Finding an Equation of a Tangent Line

Find an equation of the tangent line to the curve of intersection of the surface \(z=f(x,y)=16-x^{2}-y^{2}\):

1. With the plane \(y=2\) at the point \((1,2,11)\)
2. With the plane \(x=1\) at the point \((1,2,11)\)

Solution  (a) The slope of the tangent line to the curve of intersection of \(z=16-x^{2}-y^{2}\) and the plane \(y=2\) at any point is \( f_{x}(x,2)=-2x\). At the point \((1,2,11)\), the slope is \(f_{x}( 1,2) =-2(1) =\) \(-2\). This tangent line lies on the plane \( y=2\). Symmetric equations of the tangent line are \[ z-11= \frac{x-1}{-1/2}\qquad y=2 \]

(b) The slope of the tangent line to the curve of intersection of \(z=16-x^{2}-y^{2}\) and the plane \(x=1\) at any point is \(f_{y}(1,y)=-2y\). At the point \((1,2,11)\), the slope is \(f_{y}( 1,2) =-2( 2) =-4\). This tangent line lies on the plane \(x=1\). Symmetric equations of the tangent line are \[ z-11= \frac{y-2}{-1/4}\qquad x=1 \]