Can someone help me understand what I am doing wrong on number 10 please? V145745tor the function f(x,u)= xf+ Zxy + Sy?+ 4x - 24 +1, find a unit tangentNector to the level curve at the' poin't (-3,-4) that has ao componentpositive2x +104-2>NF(-3;4) = < 2(-3) + 2(-4)+4, 2(-3)+ l0 (-A)-2)く-b -8+43-10b-40-2 >48)normal vector48t+ (-10) =10ZA64K.9790.2040482404124040. for the function f(x,4) = 3-x*-ytat the point (3,5) that'finda unit tangent vector to the level curve1.+3has a positive x com ponent(-3)?t(3,5) =/3-3} - (s})-o -(2(3).-3) 3-(3)8-5P-0 E (-2,(5)--3)\9norma I vector =- 3093,8973514519.23.3333 =.8575

Answered: 0. for the funchion f (x,4) = 3-x?- u?,…

0. for the funchion f (x,4) = 3-x?- u?, find a unit tangent vector to the level curve at the point (3,5) that' -3 has a positive x component

Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)

6th Edition

ISBN:9781337111348

Author:Bruce Crauder, Benny Evans, Alan Noell

Publisher:Bruce Crauder, Benny Evans, Alan Noell

Chapter6: Rates Of Change

Section6.1: Velocity

Problem 12SBE

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Question

Can someone help me understand what I am doing wrong on number 10 please?

V145
745
tor the function f(x,u)= xf+ Zxy + Sy?+ 4x - 24 +1, find a unit tangent
Nector to the level curve at the' poin't (-3,-4) that has a
o component
positive
2x +
104-2>
NF(-3;4) = < 2(-3) + 2(-4)+4, 2(-3)+ l0 (-A)-2)
く-b -8+4
3-10
b-40-2 >
48)
normal vector
48t+ (-10) =
10
ZA64
K.9790
.2040
48
2404
12404
0. for the function f(x,4) = 3-x*-yt
at the point (3,5) that'
find
a unit tangent vector to the level curve
1.
+3
has a positive x com ponent
(-3)?
t(3,5) =/3-3} - (s})-o -(2(3).-3) 3-(3)8-5P-0 E (-2,(5)--3)\
9
norma I vector =
- 30
9
3,8973
51451
9.
23.3333 =
.8575

Expert Solution

Step 1

Given: $f (x, y) = \frac{3 - x^{2} - y^{2}}{- 3}$ at $P (3, 5)$

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