MTH622 ASSIGNMENT NO. 1 FALL 2022 || 100% RIGHT SOLUTION || Vectors and Classical Mechanics || BY VuTech

MTH622 ASSIGNMENT NO. 1 FALL 2022

SEND WHATSAPP OR E-MAIL FOR ANY QUERY

0325-6644800

kamranhameedvu@gmail.com

MTH622 ASSIGNMENT NO. 1 FALL 2022 || 100% RIGHT SOLUTION || Vectors and Classical Mechanics || BY VuTech

For More Updates Visit Website

www.vutechofficial.blogspot.com

MTH622

VECTORS AND CLASSICAL MECHANICS

ASSIGNMENT NO. 1

FALL 2022

LAST DATE: 20-11-2022

Question No. 1

        Find the work done by the force field `\vec F( {x,y,z}) =  - \frac{1}{2}x\hat i - \frac{1}{2}y\hat j + \frac{1}{4}\hat k` on a particle as it moves along the helix `\vec r\left( t \right) = cost\hat i + sint\hat j + t\hat k` from point (1,0,0) to point `\(\left(  - 1,0,3\pi  \right).`

        Solution:

                Here

                    `\vec F\left( x,y,z \right) =  - \frac{1}{2}x\hat i - \frac{1}{2}y\hat j + \frac{1}{4}\hat k     -  -  - -  -  - \left( A \right)`

                And

                    `\vec r\left( t \right) = cost\hat i + sint\hat j + t\hat k`

                We know that

                    `\vec r\left( t \right) = x\hat i + y\hat j + z\hat k`        

                So,

                    x=cost , y = sint , z = t

                By taking derivative w.r.t t of `\vec r\left( t \right)`

                    `\vec{dr} \left( t \right) = \left( - sint\hat i + cos\hat j + \hat k \right)dt -  - - -  - \left( 1 \right)`

                By taking dot product of (A) and (1)

                    `\vec F .d\vec r\left( t \right) = ( - \frac{1}{2}cost\hat i - \frac{1}{2}sint\hat j  +  \frac{1}{4}\hat k ) . ( - sint\hat I  + cost\hat j  +  \hat k)dt`

                    `\vec F . d\vec r\left( t \right) = \left( \frac{1}{2}cost . sint - \frac{1}{2}sint . cost + \frac{1}{4} \right)dt`

                    `\vec F . d\vec r\left( t \right) = \frac{1}{4}dt`

                    `W = \smallint \vec F .d\vec r\left( t \right)`

                    Total work done = W = `\frac{1}{4}[\int_{1}^{-1}dt + \int_{0}^{3pi}dt] `

                    = `\frac{1}{4}\left[ \left( 1 + 1 \right) + \left( 0 - 3\pi  \right) \right]`

                    = `\frac{1}{4}\left[ 2 - 3\pi  \right]`

 

 For More Updates Visit Website

www.vutechofficial.blogspot.com

 

KINDLY, DON’T COPY PASTE

SUBSCRIBE, SHARE, LIKE AND COMMENTS FOR MORE UPDATES

SEND WHATSAPP OR E-MAIL FOR ANY QUERY

0325-6644800

kamranhameedvu@gmail.com