May the force (field) be with you

Calculus Level 3

What is the amount of work done by the particle moving in the force field F ( x , y ) = x 2 i + x y j F(x,y) = x^2\ \mathbf{i} + xy\ \mathbf{j} along the quartile circle r ( t ) = cos ⁡ t i + sin ⁡ t j r(t) = \cos t\ \mathbf{i} + \sin t\ \mathbf{j} from t = 0 t=0 to t = π 2 t = \frac{ \pi}{2} ?

Details and assumptions

Work done can be calculated using the formula W = ∫ C F ⋅ d r W = \int_C F\cdot dr .


The answer is 0.

Calvin Lin by Calvin Lin

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2 solutions

Lucas Tell Marchi
Jan 28, 2014

Well, we may calculate the line integral by its definition:

∫ C F ⋅ d r = ∫ a b F ( r ( t ) ) ⋅ r ′ ( t ) \int_{C} \mathbf{F} \cdot d\mathbf{r} = \int_{a}^{b} \mathbf{F(r(\mathit{t}))} \cdot \mathbf{r'(\mathit{t})}

But observe that in this case F ( r ( t ) ) ⋅ r ′ ( t ) = 0 \mathbf{F(r(\mathit{t}))} \cdot \mathbf{r'(\mathit{t})} = 0 . Therefore

∫ C F ⋅ d r = 0 \int_{C} \mathbf{F} \cdot d\mathbf{r} = 0

0 Helpful 0 Interesting 1 Brilliant 0 Confused
Ricky Escobar
Dec 18, 2013

Work is given by the line integral $$W=\int C \mathbf{F} \cdot d \mathbf{r},$$ which, by the definition of the line integral, is $$W=\int a^b \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \, dt,$$ where r ( t ) \mathbf{r}(t) is a bijective parametrization of the curve C C such that r ( a ) \mathbf{r}(a) and r ( b ) \mathbf{r}(b) give the endpoints of C C . In our case, we are given formulas for F \mathbf{F} and r ( t ) \mathbf{r}(t) , as well as the values of a a and b b , so we simply substitute these into the formula: $$W=\int 0^{\frac{\pi}{2}} \langle \cos^2 t , \cos t \, \sin t \rangle \cdot \langle -\sin t,\cos t \rangle \, dt$$ $$=\int 0^{\frac{\pi}{2}} \left( -\cos^2 t \sin t + \cos^2 t \sin t \right) \, dt$$ $$=\int_0^{\frac{\pi}{2}} 0 \, dt$$ $$=\fbox{0}.$$

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Work is given by the line integral W = ∫ C F ⋅ d r , W=\int_C \mathbf{F} \cdot d \mathbf{r}, which, by the definition of the line integral, is W = ∫ a b F ( r ( t ) ) ⋅ r ′ ( t ) d t , W=\int_a^b \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \, dt, where r ( t ) \mathbf{r}(t) is a bijective parametrization of the curve C C such that r ( a ) \mathbf{r}(a) and r ( b ) \mathbf{r}(b) give the endpoints of C C . In our case, we are given formulas for F \mathbf{F} and r ( t ) \mathbf{r}(t) , as well as the values of a a and b b , so we simply substitute these into the formula: W = ∫ 0 π 2 ⟨ cos ⁡ 2 t , cos ⁡ t sin ⁡ t ⟩ ⋅ ⟨ − sin ⁡ t , cos ⁡ t ⟩ d t W=\int_0^{\frac{\pi}{2}} \langle \cos^2 t , \cos t \, \sin t \rangle \cdot \langle -\sin t,\cos t \rangle \, dt = ∫ 0 π 2 ( − cos ⁡ 2 t sin ⁡ t + cos ⁡ 2 t sin ⁡ t ) d t =\int_0^{\frac{\pi}{2}} \left( -\cos^2 t \sin t + \cos^2 t \sin t \right) \, dt = ∫ 0 π 2 0 d t =\int_0^{\frac{\pi}{2}} 0 \, dt = 0 . =\fbox{0}.

Kenny Lau - 5 years, 8 months ago

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