Deriving Tangent Vectors From a Surface Parametrization

From a Parametrization to Tangent Directions

A parametrization describes points on a surface using two parameters. It also gives a way to describe how a point on the surface moves when either parameter changes. The resulting vectors are the basic tangent directions associated with the parametrization.

Suppose a surface patch is described by \(\mathbf r(u,v)\), where \((u,v)\) ranges over a parameter domain in the plane. The position vector \(\mathbf r(u,v)\) has three components, so changing \(u\) or \(v\) traces a curve on the surface. Differentiating those curves gives two vectors that measure the corresponding instantaneous changes in position.

Definition: The coordinate tangent vectors of a differentiable parametrization \(\mathbf r(u,v)\) are its partial derivatives \(\mathbf r_u(u,v)=\frac{\partial \mathbf r}{\partial u}(u,v)\) and \(\mathbf r_v(u,v)=\frac{\partial \mathbf r}{\partial v}(u,v)\). At a parameter pair \((u_0,v_0)\), they are evaluated at that pair to give \(\mathbf r_u(u_0,v_0)\) and \(\mathbf r_v(u_0,v_0)\).

These are vectors in three-dimensional space, based at the surface point \(\mathbf r(u_0,v_0)\). The vector \(\mathbf r_u\) describes the change in position when \(u\) varies and \(v\) is held fixed; \(\mathbf r_v\) describes the change when \(v\) varies and \(u\) is held fixed. Either vector may be zero at a particular parameter pair, even if the parametrization has nonzero tangent directions overall.

Deriving the Vectors from Coordinate Curves

Fix a parameter pair \((u_0,v_0)\). Holding \(v=v_0\) fixed gives the surface curve \(\mathbf c_u(s)=\mathbf r(u_0+s,v_0)\). Its velocity at \(s=0\), when the derivative exists, is \(\mathbf c_u'(0)=\mathbf r_u(u_0,v_0)\). Likewise, holding \(u=u_0\) fixed gives \(\mathbf c_v(s)=\mathbf r(u_0,v_0+s)\), with \(\mathbf c_v'(0)=\mathbf r_v(u_0,v_0)\).

This explains the geometric meaning of the partial derivatives: they are velocities of particular curves on the surface. Their values depend on the parametrization and its parameter speeds, not just on the surface as a set of points.

Worked Example: A Plane Patch

Consider \(\mathbf r(u,v)=(1+2u-v,\,-2+u+3v,\,4-u+2v)\). Find the coordinate tangent vectors at \((u,v)=(1,2)\), and check them using the coordinate curves.

Differentiate each component with respect to \(u\), keeping \(v\) fixed: \(\mathbf r_u(u,v)=(2,1,-1)\). Differentiating with respect to \(v\), keeping \(u\) fixed, gives \(\mathbf r_v(u,v)=(-1,3,2)\). These are constant vectors, so at \((1,2)\) they remain \(\mathbf r_u(1,2)=(2,1,-1)\) and \(\mathbf r_v(1,2)=(-1,3,2)\).

The surface point is \(\mathbf r(1,2)=(1+2-2,\,-2+1+6,\,4-1+4)=(1,5,7)\). For the first coordinate curve, \(\mathbf c_u(s)=\mathbf r(1+s,2)=(1+2s,5+s,7-s)\). Thus \(\mathbf c_u'(s)=(2,1,-1)\), confirming \(\mathbf c_u'(0)=\mathbf r_u(1,2)\).

For the second coordinate curve, \(\mathbf c_v(s)=\mathbf r(1,2+s)=(1-s,5+3s,7+2s)\). Thus \(\mathbf c_v'(s)=(-1,3,2)\), confirming \(\mathbf c_v'(0)=\mathbf r_v(1,2)\). Both vectors give tangent directions to this plane patch at \((1,5,7)\).

How the Two Vectors Describe Other Surface Motion

A curve on the parameter domain can vary both parameters at once. Write it as \((u(t),v(t))\), and map it to the surface as \(\boldsymbol\gamma(t)=\mathbf r(u(t),v(t))\). The multivariable chain rule gives the curve’s velocity. This result is important because it explains how the coordinate tangent vectors combine to describe general tangent directions.

Theorem: If \(\mathbf r\), \(u(t)\), and \(v(t)\) are differentiable, then \(\boldsymbol\gamma'(t)=\mathbf r_u(u(t),v(t))u'(t)+\mathbf r_v(u(t),v(t))v'(t)\). In particular, at \(t=t_0\), the velocity is a linear combination of the two coordinate tangent vectors evaluated at \((u(t_0),v(t_0))\).

Write the surface curve as a composition.
\(\boldsymbol\gamma(t)=\mathbf r(u(t),v(t))\), so both parameters depend on \(t\).

Differentiate each component.
For each component of \(\mathbf r\), apply the ordinary chain rule to its dependence on \(u(t)\) and \(v(t)\).

Collect the parameter rates.
The terms multiplying \(u'(t)\) form \(\mathbf r_u u'(t)\); those multiplying \(v'(t)\) form \(\mathbf r_v v'(t)\). Adding them gives the stated formula.

The derivation is the componentwise chain rule, and also shows why the formula holds. For example, if \(\mathbf r=(r_1,r_2,r_3)\), then the first component of the velocity is \(\frac{d}{dt}r_1(u(t),v(t))=(r_1)_u u'(t)+(r_1)_v v'(t)\). The same calculation applies to \(r_2\) and \(r_3\), producing precisely the two vector terms in the theorem.

At a regular patch point, the two partial derivative vectors are linearly independent. The rank condition for regularity says exactly that the derivative of \(\mathbf r\), viewed as a matrix with columns \(\mathbf r_u\) and \(\mathbf r_v\), has rank two. Consequently, their linear combinations form a two-dimensional plane of tangent directions at the surface point. This plane is called the tangent plane. The statement concerns velocities of differentiable curves mapped through that parameter point; a particular curve can still have zero velocity if both parameter rates are zero.

Worked Example: A Cylinder

Worked Example: Coordinate Directions on a Cylinder

Consider the cylinder parametrization \(\mathbf r(u,v)=(2\cos u,\,2\sin u,\,v)\). Find its coordinate tangent vectors at \((u,v)=(\frac{\pi}{2},3)\), then check their directions by differentiating the coordinate curves.

Differentiate with respect to \(u\): \(\mathbf r_u(u,v)=(-2\sin u,\,2\cos u,\,0)\). Differentiate with respect to \(v\): \(\mathbf r_v(u,v)=(0,0,1)\). Since \(\sin(\frac{\pi}{2})=1\) and \(\cos(\frac{\pi}{2})=0\), the values at the chosen parameters are \(\mathbf r_u(\frac{\pi}{2},3)=(-2,0,0)\) and \(\mathbf r_v(\frac{\pi}{2},3)=(0,0,1)\).

The surface point is \(\mathbf r(\frac{\pi}{2},3)=(0,2,3)\). Holding \(v=3\) fixed gives \(\mathbf c_u(s)=\mathbf r(\frac{\pi}{2}+s,3)=(-2\sin s,\,2\cos s,\,3)\). Differentiating gives \(\mathbf c_u'(s)=(-2\cos s,\,-2\sin s,\,0)\), so \(\mathbf c_u'(0)=(-2,0,0)\), as required.

Holding \(u=\frac{\pi}{2}\) fixed gives \(\mathbf c_v(s)=\mathbf r(\frac{\pi}{2},3+s)=(0,2,3+s)\). Its derivative is \(\mathbf c_v'(s)=(0,0,1)\), confirming the second vector. Thus changing \(u\) moves along the circular coordinate curve, while changing \(v\) moves vertically along the cylinder.

Worked Example: A Graph Patch

Worked Example: Tangent Vectors on a Graph

Write the graph \(z=x^2+2xy\) as \(\mathbf r(u,v)=(u,v,u^2+2uv)\). Find the coordinate tangent vectors at \((u,v)=(1,-1)\), and check them with the coordinate curves.

Differentiating with respect to \(u\) gives \(\mathbf r_u(u,v)=(1,0,2u+2v)\), since the derivative of \(u^2+2uv\) with respect to \(u\) is \(2u+2v\). Differentiating with respect to \(v\) gives \(\mathbf r_v(u,v)=(0,1,2u)\), since the derivative of \(u^2+2uv\) with respect to \(v\) is \(2u\).

At \((1,-1)\), \(\mathbf r_u(1,-1)=(1,0,2-2)=(1,0,0)\) and \(\mathbf r_v(1,-1)=(0,1,2)\). The corresponding surface point is \(\mathbf r(1,-1)=(1,-1,1+2(1)(-1))=(1,-1,-1)\).

For the \(u\)-curve, \(\mathbf c_u(s)=\mathbf r(1+s,-1)=(1+s,-1,(1+s)^2-2(1+s))\). Expanding the third component gives \((1+2s+s^2)-2-2s=-1+s^2\). Therefore \(\mathbf c_u(s)=(1+s,-1,-1+s^2)\) and \(\mathbf c_u'(s)=(1,0,2s)\), so \(\mathbf c_u'(0)=(1,0,0)\).

For the \(v\)-curve, \(\mathbf c_v(s)=\mathbf r(1,-1+s)=(1,-1+s,1+2(-1+s))=(1,-1+s,-1+2s)\). Hence \(\mathbf c_v'(s)=(0,1,2)\), which agrees with \(\mathbf r_v(1,-1)\). Both checks recover the partial derivatives at the specified point.

Using the Chain Rule to Find a Different Tangent Direction

The coordinate vectors are not the only possible velocities along a patch. They correspond to parameter rates \((u',v')=(1,0)\) and \((0,1)\). Other rates give other linear combinations, as the chain-rule formula requires.

Worked Example: A Curve with Both Parameters Changing

Use the graph patch from the preceding example and the parameter path \(u(t)=1+t\), \(v(t)=-1+2t\). Find the velocity of the mapped surface curve at \(t=0\), and check it by direct substitution.

At \(t=0\), the parameters are \((1,-1)\), where \(\mathbf r_u=(1,0,0)\) and \(\mathbf r_v=(0,1,2)\). The parameter rates are \(u'(0)=1\) and \(v'(0)=2\). Therefore the chain rule gives \(\boldsymbol\gamma'(0)=1(1,0,0)+2(0,1,2)=(1,2,4)\).

For a direct check, substitute the parameter functions into the parametrization: \(\boldsymbol\gamma(t)=(1+t,-1+2t,(1+t)^2+2(1+t)(-1+2t))\). The first two components are already explicit. For the third, \((1+t)^2=1+2t+t^2\), while \((1+t)(-1+2t)=-1+t+2t^2\). Thus the third component is \(1+2t+t^2-2+2t+4t^2=-1+4t+5t^2\). So \(\boldsymbol\gamma(t)=(1+t,-1+2t,-1+4t+5t^2)\), and differentiation gives \(\boldsymbol\gamma'(t)=(1,2,4+10t)\). At \(t=0\), this is \((1,2,4)\), agreeing with the chain-rule calculation.

What the Vectors Do—and Do Not—Tell You

At a regular point, the two coordinate tangent vectors provide a basis for the tangent plane: every velocity obtained from a differentiable parameter path through that point is their linear combination. A direction in the plane need not be one of the coordinate directions alone. For example, a path for which both \(u\) and \(v\) change has velocity \(\mathbf r_u u'+\mathbf r_v v'\).

The vectors also reflect how quickly the parameters move through the surface. If a parameter path is slowed down, its velocity is scaled; if its parameter rates are zero at an instant, its velocity is the zero vector at that instant. A zero velocity does not identify a nonzero direction of travel at that instant. This is why it is useful to distinguish a surface’s tangent plane from the velocity of one particular parametrized curve.

Common pitfall: Do not differentiate the entire parametrization with respect to one symbol while allowing the other parameter to vary as well. A partial derivative holds the other parameter fixed. For instance, for \(\mathbf r(u,v)=(u,v,u^2+2uv)\), the correct \(u\)-partial derivative is \((1,0,2u+2v)\). Replacing \(v\) by a path \(v(u)\) and differentiating the third component gives \(2u+2v+2u\,v'(u)\), which is the derivative along that path—not \(\mathbf r_u\). The path derivative must instead be handled by the chain rule using both partial derivatives.

To find coordinate tangent vectors reliably, differentiate each component with respect to one parameter at a time, keep the other parameter fixed, and then substitute the parameter values of interest. To find the velocity of a more general curve on the patch, first specify how both parameters depend on time and use the chain rule. These steps connect the algebra of partial derivatives to the geometry of motion along a surface.

Check Your Understanding

Try each question by writing the relevant derivatives or chain-rule expression before evaluating it.

For \(\mathbf r(u,v)=(u+v,\,u-v,\,3u+2v)\), find \(\mathbf r_u\) and \(\mathbf r_v\).
In the definition of \(\mathbf r_u(u_0,v_0)\), which parameter is held fixed? What surface curve gives this vector as its velocity?
If \(\boldsymbol\gamma(t)=\mathbf r(u(t),v(t))\), state the formula for \(\boldsymbol\gamma'(t)\).
For the cylinder parametrization \(\mathbf r(u,v)=(2\cos u,2\sin u,v)\), which parameter change produces vertical motion?
Can a differentiable curve lying on a regular surface have zero velocity at an instant? Explain using its parameter rates.