Respuesta :
Answer:
the intersection point is ( -5/4, -5/4, -9/8)
Step-by-step explanation:
F(x,y,z) = x² +y² -z= 0
Then find differential of each terms then we have
∀f(x)=2x
∀f(y)=2y
∀f(z)=-1
The partial differential which is the director vector is at F(x,y,z)= (1,1,2)
Vf(1,1,2) = (2,2,-1)
But the given point is ( 1.1,2)
Then the parametric equation of normal line will be
x= 1+2t
y= 1+2t
z= 2 -t
The parametric equation can be formed as
(1+2t)² + (1+2t)² - (2-t)= 0
If we expand the expression above we have,
2(1+4t+4t²) - 2+t= 0
0= 8t² + 9t
t= 0 or t= -9/8
Substitute the value of t into the parametric equation above
At t=0
x= 1+2t; x= 1+2(0)=1
y= 1+2t; y= 1+2(0)=1
z=2 -t ; z= 2-(0)= 2
At =0 , we have (1,1,2)
At t= -9/8
x= 1+2t; x= 1+2(-9/8)=-5/4
y= 1+2t; y= 1+2(-9/8)=-5/4
z=2 -t ; z= 2-(-9/8)= -9/8
Therefore, the intersection point is ( -5/4, -5/4, -9/8)
The normal intersects the paraboloid at [tex](-\frac{5}{4},-\frac{5}{4},\frac{25}{8} )[/tex]
Paraboloid:
A surface all of whose intersections by planes are either parabolas and ellipses or parabolas and hyperbolas.
Given function is,
[tex]z = x^2 + y^2[/tex]
We can write the given function as,
[tex]F(x,y,z)=x^2+y^2-z[/tex]
Compute the gradient [tex]F[/tex] by using the formula,
[tex]\nabla F(x,y,z)=\left\langle 2x,2y,-1\right\rangle[/tex]
At (1,1,2) we get,
[tex]\nabla F(x,y,z)=\left\langle 2,2,-1\right\rangle[/tex]
The equation of the tangent plane at (1,1,2) is,
[tex]2(x-1)+2(y-1)-(z-2)=0\\2x+2y-z-2=0\\\frac{x-1}{2}=\frac{y-1}{2}=\frac{z-2}{-1}=t\\ x=2t+1,y=2t+1,z=-t+2[/tex]
Replace the values of [tex]x,y,z[/tex] in [tex]z = x^2 + y^2[/tex] and then solve for [tex]t[/tex] then,
[tex]-t+2=(2t+1)^2+(2t+1)^2\\8t^2+9t=0\\t=0,-\frac{9}{8}[/tex]
Substituting the value into the parametric equations,
[tex]x=2(-\frac{9}{8} )+1,y=2(-\frac{9}{8} )+1,\ and \ z=-(-\frac{9}{8} )+2\\x=-\frac{5}{4}, \ y= -\frac{5}{4} ,z=\frac{25}{8}[/tex]
Therefore, the normal intersects the paraboloid at [tex](-\frac{5}{4},-\frac{5}{4},\frac{25}{8} )[/tex]
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