HW09-solutions - huynh(lth436 HW09 gilbert(56690 This...

This preview shows page 1 - 3 out of 8 pages.

huynh (lth436) – HW09 – gilbert – (56690) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points The astroid shown in x y is the curve c ( t ) = 2 cos 3 t i + 2 sin 3 t j . Determine the arc length of the astroid. 1. arc length = 2 π 2. arc length = 2 3. arc length = 6 π 4. arc length = 12 π 5. arc length = 12 correct 6. arc length = 6 Explanation: The arc length of a curve c ( t ), a t b , is given by the integral expression I = integraldisplay b a bardbl c ( t ) bardbl dt . But when c ( t ) = 2 cos 3 t i + 2 sin 3 t j , we see that c ( t ) = 6 sin t cos 2 t i + 6 cos t sin 2 t j . In this case bardbl c ( t ) bardbl = radicalBig (6 cos t sin t ) 2 (cos 2 t + sin 2 t ) , which simplifies to bardbl c ( t ) bardbl = | 6 cos t sin t | . On the other hand, c ( t ) traces out the astroid as t ranges from 0 to 2 π . Thus I = 6 integraldisplay 2 π 0 | cos t sin t | dt = 24 integraldisplay π/ 2 0 cos t sin t dt = 12 bracketleftBig sin 2 t bracketrightBig π/ 2 0 . Consequently, the astroid has arc length = 12 . keywords: arc length, parametric curve, as- troid trig functions, definite integral 002 10.0points Find the arc length of the curve r ( t ) = (4 + 2 t ) i + e t j + (3 e t ) k between r (0) and r (5). 1. arc length = ( e 5 + e 5 ) 2 2. arc length = e 5 + e 5 3. arc length = 2 e 5 4. arc length = ( e 5 e 5 ) 2 5. arc length = e 5 e 5 correct 6. arc length = 2 e 5 Explanation: The length of a curve r ( t ) between r ( t 0 ) and r ( t 1 ) is given by the integral L = integraldisplay t 1 t 0 | r ( t ) | dt .
huynh (lth436) – HW09 – gilbert – (56690) 2 Now when r ( t ) = (4 + 2 t ) i + e t j + (3 e t ) k , we see that r ( t ) = 2 i + e t j + e t k . But then | r ( t ) | = (2 + e 2 t + e 2 t ) 1 / 2 = e t + e t . Thus L = integraldisplay 5 0 ( e t + e t ) dt = bracketleftBig e t e t bracketrightBig 5 0 . Consequently, arc length = L = e 5 e 5 . 003 10.0points The curve C is parametrized by c ( t ) = (4 2 t ) i + ln(2 t ) j + (5 + t 2 ) k . Find the arc length of C between c (1) and c (4). 1. arc length = 4 + ln 8 2. arc length = 15 ln 4 3. arc length = 15 + ln 4 correct 4. arc length = 15 2 ln 4 5. arc length = 16 + 2 ln 4 6. arc length = 8 ln 4 Explanation: The arc length of C between c ( t 0 ) and c ( t 1 ) is given by the integral L = integraldisplay t 1 t 0 | c ( t ) | dt . Now when c ( t ) = (4 2 t ) i + ln(2 t ) j + (5 + t 2 ) k we see that c ( t ) = 2 i + 1 t j + 2 t k . But then | c ( t ) | = parenleftBig 4 + 1 t 2 + 4 t 2 parenrightBig 1 / 2 = 2 t 2 + 1 t . Thus L = integraldisplay 4 1 parenleftBig 2 t + 1 t parenrightBig dt = bracketleftBig t 2 + ln t bracketrightBig 4 1 . Consequently, arc length = L = 15 + ln 4 . 004 10.0points Find the unit tangent vector T ( t ) to the graph of the vector function r ( t ) = ( 3 sin t, +4 t, +3 cos t ) .

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture