Question

2) (15 points) (a) (10 points) Compute the line integral s f(x,y) ds of the scalar function over the oriented curve. [C] 0 (5 points) How does your answer change if I reverse the orientation of Cl? f(0, y) = C): The curve parameterized by r(t) = t'i + t'j, t E (1.21

Answer 1

The answer is given as:

2 (a) : we have to con - we have to compute line integral, Sofray) ds , where flory) - 3 [C] and [c]: The curve para meterized by eilt) - t² i + th j , t t [1,2] hence x = y = t3 t4 itt (1,2] As we know that, as - Vax)² + ( de ) 2 dt = √(3+²/²+14+3)2 at - rat4 +16 to at ds = te vat 16 t² at hence frary) ds Can be written as - [c] I forry) do = - Ja+16t² at PCC) th Scofox,y) ds = 5 t t da ti6t3 at 2 st Jariot dt Now let a +16 +² = u. 32t dt = du - he + Ja +16t7 dt – st du Va+16t2 dt = Ju 32 au du 25 73 37 312 Jag - 2x [733]2(25393 = 1 1 1 [(633%2_125] = US [623.71 – 125]

- 43 [498.71] - | 0 3 6 9 % 2(b) :- When we switch the orientation of ces de the line integral will not change. Remember that we are switching the direction of the curve and this will also change the Pare meterization . Hence value of line integral will not change. Now we will prove this for general case - Let suppose the curve c has parameterization - hlt), y = git). het also suppose that initial and final Points of the curre are A and B respectively. The Pare meterization u=h(t), yagit) will determine an orientation for the curve where the positive direction is the orientation Finally, eet – C be the curve with some Points as C, Howear, in this case, curve has. B as initial & A as final Point In other words, given a curve c, the weare – c is the same curore as C except the dreen has been reversed. s for,y) ds = [ frany) as - C

length who entegral so for a line integral. cor.t.as can change the direction of the and not change the value of the coc can change the SVO or sony 2000 tried bih o no loitini wote
thankyou.