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).
To find the equation of the tangent plane to the surface at the point (-9, -9, 3), we need to find the partial derivatives of the surface equation with respect to x, y, and z.
Given surface equation: x = 4y^2 + 1z^2 - 342
Partial derivative with respect to x: ∂x/∂x = 1
Partial derivative with respect to y: ∂x/∂y = 8y
Partial derivative with respect to z: ∂x/∂z = 2z
At the point (-9, -9, 3), we substitute these values into the partial derivative equations to find the slope of the tangent plane:
∂x/∂x = 1
∂x/∂y = 8(-9) = -72
∂x/∂z = 2(3) = 6
So the equation of the tangent plane, with the coefficient of x equal to 1, becomes:
1(x + 9) - 72(y + 9) + 6(z - 3) = 0
Simplifying this equation, we have:
x - 72y + 6z - 705 = 0
Therefore, the equation of the tangent plane to the surface at the point (-9, -9, 3) is:
x - 72y + 6z - 705 = 0
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To find the equation of the normal line to the surface at the point (-9, -9, 3), we need to find the gradient vector of the surface equation, evaluated at the given point.
Gradient vector: ∇F = (∂F/∂x, ∂F/∂y, ∂F/∂z)
In this case, F(x, y, z) = x - 4y^2 - z^2 + 342
∂F/∂x = 1
∂F/∂y = -8y
∂F/∂z = -2z
At the point (-9, -9, 3), we substitute these values into the gradient vector equations:
∂F/∂x = 1
∂F/∂y = -8(-9) = 72
∂F/∂z = -2(3) = -6
So the gradient vector at the given point is:
∇F = (1, 72, -6)
Therefore, the equation of the normal line to the surface at the point (-9, -9, 3) is:
(x - (-9))/1 = (y - (-9))/72 = (z - 3)/(-6)
To find the equation of the tangent plane to the surface at a given point, we need to calculate the partial derivatives of the surface equation with respect to x, y, and z.
The given surface equation is x = 4y^2 + 1z^2 - 342.
Step 1: Calculate the partial derivative with respect to x:
∂x/∂x = 1
∂x/∂y = 8y
∂x/∂z = 2z
Step 2: Evaluate the partial derivatives at the point (-9, -9, 3):
∂x/∂x = 1
∂x/∂y = 8(-9) = -72
∂x/∂z = 2(3) = 6
Step 3: Substitute the values into the general equation of a plane:
(x - x₀) + A(y - y₀) + B(z - z₀) = 0
Where (x₀, y₀, z₀) is the given point and (A, B, C) is the normal vector to the plane.
Using the point (-9, -9, 3) and the calculated partial derivatives, we have:
(x - (-9)) + 1(y - (-9)) - 72(z - 3) = 0
(x + 9) + (y + 9) - 72(z - 3) = 0
x + y - 72z - 198 = 0
Therefore, the equation of the tangent plane to the surface x = 4y^2 + z^2 - 342 at the point (-9, -9, 3) is:
x + y - 72z - 198 = 0