Explanation Given: The equation of curve is r ( t ) = 〈 1 − t 2 1 + t 2 , 2 t 1 + t 2 〉 Calculation: The arc length parameterization of the r ( t ) = 〈 1 − t 2 1 + t 2 , 2 t 1 + t 2 〉 If r (t) is a differentiable vector valued function, then the arc length function is defined by S ( t ) = ∫ 0 t | | r ′ ( t ) | | d t (I) So, for the equation r (t) is given by r ( t ) = 〈 1 − t 2 1 + t 2 , 2 t 1 + t 2 〉 (II) It is given that x ( t ) = 1 − t 2 1 + t 2 and y ( t ) = 2 t 1 + t 2 Differentiating x (t) and y (t) with respect to t x ′ ( t ) = ( 1 + t 2 ) ( − 2 t ) − ( 1 − t 2 ) ( 2 t ) ( 1 + t 2 ) 2 x ′ ( t ) = − 4 t ( 1 + t 2 ) 2 y ′ ( t ) = ( 1 + t 2 ) ( 2 ) − ( 2 t ) ( 2 t ) ( 1 + t 2 ) 2 y ′ ( t ) = 2 − 2 t 2 ( 1 + t 2 ) 2 Therefore, the differential function is given by r ′ ( t ) = 〈 − 4 t ( 1 + t 2 ) 2 , 2 − 2 t 2 ( 1 + t 2 ) 2 〉 (IiI) Calculating the magnitude of equation (III) | | r ′ ( t ) | | = ( − 4 t ( 1 + t 2 ) 2 ) 2 + ( 2 − 2 t

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Chapter 14.3, Problem 41E

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Arc length of unit circle $r (t) = 〈 \frac{1 - t^{2}}{1 + t^{2}}, \frac{2 t}{1 + t^{2}} 〉$

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Chapter 14.3, Problem 40E

Chapter 14.3, Problem 42E

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Ch. 14.1 - Prob. 1PQ Ch. 14.1 - Prob. 2PQ Ch. 14.1 - Prob. 3PQ Ch. 14.1 - Prob. 4PQ Ch. 14.1 - Prob. 5PQ Ch. 14.1 - Prob. 6PQ Ch. 14.1 - Prob. 1E Ch. 14.1 - Prob. 2E Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E

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Arc length of unit circle r ( t ) = 〈 1 − t 2 1 + t 2 , 2 t 1 + t 2 〉

Solution Summary: The author explains that the arc length of the unit circle is S ( t ) = 2 if r (t) is a differentiable vector valued function.

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Arc length of unit circle $r (t) = 〈 \frac{1 - t^{2}}{1 + t^{2}}, \frac{2 t}{1 + t^{2}} 〉$

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Chapter 14 Solutions

Arc length of unit circle r ( t ) = 〈 1 − t 2 1 + t 2 , 2 t 1 + t 2 〉

Solution Summary: The author explains that the arc length of the unit circle is S ( t ) = 2 if r (t) is a differentiable vector valued function.

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Arc length of unit circle r(t)=〈1−t21+t2,2t1+t2〉

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Chapter 14 Solutions

Arc length of unit circle $r (t) = 〈 \frac{1 - t^{2}}{1 + t^{2}}, \frac{2 t}{1 + t^{2}} 〉$