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Math  /  Calculus

QuestionWork done by a force: Calculate the work done by a force field F(x,y)=(x∧2F(x, y)=\left(x^{\wedge} 2\right. +y∧2,x−y)\left.+y^{\wedge} 2, x-y\right) on an object moving along the curve C : x=t∧2,y=t\mathrm{x}=\mathrm{t}^{\wedge} 2, \mathrm{y}=\mathrm{t} from t=0\mathrm{t}=0 to t =1=1.

Studdy Solution

STEP 1

1. The force field is given by F(x,y)=(x2+y2,x−y) \mathbf{F}(x, y) = (x^2 + y^2, x - y) .
2. The object moves along the curve C C described by the parametric equations x=t2 x = t^2 and y=t y = t for t t ranging from 0 to 1.
3. The work done by the force field along the curve can be calculated using the line integral of the force field along the given path.

STEP 2

1. Parameterize the curve using the given parametric equations.
2. Compute the differential dr d\mathbf{r} along the curve.
3. Substitute the parameterized variables into the force field.
4. Calculate the dot product F⋅dr \mathbf{F} \cdot d\mathbf{r} .
5. Integrate the dot product over the given range of t t .

STEP 3

Parameterize the curve using x=t2 x = t^2 and y=t y = t .
r(t)=(t2,t),0≤t≤1\mathbf{r}(t) = (t^2, t), \quad 0 \leq t \leq 1

STEP 4

Compute the differential dr d\mathbf{r} along the curve.
dr=(dxdt,dydt)dt=(2t,1)dtd\mathbf{r} = \left( \frac{dx}{dt}, \frac{dy}{dt} \right) dt = (2t, 1) dt

STEP 5

Substitute the parameterized variables x=t2 x = t^2 and y=t y = t into the force field F(x,y) \mathbf{F}(x, y) .
F(x,y)=F(t2,t)=((t2)2+t2,t2−t)=(t4+t2,t2−t)\mathbf{F}(x, y) = \mathbf{F}(t^2, t) = \left( (t^2)^2 + t^2, t^2 - t \right) = \left( t^4 + t^2, t^2 - t \right)

STEP 6

Calculate the dot product F⋅dr \mathbf{F} \cdot d\mathbf{r} .
F⋅dr=(t4+t2,t2−t)⋅(2t,1)dt=(t4+t2)⋅2t+(t2−t)⋅1 dt\mathbf{F} \cdot d\mathbf{r} = \left( t^4 + t^2, t^2 - t \right) \cdot (2t, 1) dt = (t^4 + t^2) \cdot 2t + (t^2 - t) \cdot 1 \, dt
F⋅dr=(2t5+2t3+t2−t) dt\mathbf{F} \cdot d\mathbf{r} = (2t^5 + 2t^3 + t^2 - t) \, dt

STEP 7

Integrate the dot product over the given range of t t .
Work=∫01(2t5+2t3+t2−t) dt\text{Work} = \int_0^1 (2t^5 + 2t^3 + t^2 - t) \, dt
Calculate each term separately:
∫012t5 dt=2[t66]01=26=13\int_0^1 2t^5 \, dt = 2 \left[ \frac{t^6}{6} \right]_0^1 = \frac{2}{6} = \frac{1}{3}
∫012t3 dt=2[t44]01=24=12\int_0^1 2t^3 \, dt = 2 \left[ \frac{t^4}{4} \right]_0^1 = \frac{2}{4} = \frac{1}{2}
∫01t2 dt=[t33]01=13\int_0^1 t^2 \, dt = \left[ \frac{t^3}{3} \right]_0^1 = \frac{1}{3}
∫01−t dt=−[t22]01=−12\int_0^1 -t \, dt = - \left[ \frac{t^2}{2} \right]_0^1 = - \frac{1}{2}
Sum up all the terms:
Work=13+12+13−12=13+13=23\text{Work} = \frac{1}{3} + \frac{1}{2} + \frac{1}{3} - \frac{1}{2} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}
Solution: The work done by the force field F(x,y)=(x2+y2,x−y) \mathbf{F}(x, y) = (x^2 + y^2, x - y) on an object moving along the curve x=t2,y=t x = t^2, y = t from t=0 t = 0 to t=1 t = 1 is 23 \frac{2}{3} .

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