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)`