Answer:
r(t) = (2t^3 + 3)i + (1/2 e^2t - 2)j + (-cos(t) + 4)k
Explanation:
Given r′(t)=6t^2i+e^2tj+sintk and r(0)=3i−2j+k.
To find r(t), we need to integrate r′(t). Integrating each component of r′(t), we get:
r(t) = ∫ r′(t) dt = ∫ (6t^2i+e^2tj+sintk) dt
Integrating the x-component, we get:
∫ 6t^2 dt = 2t^3 + C1
Integrating the y-component, we get:
∫ e^2t dt = 1/2 e^2t + C2
Integrating the z-component, we get:
∫ sin(t) dt = -cos(t) + C3
where C1, C2, and C3 are constants of integration.
Therefore, the solution for r(t) is:
r(t) = (2t^3 + C1)i + (1/2 e^2t + C2)j + (-cos(t) + C3)k
Using the initial condition, r(0)=3i−2j+k, we can find the values of the constants of integration:
r(0) = (2(0)^3 + C1)i + (1/2 e^2(0) + C2)j + (-cos(0) + C3)k
Simplifying, we get:
C1 = 3
C2 = -2
C3 = 4
Therefore, the final solution for r(t) is:
r(t) = (2t^3 + 3)i + (1/2 e^2t - 2)j + (-cos(t) + 4)k