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QuestionCalculate the work done by the force fˉ=zˉiˉ+xjˉ+yk^\bar{f}=\bar{z} \bar{i}+x \bar{j}+y \hat{k} along the curve rˉ=cos⁡ti+sin⁡tj−tkˉ\bar{r}=\cos t i+\sin t j-t \bar{k} from t=0t=0 to t=2πt=2 \pi.

Studdy Solution

STEP 1

Assumptions1. The force vector is given by fˉ=zˉiˉ+xjˉ+yk^\bar{f}=\bar{z} \bar{i}+x \bar{j}+y \hat{k} . The particle moves along the curve ∣rˉ=cos⁡ti+sin⁡tj−tkˉ\mid \bar{r}=\cos t i+\sin t j-t \bar{k}
3. The movement of the particle is from t=0t=0 to t=πt= \pi

STEP 2

First, we need to find the differential of the position vector rˉ\bar{r}. This can be found by differentiating rˉ\bar{r} with respect to tt.
drˉ=drˉdtdtd\bar{r} = \frac{d\bar{r}}{dt} dt

STEP 3

Now, differentiate the position vector rˉ\bar{r} with respect to tt.
drˉdt=ddt(cos⁡ti+sin⁡tj−tkˉ)\frac{d\bar{r}}{dt} = \frac{d}{dt} (\cos t i+\sin t j-t \bar{k})

STEP 4

Calculate the derivative.
drˉdt=−sin⁡ti+cos⁡tj−kˉ\frac{d\bar{r}}{dt} = -\sin t i+\cos t j-\bar{k}

STEP 5

Now, we can find the differential of the position vector rˉ\bar{r}.
drˉ=−sin⁡ti+cos⁡tj−kˉdtd\bar{r} = -\sin t i+\cos t j-\bar{k} dt

STEP 6

The work done by the force is given by the dot product of the force vector and the differential of the position vector.
W=∫fˉ⋅drˉW = \int \bar{f} \cdot d\bar{r}

STEP 7

Substitute the given force vector and the differential of the position vector into the equation.
W=∫(zˉiˉ+xjˉ+yk^)⋅(−sin⁡ti+cos⁡tj−kˉ)dtW = \int (\bar{z} \bar{i}+x \bar{j}+y \hat{k}) \cdot (-\sin t i+\cos t j-\bar{k}) dt

STEP 8

Substitute the expressions for xx, yy, and zz from the position vector rˉ\bar{r} into the equation.
W=∫((−t)iˉ+cos⁡tjˉ+sin⁡tk^)⋅(−sin⁡ti+cos⁡tj−kˉ)dtW = \int ((-t) \bar{i}+\cos t \bar{j}+\sin t \hat{k}) \cdot (-\sin t i+\cos t j-\bar{k}) dt

STEP 9

Calculate the dot product and integrate.
W=∫(tsin⁡t+cos⁡2t+sin⁡2t)dtW = \int (t \sin t + \cos^2 t + \sin^2 t) dt

STEP 10

implify the integrand using the identity cos⁡2t+sin⁡2t=\cos^2 t + \sin^2 t =.
W=∫(tsin⁡t+)dtW = \int (t \sin t +) dt

STEP 11

Calculate the integral from t=0t=0 to t=πt= \pi.
W=[−tcos⁡t+t+sin⁡t]0πW = \left[ -t \cos t + t + \sin t \right]0^{\pi}

STEP 12

Evaluate the definite integral.
W=−2πcos⁡(2π)+2π+sin⁡(2π)−(−0cos⁡(0)+0+sin⁡(0))W = -2\pi \cos(2\pi) +2\pi + \sin(2\pi) - (-0 \cos(0) +0 + \sin(0))

STEP 13

implify the expression using the fact that cos⁡(2π)=\cos(2\pi) = and sin⁡(2π)=0\sin(2\pi) =0.
W=−2π+2π+0−(0+0)=0W = -2\pi +2\pi +0 - (0 +0) =0The work done by the force when it moves the particle along the arc of the curve from t=0t=0 to t=2πt=2 \pi is0.

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