Dear Dr. Taylor,
The problem asked to find the solution r(t) of the differential equation with the given initial condition: r′(t)=⟨sin6t,sin4t,3t⟩,r(0)=⟨3,8,8⟩
so I took the antiderivatives of each component and got
r(t)=⟨3-1/6cos(6t),8-1/4cos(4t),3/2t^2+8⟩ but it says that's wrong. when I take the derivate of that answer I get the r'(t) because the derivative of -cos(u) is sin(u)*u'
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It looks like you figured it out in the meantime, and clearly you figured out that r(t)=∫r'(t)dt + constant vector. The problem with your original answer, as you now know, is that it didn't account for the nonzero value of ∫r'(t)dt at t=0, so you needed to correct for that in the constant vector you added.
