Gradients and Level Curves
In this section, we use the gradient and the chain rule to
investigate horizontal and vertical slices of a surface of the form z = g( x,y) . To begin with, if k is constant, then g(x,y) = k is called the level curve of g( x,y) of
level k and is the intersection of the horizontal plane z = k and the
surface z = g( x,y) :
In particular, g( x,y) = k is a curve in the xyplane.
EXAMPLE 1 Construct the level curves of g( x,y) = x^{2}+y^{2}+2 for k = 2,3,4,5, and 6.
Solution: To begin with, g( x,y) = k is of the form x^{2}+y^{2}+2 = k, which simplifies to
If k = 2, then x^{2}+y^{2} = 0 which is true only if ( x,y) = ( 0,0) . For k = 3,4,5,and 6, we have
These are circles centered at ( 0,0) with radii 1,Ö2,Ö3, and 2, respectively.
If g( x,y) is differentiable, then the level curve g( x,y) = k can be parametrized by a vectorvalued function of
the form r( t) =
á x( t) ,y(t)
ñ . If v denotes the velocity of r, then the chain
rule implies that

d
dt

g( x,y) = 
d
dt

k Þ Ñg·v = 0 

Thus, Ñg is perpendicular to the tangent line to g(x,y) = k at the point ( p,q) .
We say that Ñg( p,q) is normal to the curve g( x,y) = k at ( p,q) .
EXAMPLE 2 Show that the gradient is normal to the curve y = 12x^{2} at the point ( 1,1) .
Solution: To do so, we notice that 2x^{2}+y = 1. Thus, the curve is
of the form g( x,y) = 1 where g( x,y) = 2x^{2}+y.
The gradient of g is
Thus, at ( 1,1) , we have Ñg( 1,1) =
á 4,1
ñ . However, if y = 12x^{2}, then y' = 4x, and when x = 1, we have y'( 1) = 4. That is, a
run of 1 leads to a rise of 4, so that the slope corresponds to the
vector
m =
á run,rise
ñ =
á1,4
ñ 

Clearly, Ñg·m =
á 4,1
ñ ·
á 1,4
ñ = 0, thus implying that Ñg is
perpendicular to the tangent line to y = 12x^{2} at ( 1,1) .
Check Your Reading: Are there any level curves of g( x,y) = x^{2}+y^{2}+2 with a level lower than k = 2?