Suppose we have a scalar field $f(x, y)=-y$ and a curve $C$ that is parameterized by $\alpha(t)=(\cos (t), \sin (t)-1)$ for $0

Q: Suppose we have a scalar field $f(x, y)=-y$ and a curve $C$ that is parameterized by $\alpha(t)=(\cos (t), \sin (t)-1)$ for $0<t<2 \pi$. What is the line integral of $f$ along $C$ ? \[ \int_{C} f d s= \] $\square$

∻Solution by Steps∻ ‖ step 1 ⋮ To find the line integral of the scalar field $f(x, y) = -y$ along the curve $C$ parameterized by $\alpha(t) = (\cos(t), \sin(t) - 1)$, we first need to compute the differential arc length $ds$. ‖ ‖ step 2 ⋮ The differential arc length is given by $ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt$. We calculate $\frac{dx}{dt}$ and $\frac{dy}{dt}$ from the parameterization. ‖ ‖ step 3 ⋮ For $\alpha(t) = (\cos(t), \sin(t) - 1)$, we have $\frac{dx}{dt} = -\sin(t)$ and $\frac{dy}{dt} = \cos(t)$. ‖ ‖ step 4 ⋮ Substituting these into the expression for $ds$, we get $ds = \sqrt{(-\sin(t))^2 + (\cos(t))^2} dt = \sqrt{1} dt = dt$. ‖ ‖ step 5 ⋮ The line integral is then $\int_{C} f ds = \int_{0}^{2\pi} -(\sin(t) - 1) dt$ since $f(x, y) = -y$ and $y = \sin(t) - 1$. ‖ ‖ step 6 ⋮ Evaluating the integral, we get $\int_{0}^{2\pi} -(\sin(t) - 1) dt = [-\cos(t) - t]_{0}^{2\pi}$. ‖ ‖ step 7 ⋮ Plugging in the limits of integration, we find $[-\cos(2\pi) - 2\pi] - [-\cos(0) - 0] = [-1 - 2\pi] - [1 - 0] = -2\pi - 2$. ‖ ∻[1] Answer∻ ⚹ $-2\pi - 2$ ⚹ ∻Key Concept∻ ⚹Line Integral of a Scalar Field⚹ ∻Explanation∻ ⚹The line integral of a scalar field along a curve is found by integrating the scalar field function multiplied by the differential arc length over the parameter interval of the curve.◊What is the parametric representation of the curve C in terms of cos(t) and sin(t⍭ Generate me a similar question◊⚹

Question

Suppose we have a scalar field $f(x, y)=-y$ and a curve $C$ that is parameterized by $\alpha(t)=(\cos (t), \sin (t)-1)$ for $0<t<2 \pi$.

What is the line integral of $f$ along $C$ ?
$$
\int_{C} f d s=
$$
$\square$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the line integral of the scalar field $f(x, y) = -y$ along the curve $C$ parameterized by $\alpha(t) = (\cos(t), \sin(t) - 1)$, we first need to compute the differential arc length $ds$. ‖ ‖ step 2 ⋮ The differential arc length is given by $ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt$. We calculate $\frac{dx}{dt}$ and $\frac{dy}{dt}$ from the parameterization. ‖ ‖ step 3 ⋮ For $\alpha(t) = (\cos(t), \sin(t) - 1)$, we have $\frac{dx}{dt} = -\sin(t)$ and $\frac{dy}{dt} = \cos(t)$. ‖ ‖ step 4 ⋮ Substituting these into the expression for $ds$, we get $ds = \sqrt{(-\sin(t))^2 + (\cos(t))^2} dt = \sqrt{1} dt = dt$. ‖ ‖ step 5 ⋮ The line integral is then $\int_{C} f ds = \int_{0}^{2\pi} -(\sin(t) - 1) dt$ since $f(x, y) = -y$ and $y = \sin(t) - 1$. ‖ ‖ step 6 ⋮ Evaluating the integral, we get $\int_{0}^{2\pi} -(\sin(t) - 1) dt = [-\cos(t) - t]_{0}^{2\pi}$. ‖ ‖ step 7 ⋮ Plugging in the limits of integration, we find $[-\cos(2\pi) - 2\pi] - [-\cos(0) - 0] = [-1 - 2\pi] - [1 - 0] = -2\pi - 2$. ‖ ∻[1] Answer∻ ⚹ $-2\pi - 2$ ⚹ ∻Key Concept∻ ⚹Line Integral of a Scalar Field⚹ ∻Explanation∻ ⚹The line integral of a scalar field along a curve is found by integrating the scalar field function multiplied by the differential arc length over the parameter interval of the curve.◊What is the parametric representation of the curve C in terms of cos(t) and sin(t⍭ Generate me a similar question◊⚹

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