Circles and Tangent Arrows
A planar vector field assigns a vector to each point in its domain. In a radial field, arrows point along lines from or toward a center. In a tangential vector field, arrows at points away from the center instead point along the circles centered there.
To construct such a field, begin with the displacement from the center to a point. A tangent vector must be perpendicular to that displacement. Rotating the displacement by a quarter-turn gives a convenient tangent direction, and multiplying it by a scalar can change the arrow's size or reverse its direction.
The condition applies at points other than the center: there is no unique tangent direction to a circle at its center. A formula may still assign a vector there, commonly the zero vector.
Deriving the Component Formula
Write \(X=x-a\) and \(Y=y-b\), so the displacement from the center is \((X,Y)\). A quarter-turn counterclockwise sends this vector to \((-Y,X)\). Its dot product with the displacement is \(X(-Y)+YX=0\), so it is perpendicular to the radius and tangent to the circle.
Every tangent vector at a point other than the center is a scalar multiple of this one. To see why, let \((u,v)\) be tangent, so \(Xu+Yv=0\), with \(X^2+Y^2>0\). Set \(\lambda=(-Yu+Xv)/(X^2+Y^2)\). Then \(\lambda(-Y)=u\) and \(\lambda X=v\): the first equality follows from \(-Y(-Yu+Xv)=Y^2u-XYv=(X^2+Y^2)u\), using \(Yv=-Xu\); the second follows from \(X(-Yu+Xv)=-XYu+X^2v=(X^2+Y^2)v\), using \(Xu=-Yv\). Thus \((u,v)=\lambda(-Y,X)\).
The proof is also a direct check: the dot product of the displacement with the formula is \((x-a)[-\lambda(y-b)]+(y-b)[\lambda(x-a)]=0\). This establishes tangency wherever the field is defined away from the center. If \(\lambda\) is undefined at some point, the field is not defined there either.
Worked Example: A Counterclockwise Field Around the Origin
Worked Example: A Counterclockwise Field Around the Origin
Take the center to be \((0,0)\) and choose the constant multiplier \(\lambda=1\). The component formula gives \(\mathbf{F}(x,y)=(-y,x)\).
Evaluate the field at four points on the unit circle:
These arrows point successively upward, left, downward, and right. Their directions follow the circle counterclockwise. Check tangency by taking a dot product at each point: \(1(0)+0(1)=0\); \(0(-1)+1(0)=0\); \((-1)(0)+0(-1)=0\); and \(0(1)+(-1)(0)=0\).
To sketch the field, draw arrows tangent to circles centered at the origin, with the orientation indicated by the sample points. The formula supplies the direction at every other point as well. At the origin, it gives \(\mathbf{F}(0,0)=(0,0)\).
Moving the Center
A change of center changes the displacement components. Replace \(x\) by \(x-a\) and \(y\) by \(y-b\) inside the rotated vector; do not rotate the absolute position \((x,y)\) unless the center is the origin. A constant positive multiplier keeps the same orientation, while a negative multiplier reverses it.
Worked Example: A Field Centered at \((1,-2)\)
Construct a field with constant multiplier \(2\) around the center \((1,-2)\). The displacement is \((x-1,y+2)\), so \(\mathbf{F}(x,y)=2(-(y+2),x-1)\). In components, \(\mathbf{F}(x,y)=(-2(y+2),2(x-1))\).
At \((3,-1)\), the displacement is \((3-1,-1+2)=(2,1)\). The field value is \[ \mathbf{F}(3,-1)=(-2(1),2(2))=(-2,4). \] The tangency check is \((2,1)\mathbin{\cdot}(-2,4)=2(-2)+1(4)=-4+4=0\).
At \((1,0)\), the displacement is \((0,2)\), and \[ \mathbf{F}(1,0)=(-2(2),2(0))=(-4,0). \] The dot product is \(0(-4)+2(0)=0\). At \((-1,-2)\), the displacement is \((-2,0)\), and \[ \mathbf{F}(-1,-2)=(-2(0),2(-2))=(0,-4). \] Here the dot product is \((-2)(0)+0(-4)=0\).
These samples anchor a sketch around \((1,-2)\): each arrow is perpendicular to the line from that center to its point. Since the multiplier is positive, the orientation is counterclockwise, as in the first example.
Allowing the Multiplier to Vary
The scalar multiplier need not be constant. At each point, it scales or reverses the quarter-turned displacement without changing the fact that the resulting vector is perpendicular to the radius. If the multiplier changes sign, arrows can reverse orientation in different regions. Where it equals zero, the field vector is zero, which has no arrow direction to display.
Worked Example: A Multiplier That Changes Sign
For a field centered at the origin, choose \(\lambda(x,y)=1+x\). Then \(\mathbf{F}(x,y)=(-(1+x)y,(1+x)x)\).
At \((0,2)\), the multiplier is \(1+0=1\), so \(\mathbf{F}(0,2)=(-1(2),1(0))=(-2,0)\). The displacement dot product is \(0(-2)+2(0)=0\).
At \((1,0)\), the multiplier is \(1+1=2\), so \(\mathbf{F}(1,0)=(-2(0),2(1))=(0,2)\). The check gives \(1(0)+0(2)=0\).
At \((-1,0)\), the multiplier is \(1-1=0\), giving \(\mathbf{F}(-1,0)=(0,0)\). At \((-2,0)\), the multiplier is \(1-2=-1\), so \(\mathbf{F}(-2,0)=(-(-1)(0),(-1)(-2))=(0,2)\). The dot product there is \((-2)(0)+0(2)=0\).
For the sketch, use the sample points to see that the field has zero vectors at \((-1,0)\) and reverses orientation across the line \(x=-1\) wherever the field is nonzero. The dot-product check confirms tangency at each sampled point.
A Practical Sketching Procedure
A component formula can seem less visual than an arrow diagram. The following sequence turns the formula into a sketch without requiring a value at every point.
Identify \((a,b)\), and use it as the reference point for every displacement.
Include points on the horizontal and vertical lines through the center, and a few points between them.
Substitute each point into the component formula and record the resulting vector.
Verify that the displacement dot product is zero, then draw the vector at its point, tangent to the centered circle.
Look for places where the scalar factor is zero or changes sign; these affect whether arrows vanish or reverse.
Common Mistake: Using a Radial Vector
A frequent error is to use the displacement itself as the field vector and call it tangent. But the displacement points along the radius, not along the circle. For example, at \((2,1)\) with center at the origin, the radial vector is \((2,1)\). Its dot product with the displacement is \(2(2)+1(1)=5\), not zero, so it is not tangent there.
The quarter-turned displacement is \((-1,2)\). Its dot product with \((2,1)\) is \(2(-1)+1(2)=-2+2=0\), so it is tangent. In general, use \((-(y-b),x-a)\) for the counterclockwise quarter-turn, not \((x-a,y-b)\). Reversing the quarter-turn gives the opposite tangent direction, which is still tangent but has the opposite orientation.
Why the Component Form Is Useful
The formula lets you construct an entire field from a center and a scalar multiplier. It also gives a quick test for a proposed field: at every point other than the center, take the dot product of the displacement and the vector. A zero result confirms tangency at that point; an expression that is identically zero confirms it throughout the field's domain.
For a sketch, sample evaluations reveal how the field behaves around the center and whether the scalar factor changes the orientation or creates zero vectors. The essential geometric check remains the same: each nonzero arrow must be perpendicular to the radius through its point.
Check Your Understanding
Use the displacement from the stated center and show your calculations when evaluating a field.
- For a field centered at \((2,1)\), write a component formula with constant multiplier \(3\).
- Evaluate \(\mathbf{F}(x,y)=(-y,x)\) at \((2,-1)\), then verify tangency to the circle centered at the origin.
- For center \((-1,2)\) and multiplier \(-1\), find the field vector at \((1,3)\) and check its dot product with the displacement.
- Why does multiplying a tangent vector by a scalar preserve tangency?
- For \(\mathbf{F}(x,y)=(-(1-y)x,(1-y)y)\), identify a line where the multiplier vanishes and state what happens to the field vectors there.