MHB How to Find the Curvature of r(t)?

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To find the curvature of the vector function r(t) = 9t i + 5 sin(t) j + 5 cos(t) k, the first step is to compute the first and second derivatives of r(t). The first derivative, r'(t), gives the velocity vector, while the second derivative, r''(t), provides the acceleration vector. The curvature can then be calculated using the formula κ(t) = |r'(t) × r''(t)| / |r'(t)|^3, where × denotes the cross product. Participants are encouraged to show their calculations for better assistance. Understanding these steps is crucial for accurately determining the curvature.
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Find the curvature. r(t) = 9t i + 5 sin(t) j + 5 cos(t) k
 
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Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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