I need help with 1 **Problem 1: Particle Motion in the xy-plane**At time \( t \), a particle moving in the xy-plane is at position \( (x(t), y(t)) \), where \( x(t) \) and \( y(t) \) are not explicitly given. For \( t \geq 0 \), \(\frac{dx}{dt} = 4t + 1\) and \(\frac{dy}{dt} = \sin(t^2)\). At time \( t = 0 \), \( x(0) = 0 \) and \( y(0) = -4\).**(a)** Find the speed of the particle at time \( t = 3 \).\[ \sqrt{(4(3) + 1)^2 + (\sin(3^2))^2}= \sqrt{13^2 + (\sin(9))^2} = \sqrt{169 + 0.31}= \sqrt{169.31} = 13.0065 \]**(b)** Find the slope of the line tangent to the path of the particle at time \( t = 3 \).\[\frac{dy}{dx} = \frac{\sin(t^2)}{4t + 1} \quad \text{at } t = 3= \frac{\sin(9)}{13} \approx 0.03170\]**(c)** Find the position of the particle at time \( t = 3 \).- \( x(3) = x(0) + \int_0^3 (4t+1) \, dt \)- \( y(3) = y(0) + \int_0^3 \sin(t^2) \, dt \)Calculations indicate:\( x(3) = 21 \)\( y(3) \approx -5.4067 \)**(d)** Find the total distance traveled by the particle over the time interval \( 0 \leq t \leq 3 \).The distance requires integrating the speed:\[\int_0^3 \sqrt{(4t + 1)^2 + (\sin(t^2))^2} \, dt\]Given calculations or numerical integration would be necessary for a precise value.

Since we only answer up to 3 sub-parts, we’ll answer the first 3. Please resubmit the question and specify the other subparts (up to 3) you’d like answered. Given, dxdt=4t+1,dydt=sint2andAt time t=0,x0=0, y0=-4a) Find the speed of the particle at time t = 3. The velocity vector is given by, vt=x't i+y't j Now, x't=4t+1andy't=sint2At t=3,x'3=12+1=13andy'3=sin9 Then, vt=x't i+y't j =13 i +sin9 jThen, Speed=x't2+y't2=132+sin29=13.007 unit Find the acceleration vector of the particle at time t = 3. at=x''t i+y''t j=4×1+0 i+cost2×2t j=4 i+2tcost2 j At t = 3, a3=4 i+ 6cos9 j =4 i +-5.467 j =4 i-5.467 jb) Find the slope of the line tangent to the path line of the particle at time t = 3. Slope = dydx=dydtdxdt =sint24t+1 At t = 3, Slope=sin913 =0.0317c) Find the position of the particle at time t = 3. Since, x't=4t+1andy't=sint2 Then, xt=4×t22+t+C ∵C=Integration Constantxt=2t2+t+CAlso,x0=0⇒202+0+C=0 ⇒C=0So,xt=2t2+tand yt=∫0tsinx2dx+C ∵C=Integration Constanty0=-4⇒C=-4yt=∫0tsinx2dx-4 At t = 3, x3=2×9+3=21andy3=∫03sinx2dx-4 =-3.226 Hence, Position of the particle at time t = 3 is (21, -3.226).