Here's a parameterization of a cylinder, for $0

Question

Here's a parameterization of a cylinder, for $0<\theta<2 \pi$ :
$$
\vec{v}(\theta, x)=(x, r \cos (\theta), r \sin (\theta))
$$

What is the inward-pointing vector normal to the area element of this cylinder given $r=8, \theta=\frac{5 \pi}{6}$, and $x=-2$ ?

Choose 1 answer:
(A) $(0,-4 \sqrt{3}, 4)$
(B) $(-4 \sqrt{3}, 4,0)$
(C) $(4 \sqrt{3},-4,0)$
(D) $(0,4 \sqrt{3},-4)$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the inward-pointing normal vector to the surface of the cylinder, we first need to find the normal vector to the surface. This can be done by taking the cross product of the partial derivatives of the vector function $\vec{v}( heta, x)$ with respect to $ heta$ and $x$. ‖ ‖ step 2 ⋮ The partial derivative of $\vec{v}( heta, x)$ with respect to $ heta$ is $\frac{\partial \vec{v}}{\partial heta} = (0, -r \sin( heta), r \cos( heta))$. Substituting $r = 8$ and $ heta = \frac{5\pi}{6}$, we get $\frac{\partial \vec{v}}{\partial heta} = (0, -8 \sin(\frac{5\pi}{6}), 8 \cos(\frac{5\pi}{6}))$. ‖ ‖ step 3 ⋮ The partial derivative of $\vec{v}( heta, x)$ with respect to $x$ is $\frac{\partial \vec{v}}{\partial x} = (1, 0, 0)$. ‖ ‖ step 4 ⋮ The cross product of $\frac{\partial \vec{v}}{\partial heta}$ and $\frac{\partial \vec{v}}{\partial x}$ gives the normal vector to the surface. The cross product is $\frac{\partial \vec{v}}{\partial x} imes \frac{\partial \vec{v}}{\partial heta} = (0, -8 \cos(\frac{5\pi}{6}), -8 \sin(\frac{5\pi}{6}))$. ‖ ‖ step 5 ⋮ Simplifying the components of the normal vector using the values of sine and cosine for $ heta = \frac{5\pi}{6}$, we get $(0, -8 \cdot (-\frac{\sqrt{3}}{2}), -8 \cdot (-\frac{1}{2})) = (0, 4\sqrt{3}, 4)$. ‖ ‖ step 6 ⋮ Since we need the inward-pointing normal vector, we take the negative of the normal vector we found, which gives us $(0, -4\sqrt{3}, -4)$. ‖ ∻[question number] Answer∻ ***D*** ∻Key Concept∻ ⚹Normal Vector to a Surface⚹ ∻Explanation∻ ⚹The normal vector to a surface can be found by taking the cross product of the partial derivatives of the parameterization vector with respect to its parameters. For an inward-pointing normal vector, we take the negative of the outward normal vector.⚹◊What is the formula for the inward-pointing vector normal to the area element of a cylinder parametrized by $\vec{v}( heta, x)$?⍭ Generate me a similar question◊

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