Quiz_2 - = | r ( t ) r 00 ( t ) | | r ( t ) | 3 . Proof....

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QUIZ 2 FOR CALCULUS III Note that no book, lecture notes or calculators are allowed during the quiz. This quiz has 2 questions and takes 20 minutes. The total score is 20 points. Your Name : Question 1. A particle starts at the origin with initial velocity i - j +3 k . Its acceleration is a ( t ) = 6 t i + 12 t 2 j - 6 t k . Find its position function. (10 points) Proof. We know that r (0) = 0 and v (0) = i - j + 3 k . v ( t ) = Z a ( t ) dt = Z 6 tdt i + Z 12 t 2 dt j + Z - 6 tdt k = (3 t 2 + c 1 ) i + (4 t 3 + c 2 ) j + ( - 3 t 2 + c 3 ) k , = (3 t 2 + 1) i + (4 t 3 - 1) j + ( - 3 t 2 + 3) k , as v (0) = i - j + 3 k . r ( t ) = Z v ( t ) dt = Z (3 t 2 + 1) dt i + Z (4 t 3 - 1) dt j + Z ( - 3 t 2 + 3) dt k = ( t 3 + t + c 1 ) i + ( t 4 - t + c 2 ) j + ( - t 3 + 3 t + c 3 ) k = ( t 3 + t ) i + ( t 4 - t ) j + ( - t 3 + 3 t ) k , as r (0) = 0 . ± Date : 6:15pm – 6:35pm, August 2, 2010. 1
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2 QUIZ 2 FOR CALCULUS III Question 2. Find the curvature of the ellipse x 2 9 + y 2 4 = 1 at the point (0 , 2) . (10 points) Hint: (1) Think of the ellipse as the intersecting curve of the elliptic cylinder x 2 9 + y 2 4 = 1 and the plane z = 0 in space and find out a vector function r ( t ) = < f ( t ) ,g ( t ) ,h ( t ) > such that for every t , the point ( f ( t ) ,g ( t ) ,h ( t )) is on the ellipse. (2) Compute the curvature by using the formula
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Unformatted text preview: = | r ( t ) r 00 ( t ) | | r ( t ) | 3 . Proof. The key point is to remember that the parametric equations of the ellipse can be chosen to be x = 3 cos t,y = 2 sin t or x = 3 sin t,y = 2 cos t . Consider the vector function r ( t ) = &lt; 3 cos t, 2 sin t, &gt;,t R As (3 cos t ) 2 9 + (2 sin t ) 4 = 1 and z = 0, then r ( t ) is the right vector function representing the ellipse on the xy-plane. The point (0 , 2) corresponds to r ( / 2). r ( t ) = &lt;-3 sin t, 2 cos t, &gt;, then r ( / 2) = &lt;-3 , , &gt; . r 00 ( t ) = &lt;-3 cos t,-2 sin t, &gt;, then r 00 ( / 2) = &lt; ,-2 , &gt; . As | r ( / 2) | = 3, and r ( / 2) r 00 ( / 2) = 6 k then = 6 3 3 = 2 9 ....
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This note was uploaded on 02/27/2011 for the course MATH 101 taught by Professor Y during the Spring '11 term at Columbia College.

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Quiz_2 - = | r ( t ) r 00 ( t ) | | r ( t ) | 3 . Proof....

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