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Curvature - Calculus 3

  1. Consider a parametric curve where the speed ∣r′(t)∣|\mathbf{r}'(t)| is constant and equal to v>0v > 0. If the curvature at a point is κ\kappa, and the curve is reparameterized by arc length ss, what is the relationship between the new curvature κ~\tilde{\kappa} and the original curvature κ\kappa?
    Answer: κ~=κ\tilde{\kappa} = \kappa because curvature is invariant under reparameterization Curvature is a geometric property that is invariant under reparameterization. The formula κ=∣r′×r′′∣∣r′∣3\kappa = \frac{|\mathbf{r}' \times \mathbf{r}''|}{|\mathbf{r}'|^3} automatically accounts for the parameterization speed. When we reparameterize by arc length, we get drds=drdt⋅dtds=r′(t)⋅1v\frac{d\mathbf{r}}{ds} = \frac{d\mathbf{r}}{dt} \cdot \frac{dt}{ds} = \mathbf{r}'(t) \cdot \frac{1}{v} since dsdt=v\frac{ds}{dt} = v. The curvature formula in terms of arc length gives the same geometric curvature value. The other choices reflect common misconceptions about how curvature transforms under reparameterization.
  2. The position vector of a moving particle is given by r(t)=⟨etcos⁡t,etsin⁡t⟩\mathbf{r}(t) = \langle e^t \cos t, e^t \sin t \rangle. What happens to the curvature κ(t)\kappa(t) as t→∞t \to \infty?
    Answer: κ(t)→0\kappa(t) \to 0 because the spiral becomes increasingly straight This is a logarithmic spiral. We have r′(t)=et⟨cos⁡t−sin⁡t,sin⁡t+cos⁡t⟩\mathbf{r}'(t) = e^t\langle \cos t - \sin t, \sin t + \cos t \rangle and r′′(t)=et⟨−2sin⁡t,2cos⁡t⟩\mathbf{r}''(t) = e^t\langle -2\sin t, 2\cos t \rangle. The cross product gives r′×r′′=2e2t\mathbf{r}' \times \mathbf{r}'' = 2e^{2t}, and ∣r′∣=et2|\mathbf{r}'| = e^t\sqrt{2}. Therefore κ=2e2t(et2)3=2e2t22e3t=12et\kappa = \frac{2e^{2t}}{(e^t\sqrt{2})^3} = \frac{2e^{2t}}{2\sqrt{2}e^{3t}} = \frac{1}{\sqrt{2}e^t}. As t→∞t \to \infty, κ(t)→0\kappa(t) \to 0. The spiral indeed becomes straighter as it expands outward.
  3. A particle moves along the curve r(t)=⟨ln⁡(1+t2),arctan⁡(t)⟩\mathbf{r}(t) = \langle \ln(1+t^2), \arctan(t) \rangle for t≥0t \geq 0. Which statement about the curvature is most accurate?
    Answer: The curvature is maximized at t=0t = 0 and approaches 00 as t→∞t \to \infty We have r′(t)=⟨2t1+t2,11+t2⟩\mathbf{r}'(t) = \left\langle \frac{2t}{1+t^2}, \frac{1}{1+t^2} \right\rangle and r′′(t)=⟨2(1−t2)(1+t2)2,−2t(1+t2)2⟩\mathbf{r}''(t) = \left\langle \frac{2(1-t^2)}{(1+t^2)^2}, \frac{-2t}{(1+t^2)^2} \right\rangle. At t=0t=0: r′(0)=⟨0,1⟩\mathbf{r}'(0) = \langle 0, 1 \rangle and r′′(0)=⟨2,0⟩\mathbf{r}''(0) = \langle 2, 0 \rangle, giving κ(0)=2\kappa(0) = 2. As t→∞t \to \infty, both r′(t)\mathbf{r}'(t) and r′′(t)\mathbf{r}''(t) approach ⟨0,0⟩\langle 0, 0 \rangle, but the rate analysis shows κ(t)→0\kappa(t) \to 0. The curve becomes increasingly straight as tt increases, and the maximum curvature occurs at the origin.
  4. A curve is given by r(t)=⟨f(t),g(t)⟩\mathbf{r}(t) = \langle f(t), g(t) \rangle where f′(2)=3f'(2) = 3, f′′(2)=−1f''(2) = -1, g′(2)=4g'(2) = 4, and g′′(2)=2g''(2) = 2. What is the curvature of the curve at t=2t = 2? Give your answer in simplest form.
    Answer: 225\frac{2}{25} The curvature formula is κ=∣f′g′′−g′f′′∣(f′2+g′2)3/2\kappa = \frac{|f'g'' - g'f''|}{(f'^2 + g'^2)^{3/2}}. At t=2t=2: κ=∣3⋅2−4⋅(−1)∣(32+42)3/2=∣6+4∣(9+16)3/2=10253/2=10125\kappa = \frac{|3 \cdot 2 - 4 \cdot (-1)|}{(3^2 + 4^2)^{3/2}} = \frac{|6 + 4|}{(9 + 16)^{3/2}} = \frac{10}{25^{3/2}} = \frac{10}{125}. The other choices represent common computational errors: choice B forgets to cube the denominator, choice C uses the wrong power in the denominator, and choice D has a sign error in the numerator.

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