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Material Type: Quiz; Professor: Ellermeyer; Class: Calculus III; University: Kennesaw State University; Term: Spring 2008;
Typology: Quizzes
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MATH 2203 ñQuiz 3 (Version 2) Solution February 18, 2008
NAME________________________________ The curve r (t) = t cos (t) i + t sin (t) j, t is pictured below.
Graph of r (t) = t cos (t) i + t sin (t) j, t Show that the curvature of this curve at the point (0; 0) is 2. Solution: The point (0; 0) corresponds to t = 0. The curvature at this point is given by (0) = jr
(^0) (0) r (^00) (0)j jr^0 (0)j^3
Note that
r^0 (t) = ( t sin (t) + cos (t)) i + (t cos (t) + sin (t)) j r^00 (t) = ( t cos (t) 2 sin (t)) i + ( t sin (t) + 2 cos (t)) j
so
r^0 (0) = i r^00 (t) = 2j.
1
This gives us r^0 (0) r^00 (0) = i 2 j = 2k and hence jr^0 (0) r^00 (0)j = 2. Thus (0) = 123 = 2.