Find the curvature of the plane curve at the given value of the parameter. r(t) = ti + (1/9t^3)j, t = 2
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r(t) = ti + \frac{1}{9}t^3j r'(t) = \frac{d}{dt}(ti) + \frac{d}{dt}(\frac{1}{9}t^3j) = i + \frac{1}{3}t^2j Now, we find the second derivative: r''(t) = \frac{d}{dt}(i) + \frac{d}{dt}(\frac{1}{3}t^2j) = 0i + \frac{2}{3}tj Now, we need to evaluate r'(t) and Show more…
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