Question

Consider the curve r(t)=⟨t2−1,1/3​t^3−t,0⟩,−2⩽t⩽2. (a) Find the curvature at t=0. (b) Find N(1).

Answer 1

(a) The curvature at t=0 is undefined. (b) N(1) = ⟨sqrt(2)/2, sqrt(2)/2, 0⟩.

Step-by-step explanation:

(a) To find the curvature at t=0, we need to find the derivative of the velocity vector with respect to time and divide it by the magnitude of the velocity:

First, let's find the velocity vector by taking the derivative of the position vector:
v(t) = r'(t) = ⟨2t, t^2 - 1, 0⟩

Next, let's find the acceleration vector by taking the derivative of the velocity vector:
a(t) = v'(t) = ⟨2, 2t, 0⟩

Finally, let's find the magnitude of the velocity vector:
|v(t)| = sqrt((2t)^2 + (t^2 - 1)^2 + 0) = sqrt(4t^2 + t^4 - 2t^2 + 1) = sqrt(t^4 + 2t^2 + 1)

Now we can find the curvature at t=0:
k(0) = |a(0)| / |v(0)|^3 = |⟨2, 0, 0⟩| / |0|^3 = 2 / 0 = undefined

(b) To find N(1), we need to find the unit normal vector at t=1. The unit normal vector is given by:
N(t) = a(t) / |a(t)|

Substituting t=1 into the acceleration vector we found earlier:
a(1) = ⟨2, 2(1), 0⟩ = ⟨2, 2, 0⟩

Now we can find N(1) by dividing a(1) by its magnitude:
N(1) = ⟨2, 2, 0⟩ / |⟨2, 2, 0⟩| = ⟨2, 2, 0⟩ / sqrt(2^2 + 2^2 + 0) = ⟨2, 2, 0⟩ / sqrt(8) = ⟨2/sqrt(8), 2/sqrt(8), 0⟩ = ⟨sqrt(2)/2, sqrt(2)/2, 0⟩

Answer 2

To find the curvature of a curve
`\(r(t) = \langle t^2 - 1, (1)/(3)t^3 - t, 0 \rangle\) at \(t = 0\)`, we can follow these steps:

Step-by-step explanation:

(a) Find the curvature at
`\(t = 0\):`

The curvature
`(\(k\))` of a curve in space is given by the formula:

`\[k = \frac{\lvert \mathbf{r}'(t) * \mathbf{r}''(t) \rvert}{\lvert \mathbf{r}'(t) \rvert^3}\]`

where
`\(\mathbf{r}'(t)\)` is the first derivative of
`\(\mathbf{r}(t)\)`with respect to
`\(t\)`, and
`\(\mathbf{r}''(t)\)`he second derivative.

First, find
`\(\mathbf{r}'(t)\)`and
`\(\mathbf{r}''(t)\):`

`\[\mathbf{r}'(t) = \langle 2t, t^2 - 1, 0 \rangle\]`

`\[\mathbf{r}''(t) = \langle 2, 2t, 0 \rangle\]`

Now, evaluate these at
`\(t = 0\):`

`\[\mathbf{r}'(0) = \langle 0, -1, 0 \rangle\]`

`\[\mathbf{r}''(0) = \langle 2, 0, 0 \rangle\]`

Now, compute the cross product:

`\[\mathbf{r}'(0) * \mathbf{r}''(0) = \langle 0, 0, 2 \rangle\]`

Compute the magnitudes:

`\[\lvert \mathbf{r}'(0) \rvert = 1\]`

`\[\lvert \mathbf{r}''(0) \rvert = 2\]`

`\[\lvert \mathbf{r}'(0) * \mathbf{r}''(0) \rvert = 2\]`

Now, plug these values into the curvature formula:

`\[k = \frac{\lvert \mathbf{r}'(0) * \mathbf{r}''(0) \rvert}{\lvert \mathbf{r}'(0) \rvert^3} = (2)/(1) = 2\]`

So, the curvature at
`\(t = 0\) is \(2\).`

(b) Find
`\(N(1)\)`

The unit normal vector
`\(N(t)\)` is given by:

`\[N(t) = \frac{\mathbf{r}'(t) * \mathbf{r}''(t)}{\lvert \mathbf{r}'(t) * \mathbf{r}''(t) \rvert}\]`

Evaluate this at
`\(t = 1\)`:

`\[\mathbf{r}'(1) = \langle 2, 0, 0 \rangle\]`

`\[\mathbf{r}''(1) = \langle 2, 2, 0 \rangle\]`

Compute the cross product:

`\[\mathbf{r}'(1) * \mathbf{r}''(1) = \langle 0, 0, 4 \rangle\]`

Compute the magnitude:

`\[\lvert \mathbf{r}'(1) * \mathbf{r}''(1) \rvert = 4\]`

Now, plug these values into the formula for
`\(N(1)\):`

`\[N(1) = \frac{\mathbf{r}'(1) * \mathbf{r}''(1)}{\lvert \mathbf{r}'(1) * \mathbf{r}''(1) \rvert} = (\langle 0, 0, 4 \rangle)/(4) = \langle 0, 0, 1 \rangle\]`

So,
`\(N(1)\)` is the unit vector in the positive z-direction, which is
`\(\langle 0, 0, 1 \rangle\).`