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A contour plot displays several level curves of a real-valued function of two variables, usually with labels indicating their values. It represents a surface's height on a flat page. Closely spaced level curves indicate a large change per unit distance only when the gaps between their labeled values are comparable.

A partial derivative measures the local response of a function to one input while its other inputs stay fixed. For a function $f$ on an open subset of ${\mathbb{R}}^2$, its partial derivatives at $(a,b)$ are the limitswhen the respective limits exist. Each is an ordinary derivative of a one-variable slice: $x\mapsto f(x,b)$ or $y\mapsto f(a,y)$. This is why the other variable is treated as a constant during partial differentiation.^[ref]

For an original example, take the polynomial functionHolding $y$ fixed gives $f_x=2x+y$, while holding $x$ fixed gives $f_y=x+4y$. At $(1,1)$, the height is $f=4$ and the two partial derivatives are $f_x=3$ and $f_y=5$. The values answer different experiments on the same mathematical model: change $x$ alone, or change $y$ alone.

The two exact slices through that point areDividing either height change by $t\ne 0$ gives $3+t$ or $5+2t$; taking the limit $t\to 0$ recovers the two slopes. At $t=0.1$, the finite height changes are $0.31$ and $0.52$, although their linear approximations are $0.3$ and $0.5$. A derivative measures the limiting response, rather than an exact finite change.

For a function of $n$ variables, ${\partial }_{j}f(a)$ varies only its $j$th coordinate: it is the same difference quotient along $e_j$, the $j$th standard basis vector. This gives a practical sensitivity analysis: one can compare how a mathematical model responds to changes in its individual inputs, provided the chosen system of units and the quantities held fixed are stated. Comparing the numerical sizes of two partial derivatives in different units does not by itself identify a physically preferable input.

Existence of the partial derivatives at one point is weaker than differentiability. The example under partial derivatives do not imply continuity shows how both coordinate tests can miss a bad diagonal approach. When the first partial derivatives are continuous in a neighborhood, they do give differentiability there, allowing the total differential to combine all small input changes.

The gradient equations are $f_x=2x(1-y)=0$ and $f_y=4y-x^2=0$. The critical points are $(0,0)$ and $(\pm 2,1)$. Their Hessian matrix isAt the origin $H=\mathrm{diag}(2,4)$ is positive definite. At either other point its determinant is $-16$, so its eigenvalues have opposite signs. ThusThe minimum value is zero and both saddle values are two. The origin is not a global minimum: at any fixed $y>1$, the term $(1-y)x^2$ makes $f$ arbitrarily negative as $|x|$ increases.

For the contour plot, the local contours around the origin are approximately $x^2+2y^2=c$. The exact saddle contour factors asso it is the union of $y=1$ and $y=x^2/2-1$, crossing at the two saddles. Other levels obey $x^2=(c-2y^2)/(1-y)$ where $y\ne 1$; the right panel of the original figure in part 3(a) shows how these branches join around the minimum and pass through the saddle contour.

Near a nondegenerate stationary point, the quadratic part of the Taylor expansion determines the local contour plot. For a strict local maximum, the model $f=f_0-X^2-Y^2$ gives nested closed level sets, with values decreasing as one moves outward. For a strict local minimum, the model $f=f_0+X^2+Y^2$ gives the same nested shape with values increasing outward. Unequal nonzero quadratic coefficients stretch the circles into ellipses after a rotation to the Hessian matrix's eigenvectors.

For a nondegenerate saddle point of a scalar function, the model $f=f_0+X^2-Y^2$ gives hyperbolas. Levels above $f_0$ open along the positive-curvature direction; levels below $f_0$ open along the negative-curvature direction. The critical level set is the two crossing lines $Y=\pm X$. Degenerate stationary points can have different local geometry, so these are the standard nondegenerate sketches.

Set both first partial derivatives equal to zero:Thus $x=0,\pm 1/\sqrt{2}$ and $y=0,\pm \sqrt{2}$, independently, giving exactly nine stationary points. The Hessian matrix is diagonal everywhere:Its diagonal entries are its eigenvalues, so the second-derivative test immediately classifies the points:

In particular,All nine stationary points are nondegenerate, so no higher-order test is needed. Neither type of local extremum is global: $f(x,0)=x^4-x^2\to +\infty$ as $|x|\to \infty$, whereas $f(0,y)=-y^4+4y^2\to -\infty$ as $|y|\to \infty$.

To organize the contour plot, complete the squares:The level set through all four outer saddle points of a scalar function factors exactly:It is therefore the union of the circle $x^2+y^2=5/2$ and the hyperbola $y^2-x^2=3/2$, which cross at those four saddle points of a scalar function. Near the origin, $f=-x^2+4y^2+O((x^2+y^2{)}^2)$, so the zero contours have tangents $y=\pm x/2$. These two exact structural features, together with the nested contours around the four local extrema, determine the requested sketch. They are an example of a separable quartic landscape with nine stationary points.

Within the requested first-quadrant square, the stationary points visible in the contour plot are the corner minimum $(0,0)$ and the saddle $(1,1)$. Near the origin, $g\simeq 4(x^2+y^2)$, so the low positive contours look like quarter-circles. Along either coordinate axis, $g=4t^2+t^4$ rises monotonically. Along the diagonal, $g(t,t)=8t^2-4t^4$ rises to $4$ at $t=1$, returns to zero at $t=\sqrt{2}$, then becomes negative.

The saddle contour $g=4$ crosses at $(1,1)$ and separates the central positive contours from the outer diagonal region. Negative contours occur toward $(2,2)$, where $g=-32$, while the axis corners $(2,0)$ and $(0,2)$ have $g=32$. These checks determine both the contour ordering and the change of connectivity at the saddle; the figure shows these level sets.

The completed contour plot is shown in the right-hand panel. Its geometry follows from the symmetries $f(-x,-y)=f(x,y)$ and $f(y,x)=f(x,y)$, together with $f\to 0$ as $x^2+y^2\to \infty$. For orientation, the two local minima have value $-2e^{-3/2}$, the two local maxima have value $2e^{-1/2}$, and the saddle point has value one.

Negative level sets between $-2e^{-3/2}$ and zero form separate closed curves around the two local minima in the first and third quadrants. Positive level sets below one form a single closed curve enclosing the origin and both local maxima. At level one, a critical level set crosses itself at the origin and surrounds the two local maxima in two lobes. Between one and $2e^{-1/2}$, the level sets split into separate closed curves around those local maxima. The zero level set remains the two unbounded branches $xy=1/4$.

Near the origin, the Taylor expansion is $f=1-x^2-4xy-y^2+O((x^2+y^2{)}^2)$. Consequently the crossing directions at the saddle height satisfy $y=(-2\pm \sqrt{3})x$, giving the local shape of the critical level set.

The most useful guide to the contour plot is the common saddle value. Direct polynomial factorization givesThus the level set $f=1/6$ is exactly the three complete linesTheir intersections are the three saddle points, and they enclose the origin in an equilateral triangle. Near the local minimum, the level sets are nearly circles because the leading term is $(x^2+y^2)/2$. For $0<c<1/6$, a closed contour surrounds the origin, while three separate open branches lie outside the triangular barrier. At the saddle level these components meet at the saddle points; above it the central closed contour has opened into the three outer channels.

The identity in polar coordinatesshows the rotation symmetry through $2\pi /3$. It identifies the function as the Hénon-Heiles potential and helps organize all three channels in the original sketch below. The dashed straight lines are the saddle level; labels on the other contours give $f$.

At each regular point, the gradientis normal to the level set and points towards increasing $f$. The arrows in the contour plot above show this direction; their lengths have been normalized to keep the outer arrows legible. Near the origin they point outwards, as expected for a local minimum. Near $(0,1)$, write $\eta =y-1$; the linear approximation is $\nabla f\simeq (3x,-\eta )$. Thus the arrows point away along the horizontal direction and towards the point along the vertical direction. The other two saddle points have the same behaviour after rotations through $2\pi /3$. All four stationary points have zero gradient, so no direction arrow is assigned at the points themselves.

The contour plot is symmetric under either $x\mapsto -x$ or $y\mapsto -y$, because $f$ depends only on their squares. Denote the three critical heights by $m=\frac{7}{2}\ln 3$, $h=\frac{1}{4}+\frac{7}{2}\ln (7/2)$ and $M=1+\frac{7}{2}\ln 3$, where $m<h<M$.

For $m<c<h$, the level set $f=c$ consists of separate closed curves around the two minima. At $c=h$ the branches meet at the two saddle points. For $h<c<M$, there is an inner closed curve surrounding the local maximum and an outer closed curve surrounding all five stationary points. At $c=M$ the inner curve shrinks to the origin; for $c>M$ only the outer closed curve remains. At each saddle point the critical level set crosses in the two directions determined byThe figure plots representative level sets, including the critical saddle height; its second panel adds the gradient directions requested below.

At a differentiable interior stationary point, all first-order changes of the scalar field vanish. In particular, the directional derivative along each coordinate axis vanishes, so $f_x=f_y=0$ and hence $\nabla f=0$. Equivalently, the first-order differential $df=\nabla f\cdot d\mathbf{r}$ vanishes for every displacement. A differentiable interior maximum or minimum must satisfy this condition; a stationary point need not be an extremum.

The standard nondegenerate local contour plot models are $-x^2-y^2$ for a maximum, $x^2+y^2$ for a minimum, and $x^2-y^2$ for a saddle point of a scalar function. Near a maximum, nested contours have lower values farther from the centre; near a minimum they have higher values. Near a saddle, hyperbolas occur on opposite sides of the critical level, whose contour has two intersecting tangent directions. More general nondegenerate quadratic forms deform these circles into ellipses; degenerate stationary points can have more complicated contours.

In the translated coordinates $u=x-1$, $v=y+1$, the zero level set is the pair of lines $u=0$ and $v=0$. The function is positive in the quadrants $uv>0$ and negative in those with $uv<0$, and tends to zero as $u^2+v^2\to \infty$. Thus its nonzero contours form closed curves around the four off-centre stationary points.

For an explicit description of the contour plot, use polar coordinates $u=r\cos \theta$, $v=r\sin \theta$:At fixed direction within a quadrant, the magnitude increases until $r=\sqrt{2}$ and then decreases towards zero. Also $|u|e^{-u^2/2}\le e^{-1/2}$, with equality exactly when $|u|=1$, and similarly for $v$. Consequently $|f|\le e^{-1}$; each positive level below $e^{-1}$ encloses a positive peak in each positive quadrant, with the analogous negative contours in the negative quadrants.

To classify the stationary points, the Hessian matrix isAt $u=v=0$ it is $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, with eigenvalues $1$ and $-1$, so $(1,-1)$ is a saddle point. At $u,v\in \{-1,1\}$ it is $-2fI$. Therefore $(2,0)$ and $(0,-2)$ are strict local maxima, and $(2,-2)$ and $(0,0)$ are strict local minima. These are also global maxima and minima by the bound just proved.

The gradient of this polynomial vanishes exactly when $x=0,\pm \sqrt{a/2}$ and $y=0,\pm \sqrt{b/2}$, giving nine stationary points. Its diagonal Hessian matrix has entries $12x^2-2a$ and $2b-12y^2$. The second-derivative test therefore gives two strict local maxima at $(0,\pm \sqrt{b/2})$, with value $b^2/4$, and two strict local minima at $(\pm \sqrt{a/2},0)$, with value $-a^2/4$. The origin and the four points $(\pm \sqrt{a/2},\pm \sqrt{b/2})$ are saddle points of a scalar function, with values zero and $(b^2-a^2)/4$, respectively. Every stationary point is nondegenerate.

Completing the squares organizes the contour plot:The level set through the four outer saddle points of a scalar function factors asIt is a circle together with a hyperbola when $a\ne b$; for $a=b$, the latter factor's complementary zero set is the pair of lines $y=\pm x$. Neither the local maxima nor the local minima are global, since the restrictions to the two coordinate axes are unbounded in opposite directions. The case $a=1$, $b=4$ gives four outer saddle values $15/4$ and an exact circle-hyperbola contour crossing at all four outer saddle points of a scalar function.