Find equations of the tangent plane and normal line to the surface x=4y2+1z2−342
at the point (-9, -9, 3).
Tangent Plane: (make the coefficient of x equal to 1). =0
Normal line: ⟨−9, , ⟩ +t⟨1, , ⟩
To find the equation of the tangent plane, we need to find the gradient vector at the given point (-9, -9, 3).
The equation of the given surface is x = 4y^2 + z^2 - 342.
Taking partial derivatives with respect to x, y, and z, we have:
∂x/∂x = 1
∂x/∂y = 8y
∂x/∂z = 2z
At the point (-9, -9, 3), we have:
∂x/∂x = 1
∂x/∂y = 8(-9) = -72
∂x/∂z = 2(3) = 6
So the gradient vector at (-9, -9, 3) is ⟨1, -72, 6⟩.
The equation of the tangent plane can be written as Ax + By + Cz = D, where the coefficients A, B, and C are the components of the gradient vector.
Plugging in the values, we have:
1(x - (-9)) - 72(y - (-9)) + 6(z - 3) = 0
x + 9 + 72y + 648 + 6z - 18 = 0
x + 72y + 6z + 639 = 0
To make the coefficient of x equal to 1, we divide the equation by 72:
(x/72) + (y/1) + (z/12) + (639/72) = 0
So the equation of the tangent plane is:
(x/72) + y + (z/12) + (639/72) = 0
For the normal line, we use the direction vector of the gradient vector, which is ⟨1, -72, 6⟩. We can write the equation of the normal line as:
⟨-9, -9, 3⟩ + t⟨1, -72, 6⟩
So the equation of the normal line is:
x = -9 + t
y = -9 - 72t
z = 3 + 6t
To find the equation of the tangent plane at the point (-9, -9, 3), we need to compute the partial derivatives of the given surface equation with respect to x, y, and z.
Given surface equation: x = 4y^2 + z^2 - 342
Partial derivative with respect to x (keeping y and z constant):
∂x/∂x = 1
Partial derivative with respect to y (keeping x and z constant):
∂x/∂y = 8y
Partial derivative with respect to z (keeping x and y constant):
∂x/∂z = 2z
Now, let's substitute the coordinates of the given point (-9, -9, 3) into the partial derivatives:
∂x/∂x = 1
∂x/∂y = 8*(-9) = -72
∂x/∂z = 2*3 = 6
The equation of the tangent plane can be written as:
1*(x - (-9)) + (-72)*(y - (-9)) + 6*(z - 3) = 0
Simplifying:
x + 9 - 72y + 9 + 6z - 18 = 0
x - 72y + 6z = -6
Thus, the equation of the tangent plane to the surface x = 4y^2 + z^2 - 342 at the point (-9, -9, 3) is x - 72y + 6z = -6.
To find the equation of the normal line, we use the normal vector of the tangent plane as the direction vector for the line.
Using the coefficients of x, y, and z from the equation of the tangent plane, we get the normal vector N⟨1, -72, 6⟩.
The equation of the normal line can be written as:
⟨-9, -9, 3⟩ + t⟨1, -72, 6⟩
So, the equation of the normal line to the surface at the point (-9, -9, 3) is ⟨-9, -9, 3⟩ + t⟨1, -72, 6⟩, where t is a parameter.