   Chapter 12.4, Problem 17E

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# 57095-12.4-17E-Question-Digital.docxFinding the Principal Unit Normal Vector In Exercises 13-20, find the principal unit normal vector to the curve at the specified value of the parameter. r ( t ) = t i + t 2 j + ln   t k , t = 1

To determine

To calculate: The principal unit normal vector to the curve r(t)=ti+t2j+logtk at t=1.

Explanation

Given:

The vector, r(t)=ti+t2j+logtk and a point t=1.

Formula used:

Unit normal vector N(t) to the curve, r(t) is given by,

N(t)=T(t)T(t) where T(t)=r(t)r(t) is the unit tangent vector.

Calculation:

Consider the vector, r(t)=logti+(t+1)j.

Now, use the formula, T(t)=r(t)r(t) to find the tangent vector to the curve r(t)=ti+t2j+logtk.

First calculate, r(t).

So, r(t)=i+2tj+1tk.

Now calculate, r(t).

So,

r(t)=12+(2t)2+(1t)2=t2+4t4+1t2=4t4+t2+1t

Now put, r(t)=1ti+j and r(t) in the formula, T(t)=r(t)r(t).

So,

T(t)=i+2tj+1tk4t4+t2+1t=t4t4+t2+1i+2t24t4+t2+1j+14t4+t2+1k

Now, calculate, T(t).

So,

T(t)=[(4t4+t2+1(1)t24t4+t2+1(16t3+2t)(4t4+t2+1)2)i+(4t4+t2+1(4t)2t224t4+t2+1(16t3+2t)(4t4+t2+1)2)j(16t3+2t2(4t4+t2+1)32)k]=(8t4+22(4t4+t2+1)32)i+(4t3+8t2(4t4+t2+1)32)j(16t3+2t2(4t4+t2+1)32)k

So,

T(t)=(8t4+22(4t4+t2+1)32)2+(4t3+8t2(4t4+t2+1)32)2+(16t3+2t2(4t4+t2+1)32)2

Now use the formula, N(t)=T(t)T(t) to find the unit normal vector

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