Tutorials › Calculus III › Lifting a Planar Curve into Space

Representing Space Curves · Tutorial 117 of 999

Lifting a Planar Curve into Space

Learn to turn a planar parametrization into a space curve by adding a height component, and to place the resulting curve on a surface given as a graph.

Beginner 9 min read

What You'll Learn

  • Identify the planar projection and height component of a space curve
  • Construct a lift by adding a height function to a planar parametrization
  • Build a curve on a graph by composing the graph function with the planar coordinates
  • Verify a lifted curve by checking its projection and height
  • Avoid substituting the parameter where the graph requires two coordinate inputs

From a Planar Path to a Space Curve

A parametrized planar curve records two coordinates as a parameter varies. A space curve records three. One direct way to create a space curve is to keep the original two coordinates and specify a height above or below the plane. The planar path determines how the curve moves horizontally; the height function determines how it moves vertically.

This construction is useful when the path must follow a prescribed route in the \(xy\)-plane but also rise, fall, or lie on a surface. It separates those choices: first describe the planar motion, then assign a height to each point of that motion.

Definition: Let \(\mathbf r(t)=(x(t),y(t))\) be a planar parametrization for \(t\) in an interval \(I\), and let \(h:I\to\mathbb R\) be a height function. The lifted space curve is the parametrization \(\mathbf R(t)=(x(t),y(t),h(t))\), for \(t\in I\). Its planar projection is the original parametrization \(\mathbf r(t)\).

The height is attached to each parameter value, not merely to each location in the plane. If the planar path visits the same point more than once, the height function may assign different heights on different visits. Thus a lifted curve can retain information about the order in which the path is traced.

A particularly important choice of height comes from a surface that is the graph of a function. If \(g:D\to\mathbb R\), where \(D\) is a subset of the plane, its graph consists of points \((x,y,z)\) with \((x,y)\in D\) and \(z=g(x,y)\). For a planar curve to be lifted onto this graph, its planar coordinates must lie in \(D\).

Definition: If \(\mathbf r(t)=(x(t),y(t))\) lies in \(D\) for every \(t\in I\), the lift of \(\mathbf r\) onto the graph \(z=g(x,y)\) is \(\mathbf R(t)=(x(t),y(t),g(x(t),y(t)))\), for \(t\in I\).

The key step is the composition \(g(x(t),y(t))\): at each parameter value, evaluate the graph function at the current planar point. The graph then determines the height rather than leaving it as a separate choice.

Choosing a Height Function

For a general lift, any real-valued function \(h(t)\) on the parameter interval gives a height component. Choosing \(h(t)=0\) leaves the curve in the \(xy\)-plane. A positive height places a point above that plane, and a negative height places it below. A varying height can make the space curve rise and fall while its planar projection remains unchanged.

Different height functions can produce different space curves from the same planar parametrization. For instance, the choices \(h_1(t)=t\) and \(h_2(t)=t^2\) usually put the curve at different heights at the same parameter value. Both are valid lifts, but only a height given by \(g(x(t),y(t))\) places the curve on the specified graph \(z=g(x,y)\).

Worked Example: Adding a Prescribed Height

Start with the planar parametrization \(\mathbf r(t)=(t,2t-1)\), for \(-1\leq t\leq 2\). Choose the height \(h(t)=t^2+1\). The lifted curve is formed by placing this height after the two planar coordinates:

$$ \mathbf R(t)=(t,2t-1,t^2+1),\qquad -1\leq t\leq 2. $$

At \(t=-1\), the coordinates are \(x=-1\), \(y=2(-1)-1=-3\), and \(z=(-1)^2+1=2\), so \(\mathbf R(-1)=(-1,-3,2)\). At \(t=0\), they are \(x=0\), \(y=-1\), and \(z=1\), so \(\mathbf R(0)=(0,-1,1)\). At \(t=2\), they are \(x=2\), \(y=4-1=3\), and \(z=2^2+1=5\), so \(\mathbf R(2)=(2,3,5)\).

To check the construction, discard the third coordinate of \(\mathbf R(t)\). The remaining pair is \((t,2t-1)=\mathbf r(t)\), as required. The third coordinate is \(t^2+1=h(t)\). Thus the curve has the stated planar projection and height throughout the interval, not only at the sample values.

Lifting a Curve onto a Graph

When the surface is a graph \(z=g(x,y)\), there is a systematic procedure: substitute the planar coordinate functions into \(g\), then use the result as the height component. The original parameter interval is retained, provided the planar points stay in the domain of \(g\). This is a direct way to construct a space curve that lies on the graph.

Worked Example: A Parabola Lifted onto a Graph

Consider the planar parabola \(\mathbf r(t)=(t,t^2)\), for \(-1\leq t\leq 1\), and the graph \(z=g(x,y)=x^2+y\). The planar points have coordinates \(x(t)=t\) and \(y(t)=t^2\). Substitute both into the graph function:

$$ g(x(t),y(t))=g(t,t^2)=t^2+t^2=2t^2. $$

Therefore, the lift is

$$ \mathbf R(t)=(t,t^2,2t^2),\qquad -1\leq t\leq 1. $$

Check selected values. At \(t=-1\), the point is \((-1,1,2)\), and the graph gives \(x^2+y=(-1)^2+1=2\). At \(t=0\), the point is \((0,0,0)\), and \(x^2+y=0^2+0=0\). At \(t=1\), the point is \((1,1,2)\), and \(x^2+y=1^2+1=2\).

For every parameter value, the third coordinate of \(\mathbf R(t)\) is \(2t^2\), while the graph height evaluated at its first two coordinates is \(t^2+t^2=2t^2\). The curve therefore lies on the graph, and its projection is \((t,t^2)\), the given planar parabola.

Why the Graph Construction Works

The verification follows directly from the definition of the graph. A point belongs to the graph of \(g\) exactly when its first two coordinates are in \(D\) and its third coordinate equals \(g\) evaluated at those first two coordinates. The graph lift is built to satisfy that equality at every parameter value.

Result: Let \(\mathbf r(t)=(x(t),y(t))\) be defined for \(t\in I\), with \((x(t),y(t))\in D\) for every \(t\in I\), and let \(g:D\to\mathbb R\). Then \(\mathbf R(t)=(x(t),y(t),g(x(t),y(t)))\) lies on the graph \(z=g(x,y)\) for every \(t\in I\). Its planar projection is \(\mathbf r(t)\).
1
Check the domain.
For each \(t\in I\), the point \((x(t),y(t))\) must belong to \(D\), so \(g(x(t),y(t))\) is defined.
2
Form the height.
Set \(z(t)=g(x(t),y(t))\). This uses the two planar coordinates at the same parameter value.
3
Verify the point.
The resulting point has third coordinate \(g(x(t),y(t))\), exactly the height required by the graph equation.
4
Check the projection.
Removing the third coordinate leaves \((x(t),y(t))=\mathbf r(t)\), so the prescribed planar path is retained.

These steps prove both claims: every point of the parametrized curve lies on the graph, and the curve projects to the original planar parametrization. The domain condition matters because a formula for \(g\) may not be defined at every point of the plane.

A Second Graph Construction

The planar curve need not be a parabola. The method applies to any parametrized planar path whose points remain in the domain of the graph function. The next example uses a line segment to make the endpoint checks explicit.

Worked Example: Lifting a Line Segment onto a Graph

Let the planar path be \(\mathbf r(t)=(1+2t,2-t)\), for \(0\leq t\leq 1\), and use the graph \(z=x^2-y^2\). Its planar coordinates are \(x(t)=1+2t\) and \(y(t)=2-t\). Substitution gives

$$ \begin{aligned} z(t)&=(1+2t)^2-(2-t)^2\\ &=(1+4t+4t^2)-(4-4t+t^2)\\ &=-3+8t+3t^2. \end{aligned} $$

Thus a parametrization of the lifted curve is

$$ \mathbf R(t)=(1+2t,2-t,-3+8t+3t^2),\qquad 0\leq t\leq 1. $$

At \(t=0\), the point is \((1,2,-3)\), and the graph gives \(1^2-2^2=1-4=-3\). At \(t=1\), the point is \((3,1,8)\), and the graph gives \(3^2-1^2=9-1=8\). At \(t=\tfrac12\), the planar coordinates are \(x=1+2(\tfrac12)=2\) and \(y=2-\tfrac12=\tfrac32\). The height from the parametrization is \(-3+8(\tfrac12)+3(\tfrac12)^2=-3+4+\tfrac34=\tfrac74\). The graph gives \(2^2-(\tfrac32)^2=4-\tfrac94=\tfrac74\).

The third coordinate agrees with \(x^2-y^2\) at the sample values, and the substitution that produced it shows the agreement for every \(t\) in the interval. The first two coordinates are unchanged, so the projection is the original segment.

A Common Substitution Mistake

When evaluating a two-variable graph function, both inputs must be the planar coordinates. The parameter is not automatically either input. For the curve \(\mathbf r(t)=(t,t^2)\) and the graph \(z=x^2+y\), an incorrect substitution would be to use \(g(t,t)\), producing \(t^2+t\). But the planar coordinates are \(x=t\) and \(y=t^2\), so the required substitution is \(g(t,t^2)\).

At \(t=\tfrac12\), the incorrect height is \((\tfrac12)^2+\tfrac12=\tfrac34\). The correct height is \((\tfrac12)^2+(\tfrac12)^2=\tfrac12\). Indeed, the planar point is \((\tfrac12,\tfrac14)\), and the graph height there is \(x^2+y=\tfrac14+\tfrac14=\tfrac12\). The incorrect point would not lie on the graph at that planar location.

Common pitfall: Do not replace \(g(x(t),y(t))\) with \(g(t,t)\) unless the planar coordinates really are \(x(t)=t\) and \(y(t)=t\). Substitute the actual coordinate functions separately.

What the Lift Preserves and What It Changes

A lift preserves the parameter interval and the planar coordinates at each parameter value. Consequently, it preserves the planar projection as a parametrized path, including how that path is traced as \(t\) increases. It changes the curve’s third coordinate according to the chosen height.

A lift does not necessarily preserve distances, shapes, or lengths from the plane; adding height can change the spatial path even though the projection stays fixed. Also, the planar projection alone does not determine a unique space curve: an independent height function must be chosen, or a graph must be specified to determine that height. These observations are useful when interpreting a parametrization: the first two components describe the projected motion, and the third component says where the curve sits vertically.

To construct and check a lift, identify the planar coordinate functions and the parameter interval first. Then either choose a height function \(h(t)\), or, for a specified graph, calculate \(g(x(t),y(t))\). Finally, verify that the resulting third coordinate matches the intended height and that dropping it recovers the given planar curve.

Check Your Understanding

Use the definitions and substitution method above to answer the following questions.

  1. If \(\mathbf r(t)=(2t,1-t)\) and \(h(t)=t+3\), write the lifted parametrization.
  2. For \(\mathbf r(t)=(t,1+t)\) on \(0\leq t\leq 2\) and \(z=x^2+y\), find the height as a function of \(t\).
  3. State the condition on the planar coordinates that must hold for a lift onto \(z=g(x,y)\) to be defined throughout an interval \(I\).
  4. For \(\mathbf r(t)=(t,t^2)\) and \(z=x^2+y\), identify the incorrect substitution \(g(t,t)\) and give the correct height.
  5. What does taking the planar projection of a lifted curve recover?