- #1
JFonseka
- 117
- 0
Suppose that F(x,y) = x[tex]^{2}[/tex]+y[tex]^{2}[/tex]. By using vector geometry, find the Cartesian equation of the tangent plan to the surface z = F(x,y) at the point where (x,y,z) = (1,2,5). Find also a vector n that is normal to the surface at this point.
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Now the step by step working is given for this question, I however get confused at one part.
First they intersect the surface z = x[tex]^{2}[/tex]+y[tex]^{2}[/tex] with the plane x = 1
Therefore z = 1 + y[tex]^{2}[/tex]
The gradient F[tex]_{y}[/tex](1,2) = 4
By using the point gradient formula for a straight line, the Cartesian eq for the tangent is:
z -5 = 4(y - 2), x =1
If [tex]\lambda[/tex] = y - 2, then the equation of the tangent line in parametric vector form is
(x y z) = (1 2 5) + [tex]\lambda[/tex](0 1 4)
Now my question is, where did the (0 1 4) come from?
======================================================================
Now the step by step working is given for this question, I however get confused at one part.
First they intersect the surface z = x[tex]^{2}[/tex]+y[tex]^{2}[/tex] with the plane x = 1
Therefore z = 1 + y[tex]^{2}[/tex]
The gradient F[tex]_{y}[/tex](1,2) = 4
By using the point gradient formula for a straight line, the Cartesian eq for the tangent is:
z -5 = 4(y - 2), x =1
If [tex]\lambda[/tex] = y - 2, then the equation of the tangent line in parametric vector form is
(x y z) = (1 2 5) + [tex]\lambda[/tex](0 1 4)
Now my question is, where did the (0 1 4) come from?