F(x,y)=e^-x sin y at P0(0,pie/2,1). I know the standard form is A(x-x0)+B(y-y0)+C(z-z0) =0.
Ax+By+Cz+D=0. How would you get to use that?
Determine the standard-form equations for the tangent plane to the given surface at the prescribed P0.?
∂F/∂x = -e^-x sin y = -(1)(1) = -1 at P0
∂F/∂y = e^-x cos y = 1(0) = 0 at P0
So we have the tangent plane as
z - z0 = ∂F/∂x|P0 (x - x0) + ∂F/∂y|P0 (y - y0)
i.e. z - 1 = -1(x - 0) + 0 (y - π/2)
%26lt;=%26gt; x + (z-1) = 0, or equivalently x + z = 1.
verbena
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