Exam2Practice1Sol

# Exam2Practice1Sol - Math 32A Practice Problems for Exam 2...

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Unformatted text preview: Math 32A: Practice Problems for Exam 2 Problem 1. A particle travels along a path r ( t ) with acceleration vector ( t 4 , 4 − t ) . Find r ( t ) if at t = 0, the particle is located at the origin and has initial velocity v = ( 2 , 3 ) . Solution: We have r ′ ( t ) = integraldisplay ( t 4 , 4 − t ) dt = ( t 5 / 5 , 4 t − 1 2 t 2 ) + v = ( t 5 / 5 + 2 , 4 t − 1 2 t 2 + 3 ) Therefore, r ( t ) = integraldisplay ( t 5 / 5 + 2 , 4 t − 1 2 t 2 + 3 ) dt = ( t 6 / 30 + 2 t, 2 t 2 − 1 6 t 3 + 3 t ) + r But r = , so r ( t ) = ( t 6 / 30 + 2 t, 2 t 2 − 1 6 t 3 + 3 t ) . Problem 2. Calculate f xy and f zxzw where f ( x, y, z, w ) = x 2 + ze w y sin( w 2 + w − 1 ) Solution: Since x appears only in the numerator, f x = 2 x y sin( w 2 + w − 1 ) = 2 x sin( w 2 + w − 1 ) parenleftbigg 1 y parenrightbigg Therefore f xy = 2 x sin( w 2 + w − 1 ) parenleftbigg − 1 y 2 parenrightbigg = − 2 x y 2 sin( w 2 + w − 1 ) On the other hand, f z = e w y sin( w 2 + w − 1 ) ⇒ f zz = 0 It follows that f zxzw = f zzxw = 0. Problem 3. r ( s ) be a vector-valued function. What condition must be satisfied in order that r ( s ) be an arc length parametrization? Solution: r ( s ) is an arc length parametrization if || r ′ ( s ) || = 1 for all s . 2 Problem 4. Define the curvature of a path. Solution: Let r ( s ) be an arc length parametrization of the path. Then the curvature at r ( s ) is κ ( s ) = || d T ds || = || r ′′ ( s ) || Problem 5. A particle moves along the spiral path described by r ( t ) = t ( cos t, sin t ) The acceleration vector decomposes as a sum of tangential and normal...
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Exam2Practice1Sol - Math 32A Practice Problems for Exam 2...

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