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91Ó°ÊÓ

Find the velocity and acceleration vectors for the position vectors given in Exercises 30–34

r(t)=cos3ti+sin4tj+tk

Short Answer

Expert verified

The velocity and acceleration vectors are:

v(t)=⟨−3sin3t,4cos4t,1⟩a(t)=⟨−9cos3t−16sin4t,0⟩

Step by step solution

01

Step 1. Given data

The given position vector isr(t)=cos3ti+sin4tj+tk

We have to find the velocity and acceleration vectors.

02

Step 2. Velocity vector

The velocity vector v(t) is given by

v(t)=ddtr(t)=x′(t)i+y′(t)j+z′(t)k=⟨x′(t),y′(t),z′(t)⟩

Therefore,

v(t)=ddt(r(t))=ddt(cos3ti+sin4tj+tk)=ddt(cos3t)i+ddt(sin4t)j+ddt(t)k=−3sinti+4cos4tj+k=⟨−3sin3t,4cos4t,1⟩

Hence the velocity vectorv(t)=⟨−3sin3t,4cos4t,1⟩.

03

Step 3. Acceleration vector

The acceleration vector a(t) is given by

a(t)=ddtv(t)=x′â¶Ä²(t)i+y′â¶Ä²(t)j+z"(t)k=⟨x′â¶Ä²(t),y′â¶Ä²(t),z′â¶Ä²(t)⟩

Therefore,

a(t)=ddt(v(t))=ddt⟨−3sin3t,4cos4t,1⟩=⟨ddt(−3sin3t),ddt(4cos4t),ddt(1)⟩=⟨−9cos3t,−16sin4t,0⟩

Hence the acceleration vectora(t)=⟨−9cos3t−16sin4t,0⟩

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