Chapter 2: Space Curves
Section 2.3: Tangent Vectors
Table 2.3.1 lists the essential tangent-vector facts for a curve described by the position vector Rp.
The derivative R′p=ddpRp is taken componentwise.
The vector R′p is tangent to the curve defined by Rp.
If p is arc length s, R′s=ddsRs is the unit tangent vector T.
If T is the unit tangent vector, T·T′=0, that is, T is orthogonal to its derivative.
The length of R′p is given by ρ=∥R′∥ = dxdp2+dydp2+dzdp2 = dssp
If the parameter p is the time t, then R′t is the velocity vector V; and ρ=v, the speed.
Table 2.3.1 Essential tangent-vector facts for a curve defined by the position vector Rp.
The generic parameter along a curve defined parametrically by the position vector Rp is taken as p. The derivative of R with respect to its parameter is always a tangent vector. This can be inferred from the definition of the derivative of R.
R′p=ddpRp = limh→0Rp+h−Rph
This definition also suggests why the derivative is taken componentwise. The difference Rp+h−Rp is a vector whose components are each divided by h. As h→0, each component of the fraction Rp+h−Rph becomes the derivative of that component.
Figure 2.3.1 contains an animation that illustrates why this derivative results in a vector tangent to the curve described by Rp.
The blue vector is Rp+h−Rp and can be considered a "secant vector" that shrinks in length as h→0.
The green vector is Rp+h−Rph, and has the same direction as the secant vector, but does not shrink in length. In the limit as h→0, this green vector becomes tangent to the curve R.
use plots in
p1 := plot([2+4*cos(t),2+2*sin(t),t=0..1.8],color=black):
p2 := arrow([0,0],[2,4],color=black):
f1 := h -> arrow([0,0],[2+4*cos(Pi/2-h),2+2*sin(Pi/2-h)],color=red):
f2 := h -> arrow([2,4],[4*sin(h),2*cos(h)-2],color=blue):
f3 := h-> arrow([2,4], [4*sin(h)/h,(2*cos(h)-2)/h], color=green):
f4 := h -> display([f1(h),f2(h),f3(h)]):
p3 := display([seq(f4(1.5-k/10),k=0..14),seq(f4(.1-k/100),k=1..9)], insequence=true):
p4 := display([p1,p2,p3],scaling=constrained);
Figure 2.3.1 Animation: R′ is a tangent vector
If Rp is the position-vector representation of the parametric curve x=p2, y=p3, p≥0, show that limh→0Rp+h−Rph, denoted by R′p=dRdp,
is the vector x′p i+y′p j. Thus, the differentiation operator ddp is applied to R by applying it to each component of R.
If Rp is the position-vector representation of C, the parametric curve x=p cosp, y=p sinp, p∈0,3 π,
Obtain ρ=R′p and the unit tangent vector T=R′/ρ.
Graph R and the vectors R′1,R′5,R′9 along C.
Graph R and the vectors T1,T5,T9 along C.
Show that T·T′p=0, thus verifying that a unit vector is necessarily orthogonal to its derivative.
To the graph in Part (c), add the vectors T′1,T′5,T′9.
Let Rp be the position-vector representation of the parametric curve x=p2−p/2,y=4/3 p3/2, p≥0, and let Rs=Rps be the reparametrization obtained in Example 2.2.6. (Recall that s is the arc length along the curve.)
Obtain the unit tangent vector Tp=R′p/ρ.
Show that Tps=ddsRs, thus verifying that R′s is automatically a unit tangent vector.
If Rp is the position vector for the general parametric curve whose components are xp,yp,zp, and s≥0 is arc length, show that ddsRps is necessarily the unit (tangent) vector T.
If Rs is the position vector for C, a curve parametrized by s, the arc length, and Ts=R′s is the unit tangent vector along C, show that T·T′=0, thereby proving that the unit tangent vector is necessarily orthogonal to its derivative.
If Rp=xp i+yp j+zp k is a position vector and Rp=Rp is its length, show that R·dRdp = R dRdp.
Given the plane curve C defined by yx=1−x21+x2,
Obtain R⁡p, the radius-vector form of the curve, by the parametrization x=p,y=y⁡p.
Obtain R′p,ρp, and Tp, where ρ=R′ and T=R′/ρ.
Graph C and the vectors R1 and T1.
Graph ρp,p∈0,3, and determine the point x,y at which ρ is a maximum in this interval.
If Rp is the position-vector form of the curve C defined parametrically by the equations x⁡p=2+3⁢p− p2,y⁡p=cos⁡p, p∈0,π,
Obtain R′p,ρp and Tp, where ρ=R′ and T=R′/ρ.
Graph C and the vectors T0,T1,T2.
On the given interval, graph ρ and determine its absolute minimum and the point on the curve where this minimum occurs.
Given the two plane curves fx=x2,gx=8−x42,x⁢≥0,
At x=1, obtain the equation of the line tangent to y=f⁡x.
Find the coordinates of the intersection of y=g⁡x and the tangent line found in Part (a).
Construct a vector from ⁡1,f1 to the point found in Part (b).
Obtain R′⁡1, the natural tangent vector at ⁡1,f1.
Show that the vectors in Parts (c) and (d) are parallel.
(Hint: Show their components are proportional.)
Draw both curves, the tangent line (Part (a)), and the tangent vector (Part (d)).
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