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Quiz 3 Version B with Resolution - Calculus III - Fall 2008 | MATH 2203, Quizzes of Advanced Calculus

Material Type: Quiz; Professor: Ellermeyer; Class: Calculus III; University: Kennesaw State University; Term: Spring 2008;

Typology: Quizzes

2010/2011

Uploaded on 06/03/2011

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MATH 2203 –Quiz 3 (Version 2) Solution
February 18, 2008
NAME________________________________
The curve r(t) = tcos (t)i+tsin (t)j,tis pictured below.
Graph of r(t) = tcos (t)i+tsin (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) = jr0(0) r00 (0)j
jr0(0)j3.
Note that
r0(t) = (tsin (t) + cos (t)) i+ (tcos (t) + sin(t)) j
r00 (t) = (tcos (t)2 sin (t)) i+ (tsin (t) + 2 cos(t)) j
so
r0(0) = i
r00 (t) = 2j.
1
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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.