   Chapter 13.4, Problem 40E

Chapter
Section
Textbook Problem

# Find the tangential and normal components of the acceleration vector.40. r(t) = t i + 2et j + e2t k

To determine

To find: The tangential components of the acceleration vector and normal components of the acceleration vector.

Explanation

Given:

r(t)=ti+2etj+e2tk

Formula used:

Write the expression for tangential component.

aT=r(t)r(t)|r(t)| (1)

Write the expression for normal component.

aN=|r(t)×r(t)||r(t)| (2)

Find r(t)

r(t)=ddt[r(t)]

Substitute ti+2etj+e2tk for r(t) ,

r(t)=ddt[ti+2etj+e2tk]=ddt(ti)+ddt(2etj)+ddt(e2tk)=i+2etj+2e2tk

Find r(t) .

r(t)=ddt[r(t)]

Substitute i+2etj+2e2tk for r(t) ,

r(t)=ddt[i+2etj+2e2tk]=ddt(i)+ddt(2etj)+ddt(2e2tk)=(0)i+2etj+4e2tk=2etj+4e2tk

Find |r(t)| .

|r(t)|=(1)2+(2et)2+(2e2t)2=1+4e2t+4e4t=(1+2e2t)2=1+2e2t

Substitute i+2etj+2e2tk for r(t) , 2etj+4e2tk for r(t) and 1+2e2t for |r(t)| in equation (1),

aT=(i+2etj+2e2tk)(2etj+4e2tk)1+2e2t=(1)(0)+(2et)(2et)+(2e2t)(4e2t)1+2e2t=4e2t+8e4t1+2e2t=4e2t(1+2e2t)1+2e2t

aT=4e2t

Thus, the tangential components of the acceleration vector is 4e2t_

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