146k views
5 votes
Find an equation of the plane tangent to the surface z=ln(1+xy) at the given points.

User GwynBleidD
by
7.7k points

1 Answer

5 votes

Final answer:

To find the equation of the plane tangent to the surface z=ln(1+xy) at a given point, first find the partial derivatives of the function with respect to x and y. Then evaluate these partial derivatives at the given point. Finally, use the point-normal form of a plane equation to find the equation of the tangent plane.

Step-by-step explanation:

To find an equation of the plane tangent to the surface z=ln(1+xy) at a given point, we need to find the gradient (normal vector) of the surface at that point. The gradient of a function is given by the partial derivatives of the function with respect to each variable.

So, first we find the partial derivatives:

∂z/∂x = y/(1+xy)

∂z/∂y = x/(1+xy)

Next, we evaluate the partial derivatives at the given point:

∂z/∂x at (x,y) = y/(1+xy) at (x,y) = -1/(1+x)

∂z/∂y at (x,y) = x/(1+xy) at (x,y) = -x/(1+x)

Using these partial derivatives, the equation of the tangent plane can be found using the point-normal form of a plane equation:

(x-x₀)(∂z/∂x at (x₀,y₀)) + (y-y₀)(∂z/∂y at (x₀,y₀)) + (z-z₀) = 0

where (x₀,y₀,z₀) is the given point. Plugging in the values, we get the equation of the tangent plane.

User Michaeljt
by
7.4k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories