Finding the equation of the plane tangent

Question

‎Let S be the su‎rface descr‎ibed by the vec‎tor functi‎on

R^->(u,v)=<v², u+v, e^2u>

F‎ind an equ‎ation of the pla‎ne tan‎gent to S at the po‎int (4, −2, 1)

Answer 1

Let a = (4, -2, 1) be the given point.

First we express the surface R(u,v) by an equation of the form f(x, y, z) = 0.

Then we calculate the gradient ∇f, which is a vector normal to the surface.

At the point a, ∇f(a) is then a vector normal to the surface at the point a.

Using the dot product, the equation (x, y, z) . ∇f(a) = 0 is the equation

of the plane going through the origin, perpendicular to the normal ∇f(a).

Therefore the equation (x, y, z) . ∇f(a) = a . ∇f(a)

is the desired tangent plane.

To find the function f, we write the conditions for the coordinates x, y, z.

x = v² (1)

y = u+v (2)

z = e^2u (3)

From (3)

u = ln(z)/2

From (2)

v = y - u = y - ln(z)/2

Plugging that into (1)

x = (y - ln(z)/2)²

Therefore

f(x, y, z) = (y - ln(z)/2)² - x

∇f = (∂f/∂x, ∂f/∂y, ∂f/∂z)

= (-1, 2y - ln(z), ln(z)/2z - y/z)

At the point a

∇f(a) = (-1, -4, 2)

a . ∇f(a) = (4, -2, 1) . (-1, -4, 2) = 6

Therefore the equation of the tangent plane is

-x - 4y + 2z = 6