Calculus with Vector Functions — Question 4

Question 4

Find the unique vector function $\mathbf r(t)$ satisfying $$\mathbf r'(t)=\left\langle 2te^{t^2},\ \frac{3}{1+t^2},\ 4t^3-2t\right\rangle, \qquad \mathbf r(0)=\left\langle 1,-2,5\right\rangle.$$ Tasks

1. Recover $\mathbf r(t)$ by componentwise antidifferentiation.

2. Use the initial vector to determine every constant.

3. Differentiate your final expression to verify it.

Question 4 – Solution

Strategy. Antidifferentiate each component and then use the initial vector to determine three constants.

Step 1: First component $$\int 2te^{t^2}\,dt=e^{t^2}+C_1,$$ using the substitution $u=t^2$.

Step 2: Remaining components $$\int\frac 3{1+t^2}\,dt=3\arctan t+C_2,$$ and $$\int(4t^3-2t)\,dt=t^4-t^2+C_3.$$ Thus $$\mathbf r(t)=\left\langle e^{t^2}+C_1,\ 3\arctan t+C_2,\ t^4-t^2+C_3\right\rangle.$$

Step 3: Initial condition At $t=0$, $$\mathbf r(0)=\left\langle 1+C_1,C_2,C_3\right\rangle=\left\langle 1,-2,5\right\rangle.$$ Hence $C_1=0$, $C_2=-2$, and $C_3=5$. Therefore $$\boxed{\mathbf r(t)=\left\langle e^{t^2},\ 3\arctan t-2,\ t^4-t^2+5\right\rangle}.$$

Verification. Componentwise differentiation gives $\left\langle 2te^{t^2},3/(1+t^2),4t^3-2t\right\rangle$, and direct substitution gives the required initial vector. These three scalar initial-value problems each have only one integration constant, so the solution is unique.