Calculus with Vector Functions — Question 3

Question 3

A curve has velocity $\vec r'(t)=\left\langle 2t,3t^2,e^{-t}\right\rangle$ and passes through $P=(1,-2,4)$ at $t=0$. Find $\vec r(t)$ and the net displacement from $t=0$ to $t=2$.

Question 3 – Solution

Strategy Integrate each velocity component and use the initial condition to determine the constant vector.

Antiderivative $\vec r(t)=\left\langle t^2,t^3,-e^{-t}\right\rangle+\vec C$. Since $\vec r(0)=\left\langle 0,0,-1\right\rangle+\vec C=P$, $\vec C=\left\langle 1,-2,5\right\rangle$. Therefore $$\boxed{\vec r(t)=\left\langle t^2+1,t^3-2,5-e^{-t}\right\rangle}.$$

Displacement $\vec r(2)-\vec r(0)=\int_0^2\vec r'(t)dt=\boxed{\left\langle 4,8,1-e^{-2}\right\rangle}$.

Verification Differentiation recovers the given velocity and substitution gives $P$.