Answer:
-4.18i+4.08 j
Explanation:
We are given that
[tex]f(x,y)=(e^x-x)cosy[/tex]
We have to find the a vector which is perpendicular to the level curve of f through the point in the direction in which f decreases most rapidly.
Grad f(x,y),[tex]\nabla f(x,y)=(e^x-1)cosyi-(e^x-x)sinyj[/tex]
Grad f(x,y) at point (2,4) is given by
[tex]\nabla f(2,4)=(e^2-1)cos4i-(e^2-2)sin4 j[/tex]
[tex]\nabla f(2,4)=−4.18 i+4.08 j[/tex]
Hence, the vector which is perpendicular to the level curve of f through the point in the direction in which f decreases most rapidly=-4.18i+4.08 j
