written 7.7 years ago by | modified 2.2 years ago by |
Mumbai University > Mechanical Engineering > Sem 4 > Applied Mathematics IV
Marks: 5M
Year: Dec 2015
written 7.7 years ago by | modified 2.2 years ago by |
Mumbai University > Mechanical Engineering > Sem 4 > Applied Mathematics IV
Marks: 5M
Year: Dec 2015
written 7.7 years ago by |
We have
$$\begin{array}{l} {\overline{F}.\overline{dr}=(3xyi-52j+10x\overline{k})(dxi+dyi+dzk)} \\ {\, \, \, \, \, \, \, \, \, \, =3xydx-52dy+10xdz} \\ {\, \, \, \, \, \, \, \, \, \, \, =\, 3(t^{2} +1)(2t^{2} )(2tdt)-5(t^{3} )(4tdt)+10(t^{2} +1)(3t^{2} dt)} \\ {\, \, \, \, \, \, \, \, \, \, \, =(12t^{5} +12t^{3} -20t^{4} +30t^{4} +30t^{2} )dt} \\ {\, \, \, \, \, \, \, \, \, \, =(12t^{5} +10t^{4} +12t^{3} +30t^{2} )dt} \end{array}$$ $$\begin{array}{l} {Work\, \, Done=\int _{C}f.\overline{dr} =\int _{1}^{2}(12t^{5} +10t^{4} +12t^{3} +30t^{2} )dt } \\ {\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\, \left[\frac{12t^{6} }{5} +\frac{10t^{5} }{5} +\frac{12t^{4} }{4} +\frac{30t^{3} }{3} \right]_{1}^{2} } \\ {\qquad \, \, \, \, \, \, \, \, \, =\left[2t^{6} +2t^{5} +3t^{4} +10^{3} \right]_{1}^{2} } \\ {\qquad \, \, \, \, \, \, \, \, \, \, =\, \left[2(2)^{6} +2(2)^{5} +3(2)^{4} +10(2)^{3} \right]-\left[2+2+3+10\right]} \end{array} $$ $$ = 303$$