Question

2. Given the surface: f ( x, y ) = (e^ x + y) + ln( y^ 2 + x)
a. Show that the point (0,1) is located on the curve f ( x, y ) = e , and find the equation of
the tangent line to the curve, at that point.
b. Find the slope of the tangent line to the curve, that results from the intercept of the
plane x = 0 with the surface, at the point where y = 1 .
c. Find the slope of the tangent line to the curve, that results from the intercept of the
plane y = 1 with the surface, at the point where x = 0 .

how to solve this kind of questions? explanation will be very-very appreciated)

Answer 1

the point (0,1) does not lie on the curve f(x,y) =e

Answer 2

the point (1,0) lies on that curve
make sure you asked the question correctly

Answer 3

no formatting went wrong. but solution is less crucial-i am fine with this part. I am lost with theory. what steps should i use?

Answer 4

I mean algebraicly i can manage whatever, but I just dont und what is asked and what to do

Answer 5

which part? a, b, or c?

Answer 6

a-i just make it all equal to e and plug in point.
b and c-no idea

Answer 7

is this a calculus question?

Answer 8

it is as i und next course after calculus

Answer 9

we have not just x and y, but also z. so it is threedomential or how to call it

Answer 10

http://www.math.hmc.edu/calculus/tutorials/tangentplanes/

Answer 11

the plane where x is 0 is the y-z plane.

Answer 12

substitute x =0 and y =1 and find the value of z at that point.

Answer 13

i think it is one step ahead. we are not asked for plane, we asked to find line. so it is (x,y) probably

Answer 14

so you are basically trying to find the slope of the tangent plane at the point (0,1,2)

Answer 15

oh wait hold on.

Answer 16

i have surface. so i need to understand about which curve in this surface we are talking
and find tanget line to it in a specific points. that is how i understand it

Answer 17

what you suggest is different-does not look right to me((

Answer 18

the point on the curve is (0,1,2) correct?
you need to find the tangent to the curve at that point.

Answer 19

what you suggest is different-does not look right to me((

Answer 20

probably)

Answer 21

btw-tangent line in this case means that i should find partial derivative?

Answer 22

yes

Answer 23

and my next step? lets assume i already got Fx and Fy

Answer 24

my calculus is a bit rusty. let me read up on this. be back in a bit.

Answer 25

sure thank you for your time

Answer 26

did you find out \[f_x and f_y\]?

Answer 27

no yet, should I?

Answer 28

yes

Answer 29

sec

Answer 30

got it

Answer 31

Fx=(e)^(x^2+y)*2x+1/y^2+x*2y

Answer 32

I have never heard about normal vector. which means it probably different material

Answer 33

ignore that.

Answer 34

http://www.math.hmc.edu/calculus/tutorials/tangentplanes/#top

Answer 35

ok

Answer 36

see the example in that.

Answer 37

almost. but again it is different material. it is not tangent plane. we dont work with planes here really. we find curves and work with curves

Answer 38

hmmm

Answer 39

well, I suggest you wait for dumbcow or amistre64 or one of the others. They are good with calculus. Meantime, I'll try and read up on this problem and see if I can help.

Answer 40

my first step-i should understand what is my curve. then i have to find derivative. then plug in values if needed. my main problem-i dont understand how do i know about which curve they are asking and how do i find it. this step from surface to curve is totally unclear for me

Answer 41

then would be nice to understand how do i find one tangent line having two derivatives

Answer 42

your curve is
z = (e^x + y) + ln( y^2+ x)

Answer 43

sec

Answer 44

the derivative to that curve will give you a tangent plane.

Answer 45

it is e in a power of x^2+Y plus ln(y^2 +X)

Answer 46

huh?

Answer 47

lol, yes. hard to write it normally

Answer 48

well, whatever the curve is, you get the tangent plane to that curve when you find the derivative of the curve.

Answer 49

it is obvious. but steps are not

Answer 50

oh now I understand
the line they are talking about is the line created by the tangent plane to the curve.

Answer 51

no tanget plane here. just lines.

Answer 52

have you done tangent planes before? if not, its kinda hard to visualize.

Answer 53

well, thats the only interpretation I can see of the problem. Maybe Im reading it wrong. sorry couldn't be of more help

Answer 54

I know how does this space looks like. for example circle looks like a huge vase with sticks out of it

Answer 55

but such visualizing does not help me to solve the problem. does not matter. you probably did not cover this yet.

Answer 56

but let me get back to you when I've researched this a bit, but I think my interpretation is right.

Answer 57

ok i will go over notes again.

Answer 58

thank you

Answer 59

oh no, I have done all this about 10 years ago. I don't remember the specifics. Don't use math all that much these days for my job :)

Answer 60

My gosg how old are you? I though everybody here are around 15-20

Answer 61

lol, i'm 25

Answer 62

I am 26

Answer 63

no kidding! are you majoring in math?

Answer 64

It is my first year in university.

Answer 65

oh okay. great! which university?

Answer 66

RRIS, Israel

Answer 67

And as you can guess i finished school 10 years ago-so since that i did not see math at all. now everything from the beggining

Answer 68

oh okay. Yeah, thats a hassle. I don't remember much from what I studied in high school either. well, I hope someone else comes along and helps you out. good luck!

Answer 69

thank you)