Explanation Given information: The parabola is, y = x 2 (1) The constant speed ( v ) of the object is 10 units/sec. Calculation: The general expression for the acceleration vector . a = a T T + a N N (2) The general expression to calculate the tangential scalar component of acceleration. a T = d d t | v | (3) The general expression to calculate the normal scalar component of acceleration. a N = κ | v | 2 (4) Substitute 10 for v in equation (3) to find a T . a T = d d t | 10 | = 0 Substitute 10 for v in equation (4) to find a N . a N = κ | 10 | 2 = 100 κ Substitute 0 for a T and 100 κ for a N in equation (2) to find a . a = ( 0 ) T + ( 100 κ ) N = ( 100 κ ) N From equation (1), the function f ( x ) can be written as, y = f ( x ) = x 2 On successive differentiation of the above equation as follows, f ' ( x ) = 2 x f ' ' ( x ) = 2 Refer Exercises 5a (in section 12.4) in the textbook. A formula for the curvature of the graph of a function in the xy -plane is given by, κ = | f ' ' ( x ) | ( 1 + [ f ' ( x ) ] 2 ) 3 2 (5) Refer EXAMPLE 4 in section 12.4 in the textbook. The position vector of the moving mass is, r ( t ) = t i + t 2 j (6) Substitute 2 x for f ' ( x ) and 2 for f ' ' ( x ) in equation (5) to find κ . κ = | 2 | ( 1 + [ 2 x ] 2 ) 3 2 κ = 2 ( 1 + 4 x 2 ) 3 2 (7) The general expression to calculate the velocity. v = d r ( t ) d t (8) Here, d r d t is the rate of change in displacement with respect to time. Substitute t i + t 2 j for r ( t ) in equation (8) to find v . v = d d t ( t i + t 2 j ) = 1 i + 2 t j The magnitude of v is determined as follows, | v | = 1 2 + ( 2 t ) 2 = 1 + 4 t 2 The unit tangent vector is, T = v | v | (9) Substitute 1 + 4 t 2 for | v | and 1 i + 2 t j for v in equation (9) to find T

Thomas' Calculus Format: Unbound (saleable) With Access Card

14th Edition

ISBN: 9780134768762

Author: Hass, Joel R.^heil, Christopher D.^weir, Maurice D.

Publisher: Prentice Hall

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Chapter 13.5, Problem 20E

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Calculate the force on the object due to its acceleration at (0, 0) and (2^1/2, 2) in terms of i and j.

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Chapter 13 Solutions

Thomas' Calculus Format: Unbound (saleable) With Access Card

Ch. 13.1 - In Exercises 1–4, find the given limits. 1. Ch. 13.1 - In Exercises 1–4, find the given limits. 2. Ch. 13.1 - In Exercises 1–4, find the given limits. 3. Ch. 13.1 - In Exercises 1–4, find the given limits. 4. Ch. 13.1 - Motion in the Plane In Exercises 5–8, r(t) is the...Ch. 13.1 - Motion in the Plane In Exercises 5–8, r(t) is the...Ch. 13.1 - Prob. 7E Ch. 13.1 - Prob. 8E Ch. 13.1 - Prob. 9E Ch. 13.1 - Prob. 10E

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Calculate the force on the object due to its acceleration at (0, 0) and (2 1/2 , 2) in terms of i and j .

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Chapter 13 Solutions

Calculate the force on the object due to its acceleration at (0, 0) and (2^1/2, 2) in terms of i and j.