For the curve given by , find the unit normal vector.
E)None of these

Question

For the curve given by  , find the unit normal vector. 
E)None of these

Quizplus · Accepted Answer

To find the unit normal vector, we first need to find the unit tangent vector:
$$\vec{T} = \frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|}$$
where $\mathbf{r}(t)$ is the vector function that defines the curve. Taking the derivative of the given curve, we have:
$$\mathbf{r}'(t) = \langle -10\sin t, 10\cos t, 3\rangle$$
So,
$$\|\mathbf{r}'(t)\| = \sqrt{(-10\sin t)^2 + (10\cos t)^2 + 3^2} = \sqrt{100 + 9} = \sqrt{109}$$
and
$$\vec{T} = \frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|} = \frac{1}{\sqrt{109}}\langle -10\sin t, 10\cos t, 3\rangle$$
Now, to find the unit normal vector, we take the derivative of $\vec{T}$ with respect to $t$ and normalize that vector:
$$\vec{N} = \frac{\frac{d\vec{T}}{dt}}{\left\|\frac{d\vec{T}}{dt}\right\|}$$
$$\frac{d\vec{T}}{dt} = \frac{1}{\sqrt{109}}\langle -10\cos t, -10\sin t, 0\rangle$$
$$\left\|\frac{d\vec{T}}{dt}\right\| = \sqrt{(-10\cos t)^2 + (-10\sin t)^2 + 0^2} = 10$$
So,
$$\vec{N} = \frac{\frac{d\vec{T}}{dt}}{\left\|\frac{d\vec{T}}{dt}\right\|} = \frac{1}{10}\langle -\cos t, -\sin t, 0\rangle$$
The only answer choice that matches this vector is B.

For the Curve Given by , Find the Unit