VC3.49

From Exampleproblems

Since (1,-1,2) lies on the given surface $ax^2-byz=(a+2)x\,$

Therefore, $a+2b=a+2\,$ so that b=1.

Thus,the given surfaces can be written as

$\phi(x,y,z)=ax^2-yz-(a+2)x=0,\psi(x,y,z)=4x^2y+z^3=4\,$

Therefore, $\nabla\phi=[i\frac{\partial}{\partial x}+j\frac{\partial}{\partial y}+k\frac{\partial}{\partial z}](ax^2-yz-(a+2)x)=(a-2)i-2j+k\,$ at (1,-1,2).

$\nabla\psi=[i\frac{\partial}{\partial x}+j\frac{\partial}{\partial y}+k\frac{\partial}{\partial z}](4x^2y+z^3)=8xyi+4x^2j+3z^2k=-8i+4j+12k\,$ at (1,-1,2).