Question

Calculate the normal component of F to the surface. (Calc 3)

Answer 1

Let F = <z, 0, y> and
Let S be the oriented surface parametrized by G(u, v) = (u2 − v, u, v2)
for 0 ≤ u ≤ 9, −1 ≤ v ≤ 4.
Calculate the normal component of F to the surface at P = (63, 8, 1) = G(8, 1).

So I tried and got 10 / sqrt(21).

Answer 2

Steps for tangential components:
\[T_{u} = \frac{ dG }{ du } = (2u, 1, 0)\]
\[T_{v} = \frac{ dG }{ dv } = (-1, 0, 2v)\]

Normal vector:
det
\[\left[ 2u \right] \left[ 1 \right] \left[ 0 \right]\]
\[\left[ -1 \right] \left[ 0 \right] \left[ 2v \right]\]

\[n = <2v, -4uv, 1>\]

F(P) = <1, 0, 8>
n(P) = <2, -4, 1>

\[e_{n}(P) = \frac{ n(P) }{ |n(P)| } = \frac{ <2, -4, 1> }{ \sqrt{21} }\]

\[F(P) * e_{n}(P) = <1,0,8> * \frac{ 1 }{ \sqrt{21} } <2, -4, 1> = \frac{ 2 }{ \sqrt{21} } + \frac{ 8 }{ \sqrt{21} } = \frac{ 8 }{ \sqrt{21} }\]

Answer 3

Steps for tangential components:
\[T_{u} = \frac{ dG }{ du } = (2u, 1, 0)\]
\[T_{v} = \frac{ dG }{ dv } = (-1, 0, 2v)\]

Normal vector:
det
\[\left[ 2u \right] \left[ 1 \right] \left[ 0 \right]\]
\[\left[ -1 \right] \left[ 0 \right] \left[ 2v \right]\]

\[n = <2v, -4uv, 1>\]

F(P) = <1, 0, 8>
n(P) = <2, -4, 1>

\[e_{n}(P) = \frac{ n(P) }{ |n(P)| } = \frac{ <2, -4, 1> }{ \sqrt{21} }\]

\[F(P) * e_{n}(P) = <1,0,8> * \frac{ 1 }{ \sqrt{21} } <2, -4, 1> = \frac{ 2 }{ \sqrt{21} } + \frac{ 8 }{ \sqrt{21} } = \frac{ 8 }{ \sqrt{21} }\]

Answer 4

check algebra

\(n(P) = <2, \color{red}{-32}, 1>\)

Answer 5

But shouldn't the end result be regardless of the y component of n(P)?

Answer 6

@IrishBoy123

Answer 7

what do you get for \(\hat n\) now?

Answer 8

what is "Calc 3", btw??

Answer 9

Calculus 3

Answer 10

Let me compute that..

Answer 11

\[7\sqrt{21}\]

Answer 12

that would be |n(P)|

Answer 13

n hat = \[\frac{ <2,-32,1> }{ 7\sqrt{21} }\]

Answer 14

well, that's correcting the glitch i found, assuming you agree. if that does not work, we need to look at something else. but the method is sound.

Answer 15

Can you please tell me how you computed the components of n(P)? I was just guessing from the solution.

Answer 16

With the correction the final answer is \[\frac{ 10 }{ 7\sqrt{21} }\] (verified correct)

Answer 17

well, n=<2v,−4uv,1>

so, −4uv = -4(8)(1) = -32

is that what you are getting at?

Answer 18

That makes sense now. Was tripping on that haha. thank you