# Thread: Equation of the plane tangent to the surface

1. ## Equation of the plane tangent to the surface

Hi,

I tried to this way:

$\displaystyle 5u^2 = 5$
$\displaystyle u = 1$

$\displaystyle 5v^2 = 45$
$\displaystyle v = sqrt(9)$

when u = 1 and v = -3
$\displaystyle 2u - v^2 = -7$
$\displaystyle 2(1) - (-3)^2 = -7$

when u = 1 and v = 3
$\displaystyle 2u - v^2 = -7$
$\displaystyle 2(1) - (3)^2 = -7$

Problem:

Find the equation of the plane tangent to the surface

at the point P that is (approximately) P(5,-7,45).

many thanks

2. Originally Posted by althaemenes
Hi,

I tried to this way:

$\displaystyle 5u^2 = 5$
$\displaystyle u = 1$

$\displaystyle 5v^2 = 45$
$\displaystyle v = sqrt(9)$

when u = 1 and v = -3
$\displaystyle 2u - v^2 = -7$
$\displaystyle 2(1) - (-3)^2 = -7$

when u = 1 and v = 3
$\displaystyle 2u - v^2 = -7$
$\displaystyle 2(1) - (3)^2 = -7$

Problem:

Find the equation of the plane tangent to the surface

at the point P that is (approximately) P(5,-7,45).

many thanks
First we take the partial to find to vectors tangent to the surface.

$\displaystyle \vec r(u,v) = 5u^2 \vec i +(2u-v^2)\vec j+5v^2 \vec k$

$\displaystyle r_u(u,v)=10u \vec i + 2 \vec j$

$\displaystyle r_v(u,v)=-2v \vec j +10v \vec k$

$\displaystyle r_u(1,3)=10(1) \vec i + 2 \vec j =10 \vec i +2 \vec j$

$\displaystyle r_v(1,3)=-2(3) \vec j +10(3) \vec k=-6 \vec j + 30 \vec k$

Now we take the cross product of these two vectors to get a Normal vector to the surface.

Can you finish from here?

Good luck.

Originally Posted by TheEmptySet
First we take the partial to find to vectors tangent to the surface.

$\displaystyle \vec r(u,v) = 5u^2 \vec i +(2u-v^2)\vec j+5v^2 \vec k$

$\displaystyle r_u(u,v)=10u \vec i + 2 \vec j$

$\displaystyle r_v(u,v)=-2v \vec j +10v \vec k$

$\displaystyle r_u(1,3)=10(1) \vec i + 2 \vec j =10 \vec i +2 \vec j$

$\displaystyle r_v(1,3)=-2(3) \vec j +10(3) \vec k=-6 \vec j + 30 \vec k$

Now we take the cross product of these two vectors to get a Normal vector to the surface.

Can you finish from here?

Good luck.
The cross product is = 60i - 300j + 12k

Therefore, equation at (5,-7,45) =

$\displaystyle 60(x-5)-14( y-(-7))+12(z-45) = 0$
$\displaystyle 60x-300-14y+98+12z-540 = 0$
$\displaystyle 60x+14y+12z-742 = 0$

Now, how do you format it in z(x,y) form? should I take z to the other side of the equation?

4. Originally Posted by althaemenes
The cross product is = 60i - 300j + 12k

Therefore, equation at (5,-7,45) =

$\displaystyle 60(x-5)-14( y-(-7))+12(z-45) = 0$
$\displaystyle 60x-300-14y+98+12z-540 = 0$
$\displaystyle 60x+14y+12z-742 = 0$

Now, how do you format it in z(x,y) form? should I take z to the other side of the equation?