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Find the equation of the tangent plane of f(x,y)=xeˣʸ at the point (2,0,2)

User Godwin
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Final answer:

The equation of the tangent plane to the surface given by f(x,y) = x e^(xy) at the point (2,0,2) is z = x after computing partial derivatives and evaluating at the point of tangency.

Step-by-step explanation:

To find the equation of the tangent plane to the surface defined by f(x,y) = xe^{xy} at a given point (2,0,2), we need to first compute the partial derivatives of f with respect to x and y.

First, compute the partial derivative with respect to x:
f_x(x,y) = e^{xy} + xy \cdot e^{xy} \cdot y.

Then, compute the partial derivative with respect to y:
f_y(x,y) = x^{2} \cdot e^{xy} \cdot y.

Evaluate both derivatives at the point (2,0): f_x(2,0) = 1 and f_y(2,0) = 0.

The equation of the tangent plane is z = f(2,0) + f_x(2,0)(x - 2) + f_y(2,0)(y - 0). Substitute the known values to get z = 2 + 1 \cdot (x - 2) + 0 \cdot (y - 0), which simplifies to z = x.

So, the equation of the tangent plane at the point (2,0,2) is z = x.

User Rinda
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