   Chapter 13.3, Problem 56E

Chapter
Section
Textbook Problem

# Show that the osculating plane at every point on the curve r ( t ) = 〈 t + 2 , 1 − t , 1 2 t 2 〉 is the same plane. What can you conclude about the curve?

To determine

To show: The osculating plane at every point on the curve r(t)=t+2,1t,12t2 is the same plane.

Explanation

Given data:

Curve is r(t)=t+2,1t,12t2 .

Formula used:

Write the expression for tangent vector of a vector function r(t) (T(t)) .

T(t)=r(t)|r(t)| (1)

Here,

r(t) is first derivative of function r(t) .

Write the equation of osculating plane with vector r(t)=r1,r2,r3 at point (x1,y1,z1) .

r1(xx1)+r2(yy1)+r3(zz1)=0 (2)

Write the expression for binormal vector of vector function r(t) (B(t)) .

B(t)=T(t)×N(t) (3)

Consider the two three-dimensional vector functions such as u(t)=u1(t),u2(t),u3(t) and v(t)=v1(t),v2(t),v3(t) .

Cross product of vectors:

Write the expression for cross product of vectors u(t) and v(t) (u(t)×v(t)) .

u(t)×v(t)=|ijku1(t)u2(t)u3(t)v1(t)v2(t)v3(t)|=[(u2(t)v3(t)v2(t)u3(t))],[(u1(t)v3(t)v1(t)u3(t))],[(u1(t)v2(t)v1(t)u2(t))]

Write the expression for magnitude of vector a (|a|) .

|a|=a12+a22+a32

Here,

a1 , a2 and a3 are the x, y, and z-coordinates of vector respectively.

Find the value of r(t) .

r(t)=ddtt+2,1t,12t2=ddt(t+2),ddt(1t),12ddt(t2)=(1+0),(01),12(2t){ddx(xn)=nxn1,ddx(x)=1,ddx(k)=0}=1,1,t

Find the value of |r(t)| .

|r(t)|=(1)2+(1)2+(t)2=1+1+t2=2+t2

Substitute 1,1,t for r(t) and 2+t2 for |r(t)| in equation (1),

T(t)=1,1,t2+t2=12+t2,12+t2,t2+t2

Apply differentiation with respect to t on both sides of equation

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