Questions: Find the equation of the tangent plane to the surface z+5=xe^y cos z at the point (5,0,0). x+y-z=5 x+y-5 z=5 5 x+y-z=5 x+5 y-z=5 x+5 y+z=5

Solution Steps

To find the equation of the tangent plane to the surface at a given point, we need to:

1. Compute the partial derivatives of the function with respect to \(x\), \(y\), and \(z\).
2. Evaluate these partial derivatives at the given point to find the normal vector to the tangent plane.
3. Use the point-normal form of the plane equation to write the equation of the tangent plane.

The surface is defined by the equation: \[ z + 5 = x e^{y} \cos z \] We can rearrange this to express it as a function: \[ f(x, y, z) = x e^{y} \cos z - z - 5 = 0 \]

We compute the partial derivatives of \(f\) with respect to \(x\), \(y\), and \(z\): \[ \frac{\partial f}{\partial x} = e^{y} \cos z \] \[ \frac{\partial f}{\partial y} = x e^{y} \cos z \] \[ \frac{\partial f}{\partial z} = -x e^{y} \sin z - 1 \]

We evaluate the partial derivatives at the point \((5, 0, 0)\): \[ \frac{\partial f}{\partial x}(5, 0, 0) = 1 \] \[ \frac{\partial f}{\partial y}(5, 0, 0) = 5 \] \[ \frac{\partial f}{\partial z}(5, 0, 0) = -1 \] Thus, the normal vector to the tangent plane at this point is: \[ \mathbf{n} = (1, 5, -1) \]

Using the point-normal form of the plane equation, we have: \[ 1(x - 5) + 5(y - 0) - 1(z - 0) = 0 \] This simplifies to: \[ x + 5y - z - 5 = 0 \] or equivalently: \[ x + 5y - z = 5 \]

Final Answer