Calculus with Vector Functions — Question 7

Question 7 – Solution

Strategy. Continuity compares one-sided function values; differentiability compares one-sided vector difference quotients.

Step 1: Continuity The defined value from the first branch is $$\mathbf r(0)=\left\langle 0,0,0\right\rangle.$$ From the left, $\left\langle t,t^2,t^3\right\rangle\to\left\langle 0,0,0\right\rangle$. From the right, $$\left\langle t,at,t^2\right\rangle\to\left\langle 0,0,0\right\rangle$$ for every fixed real $a$. Therefore $\mathbf r$ is continuous at $0$ for every real value of $a$.

Step 2: Left derivative As $h$ approaches zero through negative values, $$\frac{\mathbf r(h)-\mathbf r(0)}h =\left\langle 1,h,h^2\right\rangle\longrightarrow\left\langle 1,0,0\right\rangle.$$

Step 3: Right derivative As $h$ approaches zero through positive values, $$\frac{\mathbf r(h)-\mathbf r(0)}h =\left\langle 1,a,h\right\rangle\longrightarrow\left\langle 1,a,0\right\rangle.$$ The one-sided derivatives agree exactly when $a=0$. Hence $$\boxed{a=0},\qquad \boxed{\mathbf r'(0)=\left\langle 1,0,0\right\rangle}.$$ This also shows explicitly why continuity alone does not guarantee differentiability.