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)
so when y=1 z = x^2 +4(1)
z' = 2x ; (-1,5) tanz = 2(-1)x -2(-1)(-1) + 5
Im confused so whats the Eq. of the tangent line??
if your confused, then what you need is clarification on the process, and not just the results ...
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!!
we determine the equation for f(x,y) when y=1 as stated in the problem z = x^2 + 4 is the outcome right?
yes no doubting that
then the derivative of z at this equation gives us the slope along any point ... z' = 2x correct?
yes so thats the equation i would think. but i saw you do some thing with tan?
the equation for the tangent uses this for slope; and is anchored to the intended point
since x=-1; the slope of the line is 2(-1) = -2
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
Ohh I see. I am not sure doesnt say. how would i make parametric
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)
x is independant so x =0+1t y = 1 so y=1 + 0t z depends on x = -2x+3 -> 2 -2t
tanz is just my notation to indicate that this is a tangent line using z as the dependant variable
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
cool thanks a bunch
yw
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