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For each of the vector-valued functions in Exercises 35–39,

find the unit tangent vector, the principal unit normal vector,

the binormal vector, and the equation of the osculating plane

at the specified value of t.

r(t)=t2,t,13t3/2att=1

Short Answer

Expert verified

The unit tangent vector at t=1is 22,12,12.

The principal unit normal vector at t=1is 0,-22,22.

The binormal vector at t=1is 22,-12,-12.

The equation of the osculating plane att=1 is 2x-y-z=-13.

Step by step solution

01

Step 1. Given information 

We have been given the vector valued function r(t)=t2,t,13t3/2.

The objective is to find the unit tangent vector, the principal unit normal vector, the binormal vector, and the equation of the osculating plane at t=1.

02

Step 2. Finding unit tangent vector 

Consider r(t)=t2,t,13t32.

Now the first derivative is given as r(t)=12,12t,12tand its magnitude is

r(t)=12+14t+1t=2t+1+t24t=t+12t

So the unit tangent vector is given as:

role="math" localid="1654166261956" Tt=r'tr't=12,12t,12tt+12t=2t12,12t,12tt+1=2,1,tt+1

At t=1we have:

T(1)=2,1,11+1=22,12,12

03

Step 3. Finding principle normal unit vector 

For the unit tangent vector Tt=2,1,tt+1the first derivative is given as: T(t)=1t2t(t+1)2,1(t+1)2,1(t+1)2

and its magnitude is given as:

T(t)=(1t)22t(t+1)4+1(t+1)4+1(t+1)4=1+t22t+2t+2t2t(t+1)4=t+12t(t+1)2=12t(t+1)

And thus the unit normal vector is given as:

role="math" localid="1654166630663" Nt=T'tT't=1t2t(t+1)2,1(t+1)2,1(t+1)212t(t+1)=2t(t+1)1t2t(t+1)2,1(t+1)2,1(t+1)2=1-tt+1,-2tt+1,2tt+1

So at t=1its value is

role="math" localid="1654166704875" N(1)=1-11+1,-2×11+1,2×11+1=0,-22,22

04

Step 4. Finding the binomial vector 

The binomial vector is given as:

B(1)=T(1)×N(1)=22,12,12×0,22,22=ijk22121202222=i24+24j24+k24=22i12j12k=22,12,12
05

Step 5. Finding osculating plane 

The equation of oscillating plane is given as
B(1)xx(1),yy(1),zz(1)=022,12,12x12,y1,z13=022x1212(y1)12z13=02x1y+1z+13=02xyz=13

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