Question

Find equations of the tangent plane and normal line to the surface x=3y^2+3z^2−274 at the point (-4, -9, -3).

Tangent Plane: (Make the coefficient of x equal to 1).
a. A. (x + 3y - 2z = -275)
b. B. (x - 3y + 2z = 275)
c. C. (x + 3y - 2z = 274)
d. D. (x - 3y + 2z = -274)

Answer 1

To find the equation of the tangent plane to the surface x=3y^2+3z^2−274 at the point (-4, -9, -3), calculate the partial derivatives and substitute the point into them. Then, use the point and the partial derivatives to form the equation of the tangent plane: x + 54y - 18z = -275.

Step-by-step explanation:

To find the equation of the tangent plane to the surface x=3y^2+3z^2−274 at the point (-4, -9, -3), we need to find the partial derivatives of the surface function with respect to x, y, and z. Then we can use these derivatives to find the equation of the tangent plane.

1. Calculate the partial derivatives: ∂x/(∂y) = 6y, ∂x/(∂z) = 6z.
2. Substitute the point (-4, -9, -3) into the partial derivatives: ∂x/(∂y) = 6(-9) = -54, ∂x/(∂z) = 6(-3) = -18.
3. Using the point and the partial derivatives, the equation of the tangent plane is: x - (-4) = -54(y - (-9)) - 18(z - (-3)), which simplifies to x + 54y - 18z = -275.