Question

Find the equation of the tangent plane to z = ex * y * y⁴ - 6?

Answer 1

The equation of the tangent plane to
`\(z = e^(x) \cdot y \cdot y^4 - 6\) at a point \((x_0, y_0)\) is \(z = (y_0^4 \cdot e^(x_0)) \cdot (x - x_0) + (5y_0^4 \cdot e^(x_0)) \cdot (y - y_0) - 6\).`

Explanation

To find the equation of the tangent plane to the given surface
`\(z = e^(x) \cdot y \cdot y^4 - 6\),` we first determine its partial derivatives with respect to \(x\) and \(y\). Then, using the point
`\((x_0, y_0)\)` where the tangent plane is to be found, substitute these values into the partial derivatives to compute the slope of the tangent plane at that point. The equation of a plane in point-normal form is
`\(z - z_0 = A(x - x_0) + B(y - y_0)\),`where \(A\) and \(B\) represent the partial derivatives evaluated at
`\((x_0, y_0)\).`Thus, after calculating the partial derivatives and evaluating them at the given point, the equation of the tangent plane can be expressed as
`\(z = (y_0^4 \cdot e^(x_0)) \cdot (x - x_0) + (5y_0^4 \cdot e^(x_0)) \cdot (y - y_0) - 6\).`This equation represents the tangent plane to the surface at the specified point.

The equation of the tangent plane to \
`(z = e^(x) \cdot y \cdot y^4 - 6\) at a point \((x_0, y_0)\) is \(z = (y_0^4 \cdot e^(x_0)) \cdot (x - x_0) + (5y_0^4 \cdot e^(x_0)) \cdot (y - y_0) - 6\).`