Calculus with Vector Functions — Question 10

Question 10 – Solution

Strategy. The vector equation is three identical scalar equations. Solve them componentwise, then prove uniqueness with an integrating factor.

Step 1: Split into components Write $\mathbf r=\left\langle x,y,z\right\rangle$. Then $$x'=2tx,\qquad y'=2ty,\qquad z'=2tz.$$ For a nonzero component $w$, separation gives $$\frac{w'}w=2t \quad\Longrightarrow\quad \ln|w|=t^2+C \quad\Longrightarrow\quad w=Ke^{t^2}.$$ The formula also includes the zero solution when $K=0$.

Step 2: Apply the initial vector Since $e^0=1$, the constants are $a,b,c$. Thus $$\boxed{\mathbf r(t)=e^{t^2}\left\langle a,b,c\right\rangle}.$$ Indeed, $$\mathbf r'(t)=2te^{t^2}\left\langle a,b,c\right\rangle=2t\mathbf r(t),$$ and $\mathbf r(0)=\left\langle a,b,c\right\rangle$.

Step 3: Direct uniqueness proof Suppose $\mathbf r_1$ and $\mathbf r_2$ share the initial value, and let $\mathbf d=\mathbf r_1-\mathbf r_2$. Then $$\mathbf d'=2t\mathbf d,\qquad \mathbf d(0)=\mathbf 0.$$ Multiply by $e^{-t^2}$ and use the product rule: $$\frac d{dt}\bigl(e^{-t^2}\mathbf d(t)\bigr) =-2te^{-t^2}\mathbf d+e^{-t^2}\mathbf d'=\mathbf 0.$$ Hence $e^{-t^2}\mathbf d$ is constant and equals its value $\mathbf 0$ at $t=0$. Therefore $\mathbf d=\mathbf 0$, proving .