Explanation Given information: The first fundamental theorem using single variable calculus for the vector valued functions. A vector valued function r ( t ) = 〈 x ( t ) , y ( t ) , z ( t ) 〉 is differentiable if and only if each component is differentiable, and in this case, r ' ( t ) = d r ( t ) d t = 〈 x ' ( t ) , y ' ( t ) , z ' ( t ) 〉 If we use single variable calculus it will be; r ' ( t ) = d r ( t ) d t = 〈 x ' ( t ) 〉 Formula Used: If we use single variable calculus

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Chapter 14.2, Problem 70E

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To Prove:The second fundamental theorem using single variable calculus for the vector valued functions.

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Chapter 14.2, Problem 69E

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5. [-/1 Points] DETAILS MY NOTES SESSCALCET2 6.5.AE.003. y y= ex² 0 Video Example x EXAMPLE 3 (a) Use the Midpoint Rule with n = 10 to approximate the integral कर L'ex² dx. (b) Give an upper bound for the error involved in this approximation. SOLUTION 8+2 1 L'ex² d (a) Since a = 0, b = 1, and n = 10, the Midpoint Rule gives the following. (Round your answer to six decimal places.) dx Ax[f(0.05) + f(0.15) + ... + f(0.85) + f(0.95)] 0.1 [0.0025 +0.0225 + + e0.0625 + 0.1225 e0.3025 + e0.4225 + e0.2025 + + e0.5625 €0.7225 +0.9025] The figure illustrates this approximation. (b) Since f(x) = ex², we have f'(x) = 0 ≤ f'(x) = < 6e. ASK YOUR TEACHER and f'(x) = Also, since 0 ≤ x ≤ 1 we have x² ≤ and so Taking K = 6e, a = 0, b = 1, and n = 10 in the error estimate, we see that an upper bound for the error is as follows. (Round your final answer to five decimal places.) 6e(1)3 e 24( = ≈

Chapter 14 Solutions

CALCULUS (CLOTH)

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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To Prove: The second fundamental theorem using single variable calculus for the vector valued functions.

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