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Find beR so that 4x² dy – 2xy dæ = 145, Nhere C is the segment (1, –5) a (-2, –3), followed by the part y =1-x² x = -2 a x = (2, —3) а (4, 6). 2, followed by the segment of %3D Choose an alternative: a. None of the alternatives b. 1 C. –5 d. -3 е. 2 f. 0

Find beR so that 4x² dy – 2xy dæ = 145, Nhere C is the segment (1, –5) a (-2, –3), followed by the part y =1-x² x = -2 a x = (2, —3) а (4, 6). 2, followed by the segment of %3D Choose an alternative: a. None of the alternatives b. 1 C. –5 d. -3 е. 2 f. 0

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Find beR so that | 4a" dy – 2zy dz = 145, Where C is the segment (1, –5) a (-2, –3), followed by the part y =1-x² x = -2 a x = 2,followed by the segment of (2, -3) а (4, 5). Choose an alternative: a. None of the alternatives b. 1 с. —5 d. –3 e. 2 f. 0
Transcribed Image Text:Find beR so that | 4a" dy – 2zy dz = 145, Where C is the segment (1, –5) a (-2, –3), followed by the part y =1-x² x = -2 a x = 2,followed by the segment of (2, -3) а (4, 5). Choose an alternative: a. None of the alternatives b. 1 с. —5 d. –3 e. 2 f. 0
Expert Solution
Step 1

Let's find the parametric form of each part of curve C.

The line segment, C1 from (1, -5) to (-2,-3):

x=(1-t)(1)+t(-2)=1-3t  and y=(1-t)(-5)+t(-3)=-5+2t with 0≤t≤1.

⇒dx=-3dt, and dy=2dt

The curve, C2 can be parametrized as follows:

x=t and y=1-t2 with -2≤t≤2.

⇒dx=dt, dy=-2t dt

The parametric form for the last line segment, C3:

x=(1-t)(2)+t(4)=1+2t and y=(1-t)(-3)+t(b)=1+(3+b)t with 0≤t≤1.

⇒dx=2dt, dy=(3+b)dt.

 

 

 

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