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Question 21 2 pts ? Consider the curve $\vec{r}(t) = \langle \frac{1}{3}cos^3t, \frac{1}{3}sin^3t, 0\rangle$ where $\frac{\pi}{6} \le t \le \frac{\pi}{2}$. Find the curvature of the curve $\kappa(t)$. (Hint: here $\vec{T}'(t)$ appears easier to compute than $\vec{r}''(t)$. 1 $\kappa(t) = \frac{1}{cos^2tsin^2t}$ $\kappa(t) = cost sint$ 1 $\kappa(t) = \frac{1}{cost sint}$ 1 $\kappa(t) = \frac{-1}{cost sint}$ Next

          Question 21
2 pts
?
Consider the curve \(\vec{r}(t) = \langle \frac{1}{3}cos^3t, \frac{1}{3}sin^3t, 0\rangle\) where \(\frac{\pi}{6} \le t \le \frac{\pi}{2}\). Find the curvature of the curve \(\kappa(t)\). (Hint: here \(\vec{T}'(t)\) appears easier
to compute than \(\vec{r}''(t)\).
1
\(\kappa(t) = \frac{1}{cos^2tsin^2t}\)
\(\kappa(t) = cost sint\)
1
\(\kappa(t) = \frac{1}{cost sint}\)
1
\(\kappa(t) = \frac{-1}{cost sint}\)
Next

Question 21
2 pts
?
Consider the curve r⃗(t) = ⟨(1)/(3)cos^3t, (1)/(3)sin^3t, 0⟩ where (π)/(6)≤ t ≤(π)/(2). Find the curvature of the curve κ(t). (Hint: here T⃗'(t) appears easier
to compute than r⃗”(t).
1
κ(t) = (1)/(cos^2tsin^2t)
κ(t) = cost sint
1
κ(t) = (1)/(cost sint)
1
κ(t) = (-1)/(cost sint)
Next

Added by Carolina L.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Madhur L

07/13/2022

Step 1

$\vec{r}'(t) = \left< -\cos^2 t \sin t, \sin^2 t \cos t, 0 \right>$ $||\vec{r}'(t)|| = \sqrt{(-\cos^2 t \sin t)^2 + (\sin^2 t \cos t)^2 + 0^2} = \sqrt{\cos^4 t \sin^2 t + \sin^4 t \cos^2 t} = \sqrt{\cos^2 t \sin^2 t (\cos^2 t + \sin^2 t)} = \cos t \sin Show more…

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Question 21 2 pts Consider the curve $\vec{r}(t) = \left< \frac{1}{3} \cos^3 t, \frac{1}{3} \sin^3 t, 0 \right>$ where $\frac{\pi}{6} \leq t \leq \frac{\pi}{2}$. Find the curvature of the curve $\kappa(t)$. (Hint: here $\vec{T}'(t)$ appears easier to compute than $\vec{r}''(t)$.) 1 Ο κ(t) = cos² tsin² t κ(t)=cost sin t 1 κ(t) = cost sin t 1 Ο κ(t) = cost sin t Next