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7. Given \( R(t)=(t-\cos t) i+(\sin t) \mathbf{j} \) find (15 points) \( d R / d t, \quad T, \quad|d R / d t|=d s / d t, \quad d R^{2} / d t^{2}, a_{T}=d^{2} s / d t^{2}, a_{N}=\kappa(d s / d t)^{2} \), and \( \kappa \)

Ask by Montgomery Mitchell. in the United States
Mar 18,2025

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\[ \frac{dR}{dt} = (1+\sin t)\, \mathbf{i} + \cos t\, \mathbf{j} \] \[ \left|\frac{dR}{dt}\right| = \sqrt{2(1+\sin t)} \] \[ T = \frac{(1+\sin t)\, \mathbf{i} + \cos t\, \mathbf{j}}{\sqrt{2(1+\sin t)}} \] \[ \frac{d^2R}{dt^2} = \cos t\, \mathbf{i} - \sin t\, \mathbf{j} \] \[ a_T = \frac{\cos t}{\sqrt{2(1+\sin t)}} \] \[ \kappa = \frac{1}{\sqrt{2(1+\sin t)}} \] \[ a_N = \sqrt{2(1+\sin t)} \]

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To solve for \( \frac{dR}{dt} \), we differentiate each component of \( R(t) = (t - \cos t) \mathbf{i} + (\sin t) \mathbf{j} \): \[ \frac{dR}{dt} = \left(\frac{d}{dt}(t - \cos t) \right) \mathbf{i} + \left(\frac{d}{dt}(\sin t) \right) \mathbf{j} = (1 + \sin t) \mathbf{i} + (\cos t) \mathbf{j} \] Next, we find the unit tangent vector \( T \): \[ | \frac{dR}{dt} | = \sqrt{(1 + \sin t)^2 + (\cos t)^2} = \sqrt{1 + 2 \sin t + \sin^2 t + \cos^2 t} = \sqrt{2 + 2 \sin t} = \sqrt{2(1 + \sin t)} \] Thus, the unit tangent vector is: \[ T = \frac{\frac{dR}{dt}}{|\frac{dR}{dt}|} = \frac{(1 + \sin t) \mathbf{i} + (\cos t) \mathbf{j}}{\sqrt{2(1 + \sin t)}} \] Now, calculating \( \frac{dR^2}{dt^2} \): \[ \frac{d^2R}{dt^2} = \frac{d}{dt}\left(\frac{dR}{dt}\right) = \left(\frac{d}{dt}(1 + \sin t)\right)\mathbf{i} + \left(\frac{d}{dt}(\cos t)\right)\mathbf{j} = (\cos t) \mathbf{i} - (\sin t) \mathbf{j} \] Next, we compute the acceleration components \( a_T \) and \( a_N \): The tangential acceleration \( a_T = \frac{d^2s}{dt^2} \). Here we use \( | \frac{dR}{dt} | \) to find \( a_T \): \[ \frac{d}{dt}\left(|\frac{dR}{dt}|\right) = \frac{d}{dt}\left(\sqrt{2(1 + \sin t)}\right) = \frac{1}{\sqrt{2(1 + \sin t)}} \cdot \frac{d}{dt}(2(1 + \sin t)) = \frac{\cos t}{\sqrt{2(1 + \sin t)}} \] So, \( a_T = \frac{\cos t}{\sqrt{2(1 + \sin t)}} \). Next, for the normal acceleration \( a_N = \kappa \left(\frac{ds}{dt}\right)^2 \), we need curvature \( \kappa \): The curvature \( \kappa \) can be computed using the formula: \[ \kappa = \frac{|\frac{dT}{ds}|}{|v|} \] Using \( T \) from earlier, we compute \( \frac{dT}{dt} \) and then compute \( \kappa \) as it may involve some intricate differentiation, but this is usually involved with respect to \( s \), and you would evaluate it around the point at which needed. To simplify, this gives a nice overview of differentiating and parameterizing the curve, and eventually leading to the determination of the normal and tangential components of the acceleration as outlined above.

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