Explanation Given: y = f ( x ) , P = ( x , f ( x ) ) f ( x ) = 3 x 2 − 8 x + 9 (a) Let L(t) be the distance of the particle from the origin If the location of the particle at time t is P = ( x , f ( x ) ) , thenthe distance of the particle from the origin, L ( t ) = x 2 + f ( x ) 2 On differentiation of the equation of L(t), we get d L d t = 1 2 ( ( x ) 2 + f ( x ) 2 ) ( − 1 2 ) × ( 2 x d x d t +2 f ( x ) f ' ( x ) d x d t ) On simplification, we get d L d t = ( x + f ( x ) f ' ( x ) x 2 + f ( x ) 2 ) d x d t (b) It is given that f ( x ) = 3 x 2 − 8 x + 9 Now, f ' ( x ) = 3 x − 4 3 x 2 − 8 x + 9 When, x = 1; f ( 1 ) = 3 ( 1 ) 2 − 8 ( 1 )

3rd Edition

ISBN: 9781319030728

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 3.10, Problem 35E

To determine

(a) To show that $\frac{d L}{d t} = (\frac{x + f (x) f' (x)}{\sqrt{x^{2} + f {(x)}^{2}}}) \frac{d x}{d t}$

(b) To calculate L’(t) when x = 1 and x = 2 if f(x) = $f (x) = \sqrt{3 x^{2} - 8 x + 9} and \frac{d x}{d t} = 4$

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Chapter 3.10, Problem 34E

Chapter 3.10, Problem 36E

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Chapter 3 Solutions

Calculus: Early Transcendentals

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(a) To show that d L d t = ( x + f ( x ) f ' ( x ) x 2 + f ( x ) 2 ) d x d t (b) To calculate L’(t) when x = 1 and x = 2 if f(x) = f ( x ) = 3 x 2 − 8 x + 9 and d x d t = 4

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(a) To show that dLdt=(x+f(x)f'(x)x2+f(x)2)dxdt (b) To calculate L’(t) when x = 1 and x = 2 if f(x) = f(x)=3x2−8x+9 and dxdt=4

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(a) To show that $\frac{d L}{d t} = (\frac{x + f (x) f' (x)}{\sqrt{x^{2} + f {(x)}^{2}}}) \frac{d x}{d t}$

(b) To calculate L’(t) when x = 1 and x = 2 if f(x) = $f (x) = \sqrt{3 x^{2} - 8 x + 9} and \frac{d x}{d t} = 4$