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LiRa [457]
3 years ago
13

Find the work done by force dield f(x,y)=x^2i+ye^xj on a particle that moves along parabola x=y^2+1 from(1,0) to (2,1)

Mathematics
1 answer:
NemiM [27]3 years ago
8 0
Call the parabola P, parameterized by \mathbf r(y)=\langle y^2+1,y\rangle with 0\le y\le 1\rangle. Then the work done by \mathbf f(x,y) along P is

\displaystyle\int_P\mathbf f(x,y)\cdot\mathrm d\mathbf r=\int_{y=0}^{y=1}\mathbf f(x(y),y)\cdot\dfrac{\mathrm d\mathbf r(y)}{\mathrm dy}\,\mathrm dy
=\displaystyle\int_0^1\langle(y^2+1)^2,ye^{y^2+1}\rangle\cdot\langle2y,1\rangle\,\mathrm dy
=\displaystyle\int_0^1(2y^5+4y^3+2y+ye^{y^2+1})\,\mathrm dy=\frac73-\frac e2+\frac{e^2}2
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