I need help with 3 (especially part a) 3.For t 2 0, a particle is moving along a curve so thadxparticle is at position (1, 5). It is known thatdt(a) Is the horizontal movement of the particle to thFind the slope of the path of the particle at tim(b) Find the x-coordinate of the particle's position(c) Find the speed of the particle at time t = 4. F%3D(d) Find the distance traveled by the particle from

Name: I need help with 3 (especially part a) 3.For t 2 0, a particle is movi
Uploaded: 2024-03-20T07:07:12.000Z
Channel: Bartleby
Description: I need help with 3 (especially part a) 3.For t 2 0, a particle is moving along a curve so thadxparticle is at position (1, 5). It is known thatdt(a) Is the horizontal movement of the particle to thFind the slope of the path of the particle at tim(b)

given dxdt=t+2etdydt=sin2t ...i at t=2 x,y=1,5 part(a) put t=2 in equation i dxdtt=2=2+2e2 =2e2>0 therefore , particle is moving in right slope dydx=dydt×dtdx substitute the values dydxt=2=dydt×dtdxt=2 =sin2t×ett+2t=2 =sin22×e22+2 =sin22×e22 ≈3.055part (b) to find the x coordinates of the particle position at t=4 using the given conditions dxdt=t+2etx2=1 integrating the function ∫dx=∫24t+2etdtintegrating using calculatorx≈1.253part (c) claim; speed of the particle we will use the formula S=x't2+y't2 =t+2e2t+sin4t =t+2e-2t+sin4t put t=4 S4=4+2e-8+sin44 =6e-8+sin44 ≈0.575 acceleration at=d2xdt2,d2ydt2 again differentiate dxdt=t+2et,dydt=sin2t with respect to x d2xdt2=et×12t+2-t+2×ete2td2ydt2=2sintcost at=d2xdt2,d2ydt2 =et×12t+2-t+2×ete2t,2sintcosta4=e4×124+2-4+2×e4e8,2sin4cos4 =126e4-6e4,2sin4cos4 ≈-0.041,0.989