Respuesta :
Answer:
-30(x-4) + 29 (y-1) -12(z-3) = 0
Step-by-step explanation:
You can find the perpendicular vector to both curves using the cross product between them, and that will give you all the information you need to find the equation of the tangent plane to the surface S.
Let's find the parameter of each curve for the specific point.
For the first curve notice that
[tex]r( 0 ) = (4,1,3)[/tex]
And for the second curve notice that
[tex]r2(1) = (4,1,3)[/tex]
Thus for the first curve we reach our point at [tex]t=0,[/tex] and for the second curve we reach our point at [tex]u = 1 .[/tex]
Then we compute the tangent vector to our curves, and it would be given by the derivative so
[tex]r1'(t) = (2,-2t,-5+2t)[/tex] and at [tex]t = 0[/tex] we have that
[tex]r1'(0) = (2,0,-5)[/tex]
Similarly
[tex]r2'(1)=(3+2(1) , 6(1)^2 ,2) = (5,6,2)[/tex]
The the cross product between the two vectors would be
[tex](5,6,2) \times (2,0,-5) = (-30,29,-12)[/tex]
Since (-30,29,-12) would be the perpendicular vector to the tangent plane and (4,1,3) would be a point of the tangent then we would have that the equation of the tangent plane is given by
-30(x-4) + 29 (y-1) -12(z-3) = 0
The equation of the tangent plane at P is expressed as [tex]-30(x-4)+29(y-1)-12(z-3)=0[/tex]
The formula for calculating the equation of the tangent plane at P is expressed as [tex]a(x-x_0)+b(y-y_0)+c(z-z_0)=0[/tex]
Given the point [tex](x_0, y_0,z_0)=(4,1,3)[/tex]
Next is to get the coordinate (a, b, c) normal to the plane
[tex](a, b,c)= r'(0) \times r'(1)[/tex]
[tex]r^1'(t)=(2,-2t,-5+2t)\\r^1'(0)=(2,-2(0),-5+2(0))\\r^1'(0)=(2,0,-5)[/tex]
[tex]r^2'(u)=(3+2u,6u^2,u)\\r^2'(1)=(3+2,6(1)^2,2)\\r^2'(1)=(5,6,2)\\[/tex]
[tex](a, b,c)= r'(0) \times r'(1)\\(a, b,c)= (2,0,-5) \times (5,6,2)\\(a,b,c)=(-30,29,-12)[/tex]
a = -30
b = 29
c = -12
Substituting the resulting parameters into the formula;
[tex]-30(x-4)+29(y-1)-12(z-3)=0[/tex]
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