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To find the equation of the tangent plane to the surface \( xy^2z^3 = 2 \) at the point \( (2, 1, 1) \), we need to follow these steps:

### Step 1: Implicit Differentiation (Gradient Calculation)
The general form for the equation of a tangent plane to a surface \( F(x, y, z) = 0 \) at a point \( (x_0, y_0, z_0) \) is:
\[
F_x(x_0, y_0, z_0) (x - x_0) + F_y(x_0, y_0, z_0) (y - y_0) + F_z(x_0, y_0, z_0) (z - z_0) = 0
\]
Where:
- \( F_x \), \( F_y \), and \( F_z \) are the partial derivatives of \( F(x, y, z) \) with respect to \( x \), \( y \), and \( z \), respectively.

Let \( F(x, y, z) = xy^2z^3 - 2 \), so we first compute the partial derivatives of \( F(x, y, z) \).

### Step 2: Partial Derivatives
- \( F_x(x, y, z) = \frac{\partial}{\partial x}(xy^2z^3) = y^2z^3 \)
- \( F_y(x, y, z) = \frac{\partial}{\partial y}(xy^2z^3) = 2xyz^3 \)
- \( F_z(x, y, z) = \frac{\partial}{\partial z}(xy^2z^3) = 3xy^2z^2 \)

### Step 3: Evaluate at Point \( (2, 1, 1) \)
Now, evaluate each of these partial derivatives at the given point \( (x_0, y_0, z_0) = (2, 1, 1) \):
- \( F_x(2, 1, 1) = (1)^2(1)^3 = 1 \)
- \( F_y(2, 1, 1) = 2(2)(1)(1)^3 = 4 \)
- \( F_z(2, 1, 1) = 3(2)(1)^2(1)^2 = 6 \)

### Step 4: Equation of the Tangent Plane
Using the formula for the tangent plane:
\[
F_x(2, 1, 1)(x - 2) + F_y(2, 1, 1)(y - 1) + F_z(2, 1, 1)(z - 1) = 0
\]
Substitute the values:
\[
1(x - 2) + 4(y - 1) + 6(z - 1) = 0
\]
Simplifying:
\[
x - 2 + 4y - 4 + 6z - 6 = 0
\]
\[
x + 4y + 6z = 12
\]

### Final Answer:
The equation of the tangent plane is:
\[
x + 4y + 6z = 12
\]

Would you like a further breakdown or explanation of any part of this solution? Here are five questions to consider:

1. What is the general formula for the tangent plane to a surface?
2. How are partial derivatives used to find the tangent plane?
3. What would happen if the point of tangency were different?
4. How do partial derivatives give us the normal vector to the plane?
5. How could this method be applied to more complex surfaces?

**Tip:** The gradient vector (formed from the partial derivatives) is always perpendicular to the tangent plane at any point on the surface.

Find the equation of the tangent plane to the surface xy^2z^3 = 2 at the point (2, 1, 1).

Learn how to find the equation of the tangent plane to the surface xy^2z^3 = 2 at the point (2, 1, 1). This problem covers concepts of calculus such as partial derivatives, tangent planes, and gradients. Suitable for college-level calculus students.

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