Curvature of a plane or space curve at a point, κ = ‖r′ × r″‖ / ‖r′‖³. 39 problems with worked solutions; in 39 of them every
equation is proved by a computer algebra system.
Try each one before opening the solution. Every line is labeled with what checked it. How we verify
Find the curvature of \( \displaystyle \mathbf r(t) = \left\langle t, t^{2}, 0 \right\rangle \) at \( \displaystyle t = -1 \).
Show the first step
Show the whole solution
\[ \left[\begin{matrix}\left. \frac{d}{d t} t \right|_{\substack{ t=-1 }}\\\left. \frac{d}{d t} t^{2} \right|_{\substack{ t=-1 }}\\\left. \frac{d}{d t} 0 \right|_{\substack{ t=-1 }}\end{matrix}\right] = \left[\begin{matrix}1\\-2\\0\end{matrix}\right] \]
r′(-1).✓ Proved
\[ \left[\begin{matrix}\left. \frac{d^{2}}{d t^{2}} t \right|_{\substack{ t=-1 }}\\\left. \frac{d^{2}}{d t^{2}} t^{2} \right|_{\substack{ t=-1 }}\\\left. \frac{d^{2}}{d t^{2}} 0 \right|_{\substack{ t=-1 }}\end{matrix}\right] = \left[\begin{matrix}0\\2\\0\end{matrix}\right] \]
r″(-1).✓ Proved
\[ \left[\begin{matrix}\left(-2\right) 0 + 2 \cdot 0\\0 \cdot 0 + 0\\0 \cdot 2 + 1 \cdot 2\end{matrix}\right] = \left[\begin{matrix}0\\0\\2\end{matrix}\right] \]
r′ × r″.✓ Proved
\[ \frac{\sqrt{0^{2} + 0^{2} + 2^{2}}}{\left(\sqrt{0^{2} + 1^{2} + \left(-2\right)^{2}}\right)^{3}} = \frac{2 \sqrt{5}}{25} \]
κ = ‖r′ × r″‖ / ‖r′‖³.✓ Proved
Answer \( \kappa = \frac{2 \sqrt{5}}{25} \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find the curvature of \( \displaystyle \mathbf r(t) = \left\langle 2 \cos{\left(t \right)}, 2 \sin{\left(t \right)}, t \right\rangle \) at \( \displaystyle t = \frac{\pi}{2} \).
Show the first step
Show the whole solution
\[ \left[\begin{matrix}\left. \frac{d}{d t} 2 \cos{\left(t \right)} \right|_{\substack{ t=\frac{\pi}{2} }}\\\left. \frac{d}{d t} 2 \sin{\left(t \right)} \right|_{\substack{ t=\frac{\pi}{2} }}\\\left. \frac{d}{d t} t \right|_{\substack{ t=\frac{\pi}{2} }}\end{matrix}\right] = \left[\begin{matrix}-2\\0\\1\end{matrix}\right] \]
r′(pi/2).✓ Proved
\[ \left[\begin{matrix}\left. \frac{d^{2}}{d t^{2}} 2 \cos{\left(t \right)} \right|_{\substack{ t=\frac{\pi}{2} }}\\\left. \frac{d^{2}}{d t^{2}} 2 \sin{\left(t \right)} \right|_{\substack{ t=\frac{\pi}{2} }}\\\left. \frac{d^{2}}{d t^{2}} t \right|_{\substack{ t=\frac{\pi}{2} }}\end{matrix}\right] = \left[\begin{matrix}0\\-2\\0\end{matrix}\right] \]
r″(pi/2).✓ Proved
\[ \left[\begin{matrix}0 \cdot 0 + 2\\1 \cdot 0 + 0 \cdot 2\\0 \cdot 0 - 2 \left(-2\right)\end{matrix}\right] = \left[\begin{matrix}2\\0\\4\end{matrix}\right] \]
r′ × r″.✓ Proved
\[ \frac{\sqrt{0^{2} + 2^{2} + 4^{2}}}{\left(\sqrt{0^{2} + 1^{2} + \left(-2\right)^{2}}\right)^{3}} = \frac{2}{5} \]
κ = ‖r′ × r″‖ / ‖r′‖³.✓ Proved
Answer \( \kappa = \frac{2}{5} \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find the curvature of \( \displaystyle \mathbf r(t) = \left\langle 2 \cos{\left(t \right)}, 2 \sin{\left(t \right)}, 0 \right\rangle \) at \( \displaystyle t = 1 \).
Show the first step
Show the whole solution
\[ \left[\begin{matrix}\left. \frac{d}{d t} 2 \cos{\left(t \right)} \right|_{\substack{ t=1 }}\\\left. \frac{d}{d t} 2 \sin{\left(t \right)} \right|_{\substack{ t=1 }}\\\left. \frac{d}{d t} 0 \right|_{\substack{ t=1 }}\end{matrix}\right] = \left[\begin{matrix}- 2 \sin{\left(1 \right)}\\2 \cos{\left(1 \right)}\\0\end{matrix}\right] \]
r′(1).✓ Proved
\[ \left[\begin{matrix}\left. \frac{d^{2}}{d t^{2}} 2 \cos{\left(t \right)} \right|_{\substack{ t=1 }}\\\left. \frac{d^{2}}{d t^{2}} 2 \sin{\left(t \right)} \right|_{\substack{ t=1 }}\\\left. \frac{d^{2}}{d t^{2}} 0 \right|_{\substack{ t=1 }}\end{matrix}\right] = \left[\begin{matrix}- 2 \cos{\left(1 \right)}\\- 2 \sin{\left(1 \right)}\\0\end{matrix}\right] \]
r″(1).✓ Proved
\[ \left[\begin{matrix}2 \cos{\left(1 \right)} 0 + 0 \left(-2\right) \sin{\left(1 \right)}\\0 \left(-2\right) \cos{\left(1 \right)} - 1 \left(-2\right) \sin{\left(1 \right)} 0\\2 \cos{\left(1 \right)} 2 \cos{\left(1 \right)} - 2 \sin{\left(1 \right)} \left(-2\right) \sin{\left(1 \right)}\end{matrix}\right] = \left[\begin{matrix}0\\0\\4 \cos^{2}{\left(1 \right)} + 4 \sin^{2}{\left(1 \right)}\end{matrix}\right] \]
r′ × r″.✓ Proved
\[ \frac{1}{\sqrt{4 \cos^{2}{\left(1 \right)} + 4 \sin^{2}{\left(1 \right)}}} = \frac{1}{2} \]
κ = ‖r′ × r″‖ / ‖r′‖³.✓ Proved
Answer \( \kappa = \frac{1}{2} \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find the curvature of \( \displaystyle \mathbf r(t) = \left\langle t, \sin{\left(t \right)}, 0 \right\rangle \) at \( \displaystyle t = \frac{\pi}{2} \).
Show the first step
Show the whole solution
\[ \left[\begin{matrix}\left. \frac{d}{d t} t \right|_{\substack{ t=\frac{\pi}{2} }}\\\left. \frac{d}{d t} \sin{\left(t \right)} \right|_{\substack{ t=\frac{\pi}{2} }}\\\left. \frac{d}{d t} 0 \right|_{\substack{ t=\frac{\pi}{2} }}\end{matrix}\right] = \left[\begin{matrix}1\\0\\0\end{matrix}\right] \]
r′(pi/2).✓ Proved
\[ \left[\begin{matrix}\left. \frac{d^{2}}{d t^{2}} t \right|_{\substack{ t=\frac{\pi}{2} }}\\\left. \frac{d^{2}}{d t^{2}} \sin{\left(t \right)} \right|_{\substack{ t=\frac{\pi}{2} }}\\\left. \frac{d^{2}}{d t^{2}} 0 \right|_{\substack{ t=\frac{\pi}{2} }}\end{matrix}\right] = \left[\begin{matrix}0\\-1\\0\end{matrix}\right] \]
r″(pi/2).✓ Proved
\[ \left[\begin{matrix}0 \cdot 0 - 1 \cdot 0\\0 \cdot 0 + 0\\1 \left(-1\right) + 0 \cdot 0\end{matrix}\right] = \left[\begin{matrix}0\\0\\-1\end{matrix}\right] \]
r′ × r″.✓ Proved
\[ \frac{\sqrt{0^{2} + 0^{2} + \left(-1\right)^{2}}}{\left(\sqrt{0^{2} + 0^{2} + 1^{2}}\right)^{3}} = 1 \]
κ = ‖r′ × r″‖ / ‖r′‖³.✓ Proved
Answer \( \kappa = 1 \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find the curvature of \( \displaystyle \mathbf r(t) = \left\langle 4 \cos{\left(t \right)}, 4 \sin{\left(t \right)}, t \right\rangle \) at \( \displaystyle t = 1 \).
Show the first step
Show the whole solution
\[ \left[\begin{matrix}\left. \frac{d}{d t} 4 \cos{\left(t \right)} \right|_{\substack{ t=1 }}\\\left. \frac{d}{d t} 4 \sin{\left(t \right)} \right|_{\substack{ t=1 }}\\\left. \frac{d}{d t} t \right|_{\substack{ t=1 }}\end{matrix}\right] = \left[\begin{matrix}- 4 \sin{\left(1 \right)}\\4 \cos{\left(1 \right)}\\1\end{matrix}\right] \]
r′(1).✓ Proved
\[ \left[\begin{matrix}\left. \frac{d^{2}}{d t^{2}} 4 \cos{\left(t \right)} \right|_{\substack{ t=1 }}\\\left. \frac{d^{2}}{d t^{2}} 4 \sin{\left(t \right)} \right|_{\substack{ t=1 }}\\\left. \frac{d^{2}}{d t^{2}} t \right|_{\substack{ t=1 }}\end{matrix}\right] = \left[\begin{matrix}- 4 \cos{\left(1 \right)}\\- 4 \sin{\left(1 \right)}\\0\end{matrix}\right] \]
r″(1).✓ Proved
\[ \left[\begin{matrix}4 \cos{\left(1 \right)} 0 - \left(-1\right) 4 \sin{\left(1 \right)}\\1 \left(-4\right) \cos{\left(1 \right)} - 1 \left(-4\right) \sin{\left(1 \right)} 0\\4 \cos{\left(1 \right)} 4 \cos{\left(1 \right)} - 4 \sin{\left(1 \right)} \left(-4\right) \sin{\left(1 \right)}\end{matrix}\right] = \left[\begin{matrix}4 \sin{\left(1 \right)}\\- 4 \cos{\left(1 \right)}\\16 \cos^{2}{\left(1 \right)} + 16 \sin^{2}{\left(1 \right)}\end{matrix}\right] \]
r′ × r″.✓ Proved
\[ \frac{\sqrt{16 \cos^{2}{\left(1 \right)} + 16 \sin^{2}{\left(1 \right)} + \left(16 \cos^{2}{\left(1 \right)} + 16 \sin^{2}{\left(1 \right)}\right)^{2}}}{\left(1 + 16 \cos^{2}{\left(1 \right)} + 16 \sin^{2}{\left(1 \right)}\right)^{\frac{3}{2}}} = \frac{4}{17} \]
κ = ‖r′ × r″‖ / ‖r′‖³.✓ Proved
Answer \( \kappa = \frac{4}{17} \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find the curvature of \( \displaystyle \mathbf r(t) = \left\langle t, t^{2}, 0 \right\rangle \) at \( \displaystyle t = 0 \).
Show the first step
Show the whole solution
\[ \left[\begin{matrix}\left. \frac{d}{d t} t \right|_{\substack{ t=0 }}\\\left. \frac{d}{d t} t^{2} \right|_{\substack{ t=0 }}\\\left. \frac{d}{d t} 0 \right|_{\substack{ t=0 }}\end{matrix}\right] = \left[\begin{matrix}1\\0\\0\end{matrix}\right] \]
r′(0).✓ Proved
\[ \left[\begin{matrix}\left. \frac{d^{2}}{d t^{2}} t \right|_{\substack{ t=0 }}\\\left. \frac{d^{2}}{d t^{2}} t^{2} \right|_{\substack{ t=0 }}\\\left. \frac{d^{2}}{d t^{2}} 0 \right|_{\substack{ t=0 }}\end{matrix}\right] = \left[\begin{matrix}0\\2\\0\end{matrix}\right] \]
r″(0).✓ Proved
\[ \left[\begin{matrix}0 \cdot 0 + 2 \cdot 0\\0 \cdot 0 + 0\\0 \cdot 0 + 1 \cdot 2\end{matrix}\right] = \left[\begin{matrix}0\\0\\2\end{matrix}\right] \]
r′ × r″.✓ Proved
\[ \frac{\sqrt{0^{2} + 0^{2} + 2^{2}}}{\left(\sqrt{0^{2} + 0^{2} + 1^{2}}\right)^{3}} = 2 \]
κ = ‖r′ × r″‖ / ‖r′‖³.✓ Proved
Answer \( \kappa = 2 \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find the curvature of \( \displaystyle \mathbf r(t) = \left\langle t, t^{2}, t^{3} \right\rangle \) at \( \displaystyle t = 1 \).
Show the first step
Show the whole solution
\[ \left[\begin{matrix}\left. \frac{d}{d t} t \right|_{\substack{ t=1 }}\\\left. \frac{d}{d t} t^{2} \right|_{\substack{ t=1 }}\\\left. \frac{d}{d t} t^{3} \right|_{\substack{ t=1 }}\end{matrix}\right] = \left[\begin{matrix}1\\2\\3\end{matrix}\right] \]
r′(1).✓ Proved
\[ \left[\begin{matrix}\left. \frac{d^{2}}{d t^{2}} t \right|_{\substack{ t=1 }}\\\left. \frac{d^{2}}{d t^{2}} t^{2} \right|_{\substack{ t=1 }}\\\left. \frac{d^{2}}{d t^{2}} t^{3} \right|_{\substack{ t=1 }}\end{matrix}\right] = \left[\begin{matrix}0\\2\\6\end{matrix}\right] \]
r″(1).✓ Proved
\[ \left[\begin{matrix}2 \left(-3\right) + 2 \cdot 6\\-6 + 3 \cdot 0\\0 \left(-2\right) + 1 \cdot 2\end{matrix}\right] = \left[\begin{matrix}6\\-6\\2\end{matrix}\right] \]
r′ × r″.✓ Proved
\[ \frac{\sqrt{2^{2} + 6^{2} + \left(-6\right)^{2}}}{\left(\sqrt{1^{2} + 2^{2} + 3^{2}}\right)^{3}} = \frac{\sqrt{266}}{98} \]
κ = ‖r′ × r″‖ / ‖r′‖³.✓ Proved
Answer \( \kappa = \frac{\sqrt{266}}{98} \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find the curvature of \( \displaystyle \mathbf r(t) = \left\langle 3 \cos{\left(t \right)}, 3 \sin{\left(t \right)}, 0 \right\rangle \) at \( \displaystyle t = \frac{\pi}{2} \).
Show the first step
Show the whole solution
\[ \left[\begin{matrix}\left. \frac{d}{d t} 3 \cos{\left(t \right)} \right|_{\substack{ t=\frac{\pi}{2} }}\\\left. \frac{d}{d t} 3 \sin{\left(t \right)} \right|_{\substack{ t=\frac{\pi}{2} }}\\\left. \frac{d}{d t} 0 \right|_{\substack{ t=\frac{\pi}{2} }}\end{matrix}\right] = \left[\begin{matrix}-3\\0\\0\end{matrix}\right] \]
r′(pi/2).✓ Proved
\[ \left[\begin{matrix}\left. \frac{d^{2}}{d t^{2}} 3 \cos{\left(t \right)} \right|_{\substack{ t=\frac{\pi}{2} }}\\\left. \frac{d^{2}}{d t^{2}} 3 \sin{\left(t \right)} \right|_{\substack{ t=\frac{\pi}{2} }}\\\left. \frac{d^{2}}{d t^{2}} 0 \right|_{\substack{ t=\frac{\pi}{2} }}\end{matrix}\right] = \left[\begin{matrix}0\\-3\\0\end{matrix}\right] \]
r″(pi/2).✓ Proved
\[ \left[\begin{matrix}0 \cdot 0 - 3 \cdot 0\\0 \cdot 0 + 0 \cdot 3\\0 \cdot 0 - 3 \left(-3\right)\end{matrix}\right] = \left[\begin{matrix}0\\0\\9\end{matrix}\right] \]
r′ × r″.✓ Proved
\[ \frac{\sqrt{0^{2} + 0^{2} + 9^{2}}}{\left(\sqrt{0^{2} + 0^{2} + \left(-3\right)^{2}}\right)^{3}} = \frac{1}{3} \]
κ = ‖r′ × r″‖ / ‖r′‖³.✓ Proved
Answer \( \kappa = \frac{1}{3} \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find the curvature of \( \displaystyle \mathbf r(t) = \left\langle \cos{\left(t \right)}, \sin{\left(t \right)}, 0 \right\rangle \) at \( \displaystyle t = 1 \).
Show the first step
Show the whole solution
\[ \left[\begin{matrix}\left. \frac{d}{d t} \cos{\left(t \right)} \right|_{\substack{ t=1 }}\\\left. \frac{d}{d t} \sin{\left(t \right)} \right|_{\substack{ t=1 }}\\\left. \frac{d}{d t} 0 \right|_{\substack{ t=1 }}\end{matrix}\right] = \left[\begin{matrix}- \sin{\left(1 \right)}\\\cos{\left(1 \right)}\\0\end{matrix}\right] \]
r′(1).✓ Proved
\[ \left[\begin{matrix}\left. \frac{d^{2}}{d t^{2}} \cos{\left(t \right)} \right|_{\substack{ t=1 }}\\\left. \frac{d^{2}}{d t^{2}} \sin{\left(t \right)} \right|_{\substack{ t=1 }}\\\left. \frac{d^{2}}{d t^{2}} 0 \right|_{\substack{ t=1 }}\end{matrix}\right] = \left[\begin{matrix}- \cos{\left(1 \right)}\\- \sin{\left(1 \right)}\\0\end{matrix}\right] \]
r″(1).✓ Proved
\[ \left[\begin{matrix}\cos{\left(1 \right)} 0 + 0 \left(- \sin{\left(1 \right)}\right)\\0 \left(- \cos{\left(1 \right)}\right) - 1 \left(- \sin{\left(1 \right)}\right) 0\\\cos{\left(1 \right)} \cos{\left(1 \right)} + - \sin{\left(1 \right)} \left(- \sin{\left(1 \right)}\right)\end{matrix}\right] = \left[\begin{matrix}0\\0\\\cos^{2}{\left(1 \right)} + \sin^{2}{\left(1 \right)}\end{matrix}\right] \]
r′ × r″.✓ Proved
\[ \frac{1}{\sqrt{\cos^{2}{\left(1 \right)} + \sin^{2}{\left(1 \right)}}} = 1 \]
κ = ‖r′ × r″‖ / ‖r′‖³.✓ Proved
Answer \( \kappa = 1 \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find the curvature of \( \displaystyle \mathbf r(t) = \left\langle \cos{\left(t \right)}, \sin{\left(t \right)}, 0 \right\rangle \) at \( \displaystyle t = \frac{\pi}{2} \).
Show the first step
Show the whole solution
\[ \left[\begin{matrix}\left. \frac{d}{d t} \cos{\left(t \right)} \right|_{\substack{ t=\frac{\pi}{2} }}\\\left. \frac{d}{d t} \sin{\left(t \right)} \right|_{\substack{ t=\frac{\pi}{2} }}\\\left. \frac{d}{d t} 0 \right|_{\substack{ t=\frac{\pi}{2} }}\end{matrix}\right] = \left[\begin{matrix}-1\\0\\0\end{matrix}\right] \]
r′(pi/2).✓ Proved
\[ \left[\begin{matrix}\left. \frac{d^{2}}{d t^{2}} \cos{\left(t \right)} \right|_{\substack{ t=\frac{\pi}{2} }}\\\left. \frac{d^{2}}{d t^{2}} \sin{\left(t \right)} \right|_{\substack{ t=\frac{\pi}{2} }}\\\left. \frac{d^{2}}{d t^{2}} 0 \right|_{\substack{ t=\frac{\pi}{2} }}\end{matrix}\right] = \left[\begin{matrix}0\\-1\\0\end{matrix}\right] \]
r″(pi/2).✓ Proved
\[ \left[\begin{matrix}0 \cdot 0 - 1 \cdot 0\\0 \cdot 0 + 0\\0 \cdot 0 - 1 \left(-1\right)\end{matrix}\right] = \left[\begin{matrix}0\\0\\1\end{matrix}\right] \]
r′ × r″.✓ Proved
\[ \frac{\sqrt{0^{2} + 0^{2} + 1^{2}}}{\left(\sqrt{0^{2} + 0^{2} + \left(-1\right)^{2}}\right)^{3}} = 1 \]
κ = ‖r′ × r″‖ / ‖r′‖³.✓ Proved
Answer \( \kappa = 1 \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find the curvature of \( \displaystyle \mathbf r(t) = \left\langle \cos{\left(t \right)}, \sin{\left(t \right)}, t \right\rangle \) at \( \displaystyle t = \frac{\pi}{2} \).
Show the first step
Show the whole solution
\[ \left[\begin{matrix}\left. \frac{d}{d t} \cos{\left(t \right)} \right|_{\substack{ t=\frac{\pi}{2} }}\\\left. \frac{d}{d t} \sin{\left(t \right)} \right|_{\substack{ t=\frac{\pi}{2} }}\\\left. \frac{d}{d t} t \right|_{\substack{ t=\frac{\pi}{2} }}\end{matrix}\right] = \left[\begin{matrix}-1\\0\\1\end{matrix}\right] \]
r′(pi/2).✓ Proved
\[ \left[\begin{matrix}\left. \frac{d^{2}}{d t^{2}} \cos{\left(t \right)} \right|_{\substack{ t=\frac{\pi}{2} }}\\\left. \frac{d^{2}}{d t^{2}} \sin{\left(t \right)} \right|_{\substack{ t=\frac{\pi}{2} }}\\\left. \frac{d^{2}}{d t^{2}} t \right|_{\substack{ t=\frac{\pi}{2} }}\end{matrix}\right] = \left[\begin{matrix}0\\-1\\0\end{matrix}\right] \]
r″(pi/2).✓ Proved
\[ \left[\begin{matrix}0 \cdot 0 + 1\\1 \cdot 0 + 0\\0 \cdot 0 - 1 \left(-1\right)\end{matrix}\right] = \left[\begin{matrix}1\\0\\1\end{matrix}\right] \]
r′ × r″.✓ Proved
\[ \frac{\sqrt{0^{2} + 1^{2} + 1^{2}}}{\left(\sqrt{0^{2} + \left(-1\right)^{2} + 1^{2}}\right)^{3}} = \frac{1}{2} \]
κ = ‖r′ × r″‖ / ‖r′‖³.✓ Proved
Answer \( \kappa = \frac{1}{2} \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find the curvature of \( \displaystyle \mathbf r(t) = \left\langle \cos{\left(t \right)}, \sin{\left(t \right)}, t \right\rangle \) at \( \displaystyle t = 0 \).
Show the first step
Show the whole solution
\[ \left[\begin{matrix}\left. \frac{d}{d t} \cos{\left(t \right)} \right|_{\substack{ t=0 }}\\\left. \frac{d}{d t} \sin{\left(t \right)} \right|_{\substack{ t=0 }}\\\left. \frac{d}{d t} t \right|_{\substack{ t=0 }}\end{matrix}\right] = \left[\begin{matrix}0\\1\\1\end{matrix}\right] \]
r′(0).✓ Proved
\[ \left[\begin{matrix}\left. \frac{d^{2}}{d t^{2}} \cos{\left(t \right)} \right|_{\substack{ t=0 }}\\\left. \frac{d^{2}}{d t^{2}} \sin{\left(t \right)} \right|_{\substack{ t=0 }}\\\left. \frac{d^{2}}{d t^{2}} t \right|_{\substack{ t=0 }}\end{matrix}\right] = \left[\begin{matrix}-1\\0\\0\end{matrix}\right] \]
r″(0).✓ Proved
\[ \left[\begin{matrix}1 \cdot 0 + 0\\1 \left(-1\right) + 0 \cdot 0\\0 \cdot 0 + 1\end{matrix}\right] = \left[\begin{matrix}0\\-1\\1\end{matrix}\right] \]
r′ × r″.✓ Proved
\[ \frac{\sqrt{0^{2} + \left(-1\right)^{2} + 1^{2}}}{\left(\sqrt{0^{2} + 1^{2} + 1^{2}}\right)^{3}} = \frac{1}{2} \]
κ = ‖r′ × r″‖ / ‖r′‖³.✓ Proved
Answer \( \kappa = \frac{1}{2} \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify