To find the equation of the tangent plane to the surface $xy^2 + yz^2 - zx^2 = 1$ at the point $(1,1,1)$, we need to follow these steps:

Step 1: Compute the gradient of the surface function

The surface equation can be written as: $f(x, y, z) = xy^2 + yz^2 - zx^2 - 1$ The gradient of $f$ is given by: $\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right)$

Partial derivatives:

1. $\frac{\partial f}{\partial x} = y^2 - z x^2$
2. $\frac{\partial f}{\partial y} = 2xy + z^2$
3. $\frac{\partial f}{\partial z} = 2yz - x^2$

Step 2: Evaluate the gradient at the point $(1, 1, 1)$

Substituting $x = 1$, $y = 1$, and $z = 1$ into the partial derivatives:

1. $\frac{\partial f}{\partial x} = 1^2 - 1 \cdot 1^2 = 1 - 1 = 0$
2. $\frac{\partial f}{\partial y} = 2(1)(1) + 1^2 = 2 + 1 = 3$
3. $\frac{\partial f}{\partial z} = 2(1)(1) - 1^2 = 2 - 1 = 1$

Thus, the gradient at $(1, 1, 1)$ is: $\nabla f(1, 1, 1) = (0, 3, 1)$

Step 3: Equation of the tangent plane

The equation of the tangent plane to the surface at the point $(x_0, y_0, z_0)$ is given by: $f_x(x_0, y_0, z_0)(x - x_0) + f_y(x_0, y_0, z_0)(y - y_0) + f_z(x_0, y_0, z_0)(z - z_0) = 0$ Substituting $f_x(1, 1, 1) = 0$, $f_y(1, 1, 1) = 3$, and $f_z(1, 1, 1) = 1$ along with $(x_0, y_0, z_0) = (1, 1, 1)$, we get: $0(x - 1) + 3(y - 1) + 1(z - 1) = 0$ Simplifying: $3(y - 1) + (z - 1) = 0$ $3y - 3 + z - 1 = 0$ $3y + z - 4 = 0$ Thus, the equation of the tangent plane is: $3y + z = 4$

In the form shown in the image, this would correspond to: $-x + 3y + z = 4$

Thus, the correct values are:

• $a = -0$
• $b = 3$
• $c = 1$
• $d = 4$

Would you like more details on this? Here are some related questions:

1. How do you find the tangent plane of an implicitly defined surface in general?
2. Why is the gradient vector perpendicular to the tangent plane?
3. Could we use Lagrange multipliers to solve a similar problem?
4. How does this approach apply to level sets in multivariable calculus?
5. What is the geometric interpretation of the gradient at a point on a surface?

Tip: The gradient of a function points in the direction of the steepest ascent, which is always perpendicular to the level curve or surface at that point.