For intuition: Imagine that the function f(x,y) = e^x ln(y) - xyf(x,y)=exln(y)−xy describes the height of some terrain, where xx and yy are coordinates in the plane and ln(y)ln(y) is assumed to be the natural logarithm. Then all (x,y)(x,y) such that f(x,y)=af(x,y)=a (the height) equals some constant aa are called level curves . In our case the constant height aa is zero, since f(x,y)=0f(x,y)=0.
You might be familiar with topographic maps, in which the closed lines indicate lines of equal height.
Now the gradient grad f (x,y)= ( (partial f) /(partial x), (partial f) /(partial x) ) = (e^x ln(y) - y, e^x/y - x)∇f(x,y)=(∂f∂x,∂f∂x)=(exln(y)−y,exy−x) gives us the direction at a point (x,y)(x,y) in which f(x,y)f(x,y) (the height) changes the fastest. This is either straight up or straight down the hill, as long as our terrain is smooth (differentiable), and we are not on a top, in a bottom or on a plateau (an extremum point). This is in fact the normal direction to a curve of constant height, such that at (x,y)=(2,e^2)(x,y)=(2,e2):
grad f (2,e^2) = (e^2 ln(e^2) - e^2, e^2/e^2 - 2)=(e^2,-1)∇f(2,e2)=(e2ln(e2)−e2,e2e2−2)=(e2,−1).
Therefore, the normal line in that direction going through (2,e^2)(2,e2) can be described as
(x,y) = (2,e^2) + s(e^2,-1)(x,y)=(2,e2)+s(e2,−1),
where s in mathbbR is a real parameter. You can eliminate s to express y as a function of x if you prefer, to find
y=(x-2-e^4)/e^2.
The directional derivative in the tangent direction must be 0 (meaning that height does not change), so a tangent vector (u,v) must satisfy
grad f(2, e^2)cdot (u,v)=0
(e^2,-1) cdot (u,v) = 0
e^2u - v=0
v=e^2u,
where cdot means the dot product. So (u,v) = (1,e^2) is one valid choice. Therefore, the tangent line going through (2,e^2) can be described as
(x,y) = (2,e^2) + t(1,e^2), t in mathbbR.
Solving for y gives that
y = e^2x -e^2.
You should finally check that (2,e^2) lies on the curve f(x,y), on the tangent line, and on the normal line.
