Consider the surface xyz=30. How do you find the unit normal vector to the surface at (2,5,3)?

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Consider the surface xyz=30. How do you find the unit normal vector to the surface at (2,5,3)?

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Answer 1 · 2016-12-29T11:23:38

$ˆ \to n = \frac{1}{19} (15 ˆ \to i + 6 ˆ \to j + 10 ˆ \to k)$ $hatvecn=1/19(15hatveci+6hatvecj+10hatveck)$

Explanation:

$x y z = 30$ $xyz=30$

write as

$f (x, y, z) = x y z - 30 = 0$ $f(x,y,z)=xyz-30=0$

vector normal to $f (x)$ $" "f(x)" "$ is given by $\nabla f (x, y, z)$ $gradf(x,y,z)$

$\nabla f (x, y, z) = (ˆ \to i \frac{\partial}{\partial x} + ˆ \to j \frac{\partial}{\partial y} + ˆ \to k \frac{\partial}{\partial z}) (x y z - 30)$ $gradf(x,y,z)=(hatvecidel/(delx)+hatvecjdel/(dely)+hatveckdel/(delz))(xyz-30)$

$\nabla f (x, y, z) = y z ˆ \to i + x z ˆ \to j + x y ˆ \to k$ $gradf(x,y,z)=yzhatveci+xzhatvecj+xyhatveck$

$\nabla f (2, 5, 3) = (5 \times 3) ˆ \to i + (2 \times 3) ˆ \to j + (2 \times 5) ˆ \to k$ $gradf(2,5,3)=(5xx3)hatveci+(2xx3)hatvecj+(2xx5)hatveck$

$\nabla f (2, 5, 3) = 15 ˆ \to i + 6 ˆ \to j + 10 ˆ \to k$ $gradf(2,5,3)=15hatveci+6hatvecj+10hatveck$

call this normal vector $\to n$ $""vecn$

unit vector in this direction is given by

$ˆ \to n = \frac{\to n}{∣ ∣ \to n ∣ ∣}$ $hatvecn=vecn/|vecn|$

$∣ ∣ \to n ∣ ∣ = \sqrt{15^{2} + 6^{2} + 10^{2}} = 19$ $|vecn|=sqrt(15^2+6^2+10^2)=19$

$ˆ \to n = \frac{1}{19} (15 ˆ \to i + 6 ˆ \to j + 10 ˆ \to k)$ $hatvecn=1/19(15hatveci+6hatvecj+10hatveck)$

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Consider the surface xyz=30. How do you find the unit normal vector to the surface at (2,5,3)?

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