Question

Given:f(x,y) = x^2 + 4y^2...Find an equation for the tangent line to the curve which its intersection of the surface z=(x,y) and the plane y=1 through the point (-1,1,5)

Answer 1

so when y=1

z = x^2 +4(1)

Answer 2

z' = 2x ; (-1,5)

tanz = 2(-1)x -2(-1)(-1) + 5

Answer 3

Im confused so whats the Eq. of the tangent line??

Answer 4

if your confused, then what you need is clarification on the process, and not just the results ...

Answer 5

I know amistre I just dont have the book in front of me since im at work, I will look over when i get home, thanks again!!

Answer 6

we determine the equation for f(x,y) when y=1 as stated in the problem

z = x^2 + 4 is the outcome right?

Answer 7

yes no doubting that

Answer 8

then the derivative of z at this equation gives us the slope along any point ...

z' = 2x correct?

Answer 9

yes so thats the equation i would think. but i saw you do some thing with tan?

Answer 10

the equation for the tangent uses this for slope; and is anchored to the intended point

Answer 11

since x=-1; the slope of the line is 2(-1) = -2

Answer 12

tanz = mx -mPx + Py givent that (Px,Py) is (-1,5)

tanz = -2x +2(-1) + 5

tanz = -2x +3

but im not sure if they want that in parametric form or not

Answer 13

Ohh I see. I am not sure doesnt say. how would i make parametric

Answer 14

Also, would i do inverse tangent for solve for z? or do we genrally leave the equation of a tangent line in this type ofr form (tanz=-2x+3)

Answer 15

x is independant so x =0+1t

y = 1 so y=1 + 0t

z depends on x = -2x+3 -> 2 -2t

Answer 16

tanz is just my notation to indicate that this is a tangent line using z as the dependant variable

Answer 17

and if we claen up the tyops .... or at least the ones that were made previously lol

x = 0 + t
y = 1+0t
z = 3 -2t

Answer 18

cool thanks a bunch

Answer 19

yw