Question

Find the curvature k of the curve at the point p. r(t) = et cos(t)i et sin(t)j et?, p(1, 0, 1)

Answer 1

The derivative of
`\(\mathbf{T}(t)\)` is:

`\[\mathbf{T}'(t) = \frac{\mathbf{r}''(t) \lvert \mathbf{r}'(t) \rvert - \mathbf{r}'(t) \mathbf{r}''(t)}{(\lvert \mathbf{r}'(t) \rvert)^2}\]`

Plugging in the expressions for
`\(\mathbf{r}'(t)\)` and
`\(\mathbf{r}''(t)\)`, we can find
`\(\mathbf{T}'(t)\)`. Finally, we can evaluate the curvature
`\(k\)` at the point
`\(P(1, 0, 1)\)` by plugging in the appropriate value of
`\(t\).`

Explanation:

To find the curvature k of the curve at the point P, we can use the formula for curvature, which is given by:

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

where:

`\(\mathbf{r}(t)\)` is the vector-valued function that defines the curve, and

`\(\mathbf{T}(t)\)` is the unit tangent vector.

The prime notation
`\('\)` denotes the derivative with respect to
`\(t\).`

Given the vector-valued function
`\(\mathbf{r}(t) = e^t \cos(t) \mathbf{i} + e^t \sin(t) \mathbf{j} + e^t \mathbf{k}\),` we can first find the unit tangent vector
`\(\mathbf{T}(t)\)` and its derivative
`\(\mathbf{T}'(t)\).` Then we can evaluate the curvature at the point
`\(P(1, 0, 1)\)` by plugging in the appropriate values of
`\(t\).`

The unit tangent vector
`\(\mathbf{T}(t)\)` is given by:

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

The derivative of
`\(\mathbf{r}(t)\)` is:

`\[\mathbf{r}'(t) = e^t (-\sin(t) \mathbf{i} + \cos(t) \mathbf{j} + \sin(t) \mathbf{k}) + e^t \cos(t) \mathbf{i} + e^t \sin(t) \mathbf{j} + e^t \mathbf{k}\]`

`\[= e^t \cos(t) \mathbf{i} + e^t \sin(t) \mathbf{j} + (e^t - e^t \sin(t)) \mathbf{k}\]`

The magnitude of
`\(\mathbf{r}'(t)\)` is:

`\[\lvert \mathbf{r}'(t) \rvert = √((e^t \cos(t))^2 + (e^t \sin(t))^2 + (e^t - e^t \sin(t))^2)\]`

`\[= \sqrt{e^(2t) (\cos^2(t) + \sin^2(t)) + e^(2t) - 2e^(2t) \sin(t) + e^(2t) \sin^2(t)}\]`

`\[= \sqrt{e^(2t) + e^(2t) - 2e^(2t) \sin(t) + e^(2t) \sin^2(t)}\]`

`\[= \sqrt{2e^(2t) - 2e^(2t) \sin(t)}\]`

The unit tangent vector
`\(\mathbf{T}(t)\)` is then:

`\[\mathbf{T}(t) = \frac{e^t \cos(t) \mathbf{i} + e^t \sin(t) \mathbf{j} + (e^t - e^t \sin(t)) \mathbf{k}}{\sqrt{2e^(2t) - 2e^(2t) \sin(t)}}\]`

The derivative of
`\(\mathbf{T}(t)\)` is:

`\[\mathbf{T}'(t) = \frac{\mathbf{r}''(t) \lvert \mathbf{r}'(t) \rvert - \mathbf{r}'(t) \mathbf{r}''(t)}{(\lvert \mathbf{r}'(t) \rvert)^2}\]`

Plugging in the expressions for
`\(\mathbf{r}'(t)\) and \(\mathbf{r}''(t)\)`, we can find
`\(\mathbf{T}'(t)\).` Finally, we can evaluate the curvature
`\(k\)` at the point
`\(P(1, 0, 1)\)` by plugging in the appropriate value of
`\(t\).`