Question

calc 3 help!!

Answer 1

W=F*d
force is given to you as a vector.
and distance would be the arc length of <t, t^2, t^3> for 0<t<1

Answer 2

the first step is to rewrite the force vector in terms of t. as that path is the only area we care about. so
\[F=<xe^y,xyz>=<te^{t^2},t^6>\]

Answer 3

the arc length, d, of <f(t),g(t),h(t)> for a<t<b is:
\[d=\int\sqrt{f^{'}(t)^2+g^{'}(t)^2+h^{'}(t)^2}dt=\int\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2+(\frac{dz}{dt})^2}dt\]

Answer 4

\[\frac{dx}{dt}=1; \frac{dy}{dt}= 2t; \frac{dz}{dt}= 3t^2\]
\[d=\int \limits_{0}^{1}\sqrt{1+4t^2+9t^4}dt\]

Answer 5

this way will be too complicated. sorry about that. may i show a different approach?

Answer 6

\[W=\int F(r)*dr\] for 0<r<1
\[r=<x,y,z>=<t,t^2,t^3>\]
\[dr=<dx,dy,dz>=<1,2t,3t^2>dt\]
\[F(r)=<xe^y,xyz>=<te^{t^2},t^6>\]
\[F(r)*dr=<te^{t^2},t^6,0>*<1,2t, 3t^2>=te^{t^2}+2t^7\]
\[W= \int\limits_{0}^{1}(te^{t^2}+2t^7 )dt\]

Answer 7

\[W=\frac{2e-1}{4}\]

Answer 8

i'm assuming the units is Newtons