Mathematics, 20.07.2019 19:00, HaJEReMY5170

Find the work done by the force field f in moving an object from a to b. f(x, y) = 2y3/2 i + 3x y j a(1, 1), b(3, 9)

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answered: timozy95

To answer this question we need to apply the concept of work, that is

w = ∫ ₐᵇ F×ds

Solution is:

w = 523 J

w = ∫ₐᵇ F × ds

F = 2y³/² i + 3xy j and ds = dx i + dy j

Then F × ds = ( 2y³/² i + 3xy j ) × ( dx i + dy j )

F × ds = 2y³/² dx + 3xy dy

And = ∫ₐᵇ F×ds = ∫ₐᵇ2y³/² dx + 3xy dy a ( 1 , 1 ) b ( 3 , 9 )

∫ₐᵇ F×ds = 2y³/²∫dx + 3x ∫ydy

∫ₐᵇ F×ds = 2y³/²x |₁,₁³,⁹ + 3xy²/2 |₁,₁³,⁹

∫ₐᵇ F×ds = 2 × (9)³/² × (3) - 2 × (1 ) ³/²× (1) + 3 × (3)× (9)² /2 - 3 × (1)³/²/2

∫ₐᵇ F×ds = 162 - 2 + 364,5 - 1,5

∫ₐᵇ F×ds = 160 + 363

∫ₐᵇ F×ds = 523 J

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answered: missinghalloween13

Assuming the path from

is a line segment, we can parameterize it by
$\mathbf r(t)=(1-t)(1,1)+t(3,9)=(1+2t,1+8t)=(x(t),y(t))$
for $0\le t\le1$ . Then the work done by $\mathbf f(x,y)$ along this path, which I'll denote $\mathcal C$ , is
$\displaystyle\int_{\mathcal C}\mathbf f(x,y)\cdot\mathrm d\mathbf r=\int_{t=0}^{t=1}f(x(t),y(t))\cdot(2,8)\,\mathrm dt$ $=\displaystyle\int_0^1(2(1+8t)^{3/2},3(1+2t)(1+8t))\cdot(2,8)\,\mathrm dt$ $=\displaystyle4\int_0^1(1+8t)^{3/2}\,\mathrm dt+24\int_0^1(1+10t+16t^2)\,\mathrm dt$
$=\dfrac{242}5+272=\dfrac{1602}5$

answered: Guest

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answered: Guest

here is a big fat