What is the y-component of r(pi/2)?

Question

If r'(t)=<sin(t), -cos(t), 2t> and r(0)=<1, 1, 2>. What is the y-component of r(pi/2)?

r(t)=<-cos(t), -sin(t), t^2>

r(pi/2)=-1

So -1 is the answer?

Answer 1

Final value = initial value + change

y(pi/2) = y(0) + ∫{0, pi/2} -cos(t) dt = 1 - sin(pi/2) = 0

Answer 2

Let's check and correct  the components for r(t). After integration of components of r'(t) we will get:

r(t) = <-cos(t) + c₁, -sin(t) + c₂, t^2 + c₃>. As  r(0)=<1, 1, 2> so we have:

-cos(0) + c1 = 1   that is  c1 = 2

-sin(0) + c2 = 1   that is   c2 = 1

c3  =  2

and we have for r(t):   r(t)  = <-cos(t) + 2, -sin(t) + 1, t^2 + 2>.

Now we can calculate  r(pi/2) = <2, 0, (pi^2/4+2)>, so y-component of r(pi/2)  = 0