Answer to Question #197818 in Calculus for Rocky Valmores

Question

Find the total work done in moving a particle in a force field given by 𝐹 = 3𝑥𝑦𝑖 − 5𝑧𝑗 + 10𝑥𝑘 along the curve x = 𝑡² + 1, 𝑦 = 2𝑡² , 𝑧 = 𝑡³ 𝑓𝑟𝑜𝑚 𝑡 = 1 𝑡𝑜 𝑡 = 2.

Answer 1

$\begin{array}{l} {\overline{F}.\overline{dr}=(3xyi-52j+10x\overline{k})(dxi+dyi+dzk)} \\ {\, \, \, \, \, \, \, \, \, \, =3xydx-52dy+10xdz} \\ {\, \, \, \, \, \, \, \, \, \, \, =\, 3(t^{2} +1)(2t^{2} )(2tdt)-5(t^{3} )(4tdt)+10(t^{2} +1)(3t^{2} dt)} \\ {\, \, \, \, \, \, \, \, \, \, \, =(12t^{5} +12t^{3} -20t^{4} +30t^{4} +30t^{2} )dt} \\ {\, \, \, \, \, \, \, \, \, \, =(12t^{5} +10t^{4} +12t^{3} +30t^{2} )dt} \end{array}$ $\begin{array}{l} {Work\, \, Done=\int _{C}f.\overline{dr} =\int _{1}^{2}(12t^{5} +10t^{4} +12t^{3} +30t^{2} )dt } \\ {\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\, \left[\frac{12t^{6} }{5} +\frac{10t^{5} }{5} +\frac{12t^{4} }{4} +\frac{30t^{3} }{3} \right]_{1}^{2} } \\ {\qquad \, \, \, \, \, \, \, \, \, =\left[2t^{6} +2t^{5} +3t^{4} +10^{3} \right]_{1}^{2} } \\ {\qquad \, \, \, \, \, \, \, \, \, \, =\, \left[2(2)^{6} +2(2)^{5} +3(2)^{4} +10(2)^{3} \right]-\left[2+2+3+10\right]} \end{array}$ $= 303$