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Math  /  Calculus

Question9. The work done by the force field F⃗(x,y)=⟨y,−x⟩\vec{F}(x, y)=\langle y,-x\rangle acting on an object as it moves al the parabola y=x2−1y=x^{2}-1 from (1,0)(1,0) to (−2,3)(-2,3) is given by: ∫cydx−xdy\int_{c} y d x-x d y r⃗(t)=1−(t,t2−1)\vec{r}(t)=1-\left(t, t^{2}-1\right)

Studdy Solution

STEP 1

1. The force field is given by F⃗(x,y)=⟨y,−x⟩\vec{F}(x, y) = \langle y, -x \rangle.
2. The path of the object is along the parabola y=x2−1y = x^2 - 1.
3. The integral ∫cy dx−x dy\int_{c} y \, dx - x \, dy represents the work done by the force field along the curve cc.
4. The parametric representation of the path is r⃗(t)=(t,t2−1)\vec{r}(t) = (t, t^2 - 1) with tt ranging from 1 to -2.

STEP 2

1. Parametrize the curve.
2. Determine the limits of integration.
3. Express dxdx and dydy in terms of dtdt.
4. Substitute into the integral and evaluate.

STEP 3

Parametrize the curve using the given parametric equations. The curve is already parametrized as:
r⃗(t)=(t,t2−1)\vec{r}(t) = (t, t^2 - 1)

STEP 4

Determine the limits of integration. The object moves from (1,0)(1, 0) to (−2,3)(-2, 3). In terms of tt, this corresponds to:
- When t=1t = 1, the point is (1,0)(1, 0). - When t=−2t = -2, the point is (−2,3)(-2, 3).
Thus, the limits of integration are from t=1t = 1 to t=−2t = -2.

STEP 5

Express dxdx and dydy in terms of dtdt:
- From r⃗(t)=(t,t2−1)\vec{r}(t) = (t, t^2 - 1), we have x=tx = t and y=t2−1y = t^2 - 1. - Therefore, dx=dtdx = dt. - For dydy, differentiate y=t2−1y = t^2 - 1 with respect to tt: dy=2t dtdy = 2t \, dt.

STEP 6

Substitute x=tx = t, y=t2−1y = t^2 - 1, dx=dtdx = dt, and dy=2t dtdy = 2t \, dt into the integral:
∫cy dx−x dy=∫1−2((t2−1) dt−t⋅2t dt)\int_{c} y \, dx - x \, dy = \int_{1}^{-2} ((t^2 - 1) \, dt - t \cdot 2t \, dt)
Simplify the expression inside the integral:
=∫1−2(t2−1−2t2) dt= \int_{1}^{-2} (t^2 - 1 - 2t^2) \, dt =∫1−2(−t2−1) dt= \int_{1}^{-2} (-t^2 - 1) \, dt
Evaluate the integral:
=∫1−2(−t2−1) dt=[−t33−t]1−2= \int_{1}^{-2} (-t^2 - 1) \, dt = \left[ -\frac{t^3}{3} - t \right]_{1}^{-2}
Calculate the definite integral:
=(−(−2)33−(−2))−(−133−1)= \left( -\frac{(-2)^3}{3} - (-2) \right) - \left( -\frac{1^3}{3} - 1 \right) =(83+2)−(−13−1)= \left( \frac{8}{3} + 2 \right) - \left( -\frac{1}{3} - 1 \right) =(83+63)−(−13−33)= \left( \frac{8}{3} + \frac{6}{3} \right) - \left( -\frac{1}{3} - \frac{3}{3} \right) =143−(−43)= \frac{14}{3} - \left( -\frac{4}{3} \right) =143+43= \frac{14}{3} + \frac{4}{3} =183= \frac{18}{3} =6= 6
The work done by the force field is:
6 \boxed{6}

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