Answer/Step-by-step explanation:
To find the equation of the tangent plane to the surface x = h(y,z) at the point (-2,-5,0) given the partial derivatives, we can proceed as follows:
1. **Equation of the tangent plane**: The equation of a plane can be expressed in the form Ax + By + Cz = D, where (A, B, C) is the normal vector to the plane.
2. **Normal vector calculation**: The normal vector to the tangent plane is given by the gradient of the function h(y,z) at the point (-2,-5,0). Since x = h(y,z), the normal vector will be parallel to the gradient of h(y,z).
3. **Calculate the gradient**: The gradient of h(y,z) is given by the partial derivatives: ∇h = (dh/dy, dh/dz). In this case, ∇h = (3, 5) at the point (-5,0).
4. **Equation of the tangent plane**: Substituting the normal vector components into the plane equation, we get 3x + 5y + Cz = D.
5. **Using the point (-2,-5,0)**: Plug in the coordinates of the point (-2,-5,0) into the equation: 3(-2) + 5(-5) + C(0) = D.
6. **Solve for D**: Simplify the equation to find the value of D.
7. **Final equation of the tangent plane**: Substitute the values of A, B, and D back into the equation to get the final equation of the tangent plane in the form Ax + By + Cz = D.
