Questions: Find the equation for the tangent plane to the surface z=e^(2x^2+3y^2) at the point (0,0,1). z=0 z=1 z=2 z=-1

Solution Steps

To find the equation of the tangent plane to the surface \( z = e^{2x^2 + 3y^2} \) at the point \((0,0,1)\), we need to:

1. Compute the partial derivatives of \( z \) with respect to \( x \) and \( y \).
2. Evaluate these partial derivatives at the point \((0,0,1)\).
3. Use the point-slope form of the plane equation: \( z - z_0 = f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0) \).

We are given the surface defined by the equation \( z = e^{2x^2 + 3y^2} \) and the point at which we need to find the tangent plane, which is \( (0, 0, 1) \).

We compute the partial derivatives of \( z \) with respect to \( x \) and \( y \): \[ \frac{\partial z}{\partial x} = 4x e^{2x^2 + 3y^2} \] \[ \frac{\partial z}{\partial y} = 6y e^{2x^2 + 3y^2} \]

Next, we evaluate these partial derivatives at the point \( (0, 0) \): \[ \frac{\partial z}{\partial x} \bigg|_{(0,0)} = 4(0)e^{2(0)^2 + 3(0)^2} = 0 \] \[ \frac{\partial z}{\partial y} \bigg|_{(0,0)} = 6(0)e^{2(0)^2 + 3(0)^2} = 0 \]

Using the point-slope form of the plane equation, we have: \[ z - z_0 = \frac{\partial z}{\partial x} \bigg|_{(0,0)} (x - x_0) + \frac{\partial z}{\partial y} \bigg|_{(0,0)} (y - y_0) \] Substituting the values: \[ z - 1 = 0 \cdot (x - 0) + 0 \cdot (y - 0) \] This simplifies to: \[ z = 1 \]

Final Answer