Question

ok, really starting to get frustrated with math......been trying to finish this section for 2 days now but i keep running into problems....

Find equations of the tangent plane and normal line to the surface x=3y2+4z2−211 at the point (-3, 2, 7).
Tangent Plane: (make the coefficient of x equal to 1).

Normal line: ⟨−3, , ⟩
+t⟨1, , ⟩.

Answer 1

i got grad f =<-1, 6y, 8z> and grad f at the point =<-1,12,56> so should be <-1, 12, 56>*<x+3, y-2, z-7> or -1(x+3)+12(y-2)+56(z-7). and of course, webwork says it's wrong.............

Answer 2

hi

Answer 3

since they want coeff of x = 1

Answer 4

huh?

Answer 5

why are you finding distance? i'm confused.....what's d?

Answer 6

huh

Answer 7

equation of a plane is given by ax+by+cz=d
where a,b,c,d are constants

Answer 8

also if u wanna arrive at a relationship u are missing an equality here

Answer 9

<-1, 12, 56>*<x+3, y-2, z-7> =0
the dot of these 2 vectors for any given point x,y,z on the plane is 0 since the normal line is perpendicular or "normal" to the plane

Answer 10

-1(x+3)+12(y-2)+56(z-7)=0
simply this
-x-3+12y-24+56z-56*7=0
so
-x+12y+56z=3+24+56*7
x-12y-56z=-(3+24+56*7)

Answer 11

why can't i use grad f (x0, y0, z0)*(x-x0, y-y0, z-z0>?

Answer 12

cause that -1(x+3)+12(y-2)+56(z-7)=0 is exactly what i got and it's wrong.

Answer 13

looks like i made a silly error the first time
it shud be the coeff from the normal so
-x+12y+56z=d
-3+12(2)+56*7=d
x-12y-56z=-(-3+12(2)+56*7)

Answer 14

i think -1(x+3)+12(y-2)+56(z-7)=0 is right, they just want the coefficient of x to be one, so i don't know how to change that.

Answer 15

i just did it for u

Answer 16

x-12y-56z=-413

Answer 17

but that's still not the right answer

Answer 18

do they want it entered in some particular form?

Answer 19

sorry, forgot to make the screen shot .jpeg....gimme a sec

Answer 20

oh wait its not -3 hters its --3 so

Answer 21

just input it in like that it will accept it its webwork

Answer 22

oh =0 so ya
x-12y-56z+(-3+12(2)+56*7)=0

Answer 23

did it work?

Answer 24

nope

Answer 25

its not 413 u pt anymore right

Answer 26

Entered Answer Preview Result Messages
x-12*y-56*z+(-3)+12*2+56*7
x−12y−56z+(−3)+12⋅2+56⋅7
x-12*y-56*z+(-3)+12*2+56*7
incorrect This answer is equivalent to the one you just submitted.

Answer 27

hence why i'm really starting to get annoyed with math.....

Answer 28

oh change the 3 to -3

Answer 29

wanna just give up and get an F on this assignment.......

Answer 30

wtf.. dude its easy.. change the 3 to -3

Answer 31

well, like i said in the beginning of the post, i keep running into problems. this is only qeustion 11 of 20 and i've run into issues on most all of them

Answer 32

x-12y-56z+(-(-3)+12(2)+56*7)=0

Answer 33

every time i think i understand something, i'm wrong.

Answer 34

i didn't know about setting it equal to d....had no idea what you were doing in the beginning.

Answer 35

is it right yet

Answer 36

the first part, finally.

Answer 37

...

Answer 38

its not my fault.. i dont want to keep scrolling up to see the point... no time to memorize the point was -3 not and 3 for x

Answer 39

as far as the equation for d its straight forward

Answer 40

i didn't say it was your fault. you're helping.

Answer 41

you know the dot of normal * your <x-a,y-b...> vectors will leave the normal line's components as the components on x y and z

Answer 42

just saying i'm extremely frustrated is all. been working on this homework set for 2 days....

Answer 43

instead of doing the dot product which is a bit annoying, u can just calculate the normal and u aleady know like if the normal is <a,b,c>
then u can say
ax+by+cz=d
now u plug in the point x,y,z and see what d has to be

Answer 44

such that the equation of your plane is satisfied

Answer 45

k do the nromal line since they want it at -3 for x

Answer 46

and parametrized at 1 change in x

Answer 47

we can see what the other vaues have to be

Answer 48

ax+by+cz=d.....(-3)-12(2)-56(7)+419=d

Answer 49

your normal line vector is
<-1,12,56> same as
<1,-12,-56> right

Answer 50

so right away u know this part
<3,...,...> +t<1,-12,-56>

Answer 51

no....that = 0

Answer 52

now to see what the starting point must be

Answer 53

<-3*,...,...> +t<1,-12,-56>

Answer 54

ax+by+cz=d.....(-3)-12(2)-56(7)+419=d
d= 0 .........which doesn't sound helpful at all........grr......

Answer 55

<-3*,12*3,3(56)> +t<1,-12,-56>

Answer 56

see if that works

Answer 57

no

Answer 58

Entered Answer Preview Result
x-12*y-56*z+3+12*2+56*7
x−12y−56z+3+12⋅2+56⋅7
correct
36
12⋅3
incorrect
168
3⋅56
incorrect
-12
−12
correct
-56
−56
correct

Answer 59

In your very first reply you have the equation of the tangent plane correct:
-1(x+3)+12(y-2)+56(z-7) = 0
But they want the x-coefficient to be 1. So multiply throughout by -1:
(x+3) - 12(y-2) - 56(z-7) = 0
x - 12y - 56z + 419 = 0

Answer 60

im asking for the 2nd part

Answer 61

can u take screen shot that is confusing to look at

Answer 62

that was the second part....you went really fast, swissgirl. i'm really lost now. i have no idea what you did.....

Answer 63

Normal line = <-3, 2, 7> + t<-1,12,56>

Answer 64

oh oops just put the point in
<1,

Answer 65

for a second i thought they wanted u to just parametrize a line going thru the origin

Answer 66

=]

Answer 67

OH! yeah, cause <x0, y0, z0>+t,...

Answer 68

my frustration is confusing me and my ADHD is not helping. my brain hurts.

Answer 69

alright go on next question

Answer 70

Try: <-3, 2, 7> + t<1,-12,-56>

Answer 71

he already got it aum

Answer 72

Oh, ok.

Answer 73

cant u see he doesnt need ur shiet right now the man is stressed!

Answer 74

he's stressing me out too

Answer 75

wait, what? aum helped and you helped....who's stressing who out? math is stressing me. that's all. i got the answer right.

Answer 76

just jokin, lets goo next question

Answer 77

oh, ok.

Answer 78

next one is very similar. Find equations of the tangent plane and normal line to the surface z−16=xeycosz at the point (-16, 0, 0).
Tangent Plane: (make the coefficient of z equal to 1).
=0.
Normal line: ⟨−16, , ⟩
+t⟨ , , 1⟩.

Answer 79

take screnshot

Answer 80

so first step is finding partial x and partial y.

Answer 81

kk

Answer 82

yah

Answer 83

u know why the partial is working right

Answer 84

tangent is derivative or slope. yeah.

Answer 85

ok ya do the same thing
write it as G=f(x,y,z)

Answer 86

<G_x,G_y,G_z>

Answer 87

i have to get z-16 on the other side, right?

Answer 88

not hte 16 we dont care about constants when it comes to the slopes

Answer 89

i mean constants that in a term by themself that is

Answer 90

yeah, but we have to find the partial of z so it has to be on the same side?

Answer 91

http://prntscr.com/4u50nl

Answer 92

yeah, that's wht i got for partials. just a little slower to get there.