Calculus 3 Tangent

Question

Find the tangent plane to the equation z=−x2−4y2+2y at the point (3,-4,-81)

Answer 1

APPROACH ONE Using the gradian vector of the function

f(x,y,z) = x² + 4y² −2y +z

∇ f = < 2x, 8y−2 , 1>

Then the normal vector at (3, - 4, - 81) , is equal to ∇ f(3, - 4, - 81) = <6, - 34, 1>

Hence the equation of the tangent plane is given by the equation

6(x -3) − 34(y+4) +1(z+81) = 0

APPROACH TWO

If we let z= f(x,y) = - x² - 4y² +2y

then f_x = -2x and f_y = -8y +2

Then the equation of the tangent plane using the formula z−z₀= f_x(x−x₀) + f_y( y− y₀) for

( x_0, y₀, z₀) = ( 3, -4, -81)

we get z+ 81= - 6(x -3) +34 ( y +4)

APPROACH THREE ( GEOMETRIC)

If we let z= f(x,y) = - x² - 4y² +2y

then f_x = -2x and f_y = -8y +2

f_x = - 6 at ( 3, -4 ) the y's being held constant , in our case y = -4 .

That means when we cut the surface of z = f(x,y) with the plane y= -4 we get a curve C₁ which has slope (Δz)/ (Δx)= - 6 = (-6)/( 1)

This implies that ( Δx, Δy, Δz) = ( 1, 0,- 6), Δy being zero since the y's held constant , y = 4.

Which tells us that a vector u=< 1, 0,- 6 > is the directing vector of the tangent line to the surface that also lies on the plane y = -4 and also tangent to the curve C₁ at the point (3, - 4, -81)

Similarly working we obtain the vector v=< 0 , 1, 34 > as the directing vector of the tangent line to the surface that lies on the x = 3.

Consequently the normal vector of the tangent plane should be the cross product of vectors u and v.

uΧv= < 1, 0,- 6 >Χ< 0 , 1, 34 > = < 6, - 34, 1>

Hence the equation of the tangent plane is given by the equation

6(x -3) − 34(y+4) +1(z+81) = 0