Explanation Given: The position vector is r ( t ) = 3 t i ^ − 6 t j ^ and the time at which the components are to be calculated is t = 3 . Formula used: The tangential component of acceleration is given by a T = v . a ‖ v ‖ here v is the velocity of the plane curve. The normal component of acceleration is given by, a N = ‖ v × a ‖ ‖ v ‖ Consider two vectors, a = a i ^ + b j ^ + c k ^ and b → = x i ^ + y j ^ + z k ^ then their dot product is given by, a . b → = a x + b y + c z The cross product of the vectors is given by, a → × b → = | i ^ j ^ k ^ a b c x y z | The derivative of x n is given by, d d x ( x n ) = n x n − 1 The velocity vector of the curve is given by, v ( t ) = d d x r ( t ) The acceleration vector is given by, a ( t ) = d d x ( v ( t ) ) Calculation: The position vector of the curve is, r ( t ) = 3 t i ^ − 6 t j ^ . Differentiate the above vector to obtain the velocity of the curve. Thus, v ( t ) = d d x r ( t ) = d d x ( 3 t i ^ − 6 t j ^ ) = d d x ( 3 t i ^ ) − d d x ( 6 t j ^ ) Use the formula, d d x ( x n ) = n x n − 1 to obtain the above derivative, Thus, v ( t ) = ( − 3 t 2 i ^ ) − 6 j ^ Differentiate the above vector to obtain the acceleration vector of the curve, a ( t ) = d d x ( v ( t ) ) = d d x ( − 3 t 2 i ^ − 6 j ^ ) Use the formula, d d x ( x n ) = n x n − 1 to obtain the above derivative, a ( t ) = d d x ( − 3 t 2 i ^ − 6 j ^ ) = 6 t 3 i ^ The modulus of the velocity vector is, ‖ v ( t ) ‖ = ‖ ( − 3 t 2 i ^ ) − 6 j ^ ‖ = ( − 3 t 2 ) 2 + 36 = 9 + 36 t 4 t 4 Therefore, modulus of the velocity vector is, ‖ v ( t ) ‖ = 9 + 36 t 4 t 2 The tangential component of the acceleration of the curve is given by, a T = v

Calculus: Early Transcendental Functions

7th Edition

ISBN: 9781337552516

Author: Ron Larson, Bruce H. Edwards

Publisher: Cengage Learning

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Chapter 12, Problem 51RE

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To calculate: The tangential and normal components of acceleration at time t=3 when the position vector is, r(t)=3ti^−6tj^.

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Chapter 12, Problem 50RE

Chapter 12, Problem 52RE

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The tangential and normal components of acceleration at time t = 3 when the position vector is, r ( t ) = 3 t i ^ − 6 t j ^ .

Solution Summary: The author explains the formula used to calculate the tangential and normal components of acceleration at time t=3.

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To calculate: The tangential and normal components of acceleration at time t=3 when the position vector is, r(t)=3ti^−6tj^.

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Chapter 12 Solutions