1.3 Coordinates, curves and surfaces

Separate a geometric object from its description

A point can have Cartesian or polar coordinates. A circle can be an equation, a level set or the image of a parametrized curve. Each description reveals different information. Cartesian coordinates make linear algebra convenient; radial coordinates expose rotational symmetry; parameters describe motion; level sets organize a surface by height. The goal is to choose a description that makes the question easier while keeping track of the same geometric object.

A parametrization also contains information that an equation does not specify. It tells us the order in which points are visited and how rapidly a moving point passes through them. Two parametrizations may trace the same circle but describe different motions. Conversely, the same coordinate formula can describe different sets when its parameter range changes. Every curve calculation therefore begins with both a formula and its domain.

Curves and their derivatives

Definition

A parametrized curve

A curve in $\mathbb R^n$ is a continuous map $r:I\to\mathbb R^n$ from an interval, written $r(t)=(x_1(t),\ldots,x_n(t))$. For $I=[a,b]$, it is closed when $r(a)=r(b)$. It is simple when it has no repeated points, except that the two endpoints may coincide in the closed case.

The image $r(I)$ is the set traced by the curve. The map includes the parameter and its traversal information.

For instance, a circle traced once and the same circle traced twice have the same image. The second parametrization is not simple because points repeat before the final endpoint. A closed curve need not be simple, and a simple curve need not be closed. These distinctions become useful when discussing the inside and outside of a planar closed curve.

Theorem

Differentiate component by component

The derivative $r'(t_0)$ exists exactly when all component derivatives exist, and then

$$r'(t_0)=(x_1'(t_0),\ldots,x_n'(t_0)).$$

If $r'(t_0)\ne0$, the tangent line has parameter equation

$$\ell(s)=r(t_0)+s\,r'(t_0),\qquad s\in\mathbb R.$$

Proof

From secant directions to a tangent vector

The derivative is the vector limit of $[r(t_0+h)-r(t_0)]/h$. A vector tends to zero exactly when each coordinate tends to zero: each coordinate is bounded in absolute value by the norm, and the norm is at most the square root of the dimension times the largest absolute coordinate. Applying this to the difference quotient proves the component formula. Nonzero limiting velocity supplies a direction through the point $r(t_0)$, giving the tangent line.

If the parameter represents time, the first derivative is velocity, its norm is speed, and the second derivative is acceleration. Velocity is a vector, whereas speed is a nonnegative scalar. A zero derivative requires more analysis before assigning a tangent direction; it does not automatically mean that the image has no tangent. The definition above identifies the regular case in which the derivative itself provides a direction.

Ordinary product rules extend to vector operations. For a scalar function $f$ and vector curves $r,q$,

$$(fr)'=f'r+fr',\qquad \langle r,q\rangle'=\langle r',q\rangle+\langle r,q'\rangle.$$

In three dimensions, $(r\times q)'=r'\times q+r\times q'$. The cross-product order is preserved. Each identity follows by expanding the coordinates and applying the scalar product rule. For example, differentiating a constant squared length gives $\langle r,r'\rangle=0$: motion on a sphere has velocity perpendicular to its radius vector.

Speed and arc length

Theorem

Arc length of a continuously differentiable curve

If $r:[a,b]\to\mathbb R^n$ is continuously differentiable, its length, counting traversal, is

$$L=\int_a^b\|r'(t)\|\,dt.$$

A continuously differentiable monotone change of parameter with nonzero derivative preserves this length when its endpoints trace the same interval.

Proof

Why the speed is integrated

Divide the parameter interval into small pieces and join consecutive curve points by chords. On a short interval of length $\Delta t$, differentiability gives a displacement approximated by $r'(t)\Delta t$, whose length is $\|r'(t)\|\Delta t$. Uniform continuity of the derivative on the closed interval controls the total approximation error, and the chord sums converge to the integral. If $t=\phi(u)$, the chain rule gives $\|(r\circ\phi)'(u)\|=\|r'(\phi(u))\||\phi'(u)|$. Substitution proves invariance for increasing or decreasing changes of parameter.

The absolute value in the transformed speed is essential. Reversing traversal changes the velocity direction but cannot make length negative. Length is also different from endpoint distance: a closed curve has coincident endpoints yet can have positive length. If a curve repeats a loop, the integral counts each traversal. A parametrization that changes direction or repeats points is not a one-to-one monotone reparametrization of a single traversal, so length invariance does not erase repetitions.

For a regular curve, dividing velocity by speed gives the unit tangent in the direction of traversal. This normalization changes magnitude without changing direction. It should be performed only where speed is positive; the unnormalized derivative remains the more convenient quantity for computing the length integral.

Worked example

A helix: tangent, speed and one-turn length

Let $r(t)=(3\cos t,3\sin t,4t)$ for $0\le t\le2\pi$. Then

$$r'(t)=(-3\sin t,3\cos t,4),\qquad \|r'(t)\|=5.$$

At $t=0$, the point is $(3,0,0)$ and the tangent direction is $(0,3,4)$, so the tangent line is $(3,0,0)+s(0,3,4)$. The one-turn length is $\int_0^{2\pi}5\,dt=10\pi$. The endpoints are vertically separated by $8\pi$, which is smaller than the travelled length. The circle in the first two coordinates and the vertical rise both contribute to the speed.

Radial coordinate systems

Definition

Polar, cylindrical and spherical coordinates

We use nonnegative radial coordinates. In the plane,

$$x=r\cos\theta,\quad y=r\sin\theta,\quad r\ge0.$$

Cylindrical coordinates in space add the unchanged height $z$. For spherical coordinates $(\rho,\theta,\phi)$, the angle $\phi\in[0,\pi]$ is measured from the positive $z$-axis, and

$$x=\rho\sin\phi\cos\theta,\quad y=\rho\sin\phi\sin\theta,\quad z=\rho\cos\phi,\qquad \rho\ge0.$$

The planar radius is distance to the origin in the plane and distance to the vertical axis in cylindrical coordinates. The spherical radius is distance to the spatial origin. Thus the two radii are generally different. Their relation is a right-triangle relation, with the height as the remaining side. The azimuth angle is shared by both spatial systems because both use the direction of the projection onto the horizontal plane.

Angles are periodic. A convenient representative range is $0\le\theta\lt2\pi$, but a parametrized curve may allow the angle to increase through several turns. At planar radius zero the azimuth is not determined by the point. On the vertical axis in spherical coordinates the same ambiguity occurs, and at the origin neither angular coordinate is determined. These are ambiguities in the coordinate description, rather than multiple geometric points.

When converting back to an angle, determine the quadrant from the signs of the Cartesian coordinates. A principal inverse tangent alone gives only one of the possible directions. The ratio of the two coordinates cannot distinguish opposite rays, and it is unavailable when the horizontal coordinate is zero. An angle chosen in the correct quadrant, followed by substitution into the coordinate formulas, provides a direct check.

Worked example

Convert one spatial point in two ways

For $P=(-2,-2,2\sqrt2)$, the planar radius is $r=2\sqrt2$ and an azimuth is $5\pi/4$. Its cylindrical coordinates are $(2\sqrt2,5\pi/4,2\sqrt2)$. The spherical radius is $\rho=4$, and $\cos\phi=z/\rho=1/\sqrt2$, giving $(\rho,\theta,\phi)=(4,5\pi/4,\pi/4)$. Substitution recovers both negative horizontal coordinates and the positive height.

Coordinate equations describe geometric sets. A fixed positive polar radius gives a circle; a fixed positive cylindrical radius gives a cylinder; a fixed positive spherical radius gives a sphere. A fixed azimuth gives a ray in the plane and a vertical half-plane in space. A fixed spherical angle strictly between zero and a right angle describes the upper part of a cone. At a right angle it describes the horizontal plane. Recording the angle's range distinguishes a half-plane from a whole plane and one cone branch from both branches.

Worked example

A polar equation and its allowed angles

For $r=6\sin\theta$ with $r\ge0$, admissible angles satisfy $\sin\theta\ge0$. Multiplying by $r$ gives

$$x^2+y^2=6y,\qquad x^2+(y-3)^2=9.$$

This is the circle of radius $3$ centred at $(0,3)$, traced once for $0\le\theta\le\pi$ with its start and endpoint at the origin. If signed radii are introduced, the equivalence $(r,\theta)=(-r,\theta+\pi)$ must be used. That is a different coordinate convention; the Cartesian point is still given by the same sine and cosine formulas.

Graphs, level sets and traces

For a real-valued function of two variables, its graph consists of spatial points $(x,y,f(x,y))$. A level set is the set of inputs where the function equals a fixed constant. The graph therefore lives one dimension higher than the domain, while the level set stays in the input space. A horizontal section of the graph corresponds to a level set, lifted to the appropriate height. Fixing one input coordinate instead gives a vertical trace.

These descriptions answer different questions. A domain tells us where the formula is defined. A level set tells us which inputs produce the same output. A trace describes the graph's intersection with a specified plane. Before drawing a surface, it is useful to compute a few traces and inspect how they vary. A single attractive three-dimensional sketch can hide empty sections, disconnected pieces or exceptional heights that the equations reveal immediately.

Common quadric surfaces have the following forms, with positive constants $a,b,c$:

The ambient space matters. An equation involving only two coordinates can describe a cylinder in three-dimensional space because the missing coordinate is free. Completing squares locates translated centres or vertices. After translation, scaling reveals the principal lengths and the signs distinguish the types above. The constant on the other side must be checked as well: a sum of nonnegative squares cannot equal a negative number.

Worked example

Complete squares before identifying a surface

Consider $x^2+4y^2-2x+8y-2z+1=0$. Completing squares gives

$$(x-1)^2+4(y+1)^2=2(z+2).$$

It is an elliptic paraboloid with vertex $(1,-1,-2)$ opening towards increasing $z$. For $z\gt-2$, the horizontal trace is an ellipse. At $z=-2$ it is one point, and below that height it is empty. The traces $y=-1$ and $x=1$ are upward parabolas with different curvatures.

A coordinated problem-solving method

For a curve, first decide whether the question concerns its image or its traversal. To find a tangent, evaluate both the position and the velocity at the specified parameter. To find travelled length, take the norm of the velocity before integrating. Integrating the velocity itself gives displacement, which can be smaller than length and can vanish for a closed journey. This distinction can often be checked before doing any integration.

For a coordinate conversion, record the angle convention and the nonnegative radii, determine the quadrant, and verify by substitution. For a surface, identify its ambient space, isolate any domain restrictions, complete squares where appropriate, and examine several coordinate-plane and horizontal sections. Parameter formulas and implicit equations can then be compared through substitution: every parameter value must satisfy the equation, and every intended point must be reachable in the stated parameter range.

Common mistake

An equation alone does not fix a drawing

The equation $x^2+y^2=4$ is a circle in $\mathbb R^2$ and a cylinder in $\mathbb R^3$. A fixed azimuth describes only a half-plane when the radius is nonnegative. A tangent line needs the point as well as a nonzero direction. Always retain the ambient space and parameter ranges as part of the description.

Exercises and solutions

Checkpoint

1. Compute motion and a tangent.

For $r(t)=(t,t^2,t^3)$, find the velocity, acceleration, speed and tangent line at $t=1$.

Solution · 1. Differentiate the three coordinates

$r'(t)=(1,2t,3t^2)$ and $r''(t)=(0,2,6t)$. At $t=1$, the speed is $\sqrt{14}$ and the tangent line is $(1,1,1)+s(1,2,3)$.

Checkpoint

2. Choose the correct polar quadrant.

Give polar coordinates of $(-\sqrt3,-3)$ with $r\ge0$ and $0\le\theta\lt2\pi$.

Solution · 2. Both coordinates are negative

The radius is $\sqrt{3+9}=2\sqrt3$. The point lies in the third quadrant, with reference angle $\pi/3$, giving $(r,\theta)=(2\sqrt3,4\pi/3)$.

Checkpoint

3. Identify a quadric using sections.

Describe $x^2/4+y^2/9-z^2=1$ and its traces at $z=0$, $x=0$ and $y=0$.

Solution · 3. A one-sheet hyperboloid

At $z=0$ the trace is the ellipse $x^2/4+y^2/9=1$. The traces $y^2/9-z^2=1$ and $x^2/4-z^2=1$ in the other two planes are hyperbolas. Every fixed height gives the nonempty ellipse $x^2/4+y^2/9=1+z^2$, so the surface is a one-sheet hyperboloid.

Signed radii, inner loops and area bounds

For this subsection only, allow a signed radius $r$. The Cartesian point is still $(r\cos\theta,r\sin\theta)$. A negative radius means the point is on the opposite ray: its actual distance is $|r|$ and its geometric angle is $\theta+\pi$ modulo a full turn. Thus $(-1,0)$ in signed polar coordinates represents the Cartesian point $(-1,0)$, not a point one unit to the right. This temporary signed convention does not change the nonnegative-distance convention used in the earlier coordinate definitions.

Worked example

Find the inner loop before integrating

Consider the signed polar curve $r=1-2\cos\theta$. First solve for the pole:

$$1-2\cos\theta=0\quad\Longleftrightarrow\quad \theta=\pm\frac\pi3\pmod{2\pi}.$$

On $-\pi/3<\theta<\pi/3$, the radius is negative. At the endpoints it is zero, and at $\theta=0$ it is $-1$, giving the point $(-1,0)$. The actual geometric angles run from $2\pi/3$ to $4\pi/3$. The curve therefore leaves the origin into the upper-left half-plane, passes through $(-1,0)$, and returns through the lower-left half-plane. This is one complete inner loop, not a right-facing loop.

Why is it traced once? In the interior the nonzero radius has one fixed sign, and the actual angle increases strictly over an interval shorter than a full turn. Two distinct interior parameters cannot represent the same nonzero point. The endpoints both represent the origin, so the closed loop has no other repetition. On $\pi/3\le\theta\le5\pi/3$ the radius is nonnegative and traces the outer loop, passing through $(-3,0)$ at $\theta=\pi$. If nonnegative radii had been imposed on the original equation, the negative-radius branch would have been discarded and the inner loop would not be present in that interpretation.

Theorem

Area of one radial loop, with its traversal specified

For a simple loop swept once by increasing angle over an interval $[\alpha,\beta]$, with no overlapping radial sectors in its interior, its enclosed area is

$$A=\frac12\int_\alpha^\beta r(\theta)^2\,d\theta.$$

For a branch with negative radius throughout its interior, use $\rho=-r$ and $\varphi=\theta+\pi$. Then $\rho^2=r^2$ and $d\varphi=d\theta$, giving the same formula. Squaring the radius does not choose the correct loop or remove repeated traversal; the bounds must do that.

A thin sector of angle $\Delta\theta$ and distance $\rho$ has area approximately $\rho^2\Delta\theta/2$. Summing these sectors explains the factor one half. The non-overlap and one-traversal requirements explain why a blind integral over a full period may count more than the requested region.

Worked example

Evaluate and check the inner-loop area

Use exactly the inner-loop bounds found above:

$$A_{\mathrm{inner}}=\frac12\int_{-\pi/3}^{\pi/3}(1-2\cos\theta)^2\,d\theta.$$

The integrand is even, so the factor two from symmetry cancels the one half. Expanding and using $\cos^2\theta=(1+\cos2\theta)/2$ gives

$$A_{\mathrm{inner}}=\int_0^{\pi/3}(3-4\cos\theta+2\cos2\theta)\,d\theta =\left[3\theta-4\sin\theta+\sin2\theta\right]_0^{\pi/3} =\pi-\frac{3\sqrt3}{2}.$$

This is approximately $0.544$, a positive small area. The inner loop has distance at most one from the origin, so an answer larger than the unit disc's area would be impossible. Integrating over $0\le\theta\le2\pi$ gives $3\pi$: that is the sum of the two loop areas counted by traversal, not the inner-loop area and not the area of their union. The inner loop lies inside the outer loop, so its region is counted again in that sum.

Common mistake

A turn of a spiral is not automatically a closed loop

For the Archimedean spiral $r=k\theta$, where $k>0$ and $0\le\theta\le2\pi$, the endpoints are the origin and $(2\pi k,0)$. They lie on the same ray but are different points, so one full turn is not a closed curve. An area question must specify the remaining boundary, such as the radial segment joining the final point to the origin. Distinguish angular return from return to the same point.

Checkpoint

4. A scaled inner loop and a misleading full-period integral.

For the signed curve $r=2-4\cos\theta$, identify a once-traced inner-loop interval, its point at $\theta=0$, and its enclosed area. Explain why squaring the radius and integrating over a full period would not answer the question.

Solution · 4. Length scaling becomes area scaling squared

The zeros remain $\theta=\pm\pi/3$, so use $[-\pi/3,\pi/3]$. At zero the signed radius is $-2$, giving the Cartesian point $(-2,0)$. Every distance is twice that in the worked example, hence

$$A=\frac12\int_{-\pi/3}^{\pi/3}4(1-2\cos\theta)^2\,d\theta =4\pi-6\sqrt3.$$

A full-period integral would also include the outer loop, yielding $12\pi$ instead. The squared radius makes each sector contribution nonnegative; it does not label which loop the sector belongs to.

Checkpoint

5. State a spiral boundary before calculating area.

For $r=k\theta$, with $k>0$ and $0\le\theta\le2\pi$, find the area bounded by this one-turn spiral and the radial segment from its endpoint back to the origin. Why is this not an inner-loop calculation?

Solution · 5. The radial segment completes the boundary

Each interior angle meets the spiral once, and the region consists of distances from zero up to $k\theta$ on that ray. Therefore

$$A=\frac12\int_0^{2\pi}k^2\theta^2\,d\theta =\frac{k^2}{6}(2\pi)^3=\frac{4\pi^3k^2}{3}.$$

The segment closes the two different endpoint positions. There is no pole-to-pole loop of the spiral over this interval: its radius is zero only at the initial angle. Removing the specified segment would leave an open curve rather than the stated closed boundary.