Calculus with Vector Functions — Question 2

Question 2

Let $$\mathbf r(t)=\left\langle t^2-1,\ t^3-3t,\ 2t+1\right\rangle.$$ Tasks

1. Find $\mathbf r'(t)$ directly from the vector difference quotient.

2. Find the tangent line at the point corresponding to $t=1$.

3. Verify that your direction vector is nonzero.

Question 2 – Solution

Strategy. Expand $\mathbf r(t+h)-\mathbf r(t)$ componentwise, divide by $h$, and then let $h\to 0$.

Step 1: Difference quotient $$\begin{align*} \frac{\mathbf r(t+h)-\mathbf r(t)}h &=\left\langle \frac{(t+h)^2-t^2}{h}, \frac{(t+h)^3-3(t+h)-(t^3-3t)}h, \frac{2h}{h}\right\rangle\\ &=\left\langle 2t+h,\ 3t^2+3th+h^2-3,\ 2\right\rangle. \end{align*}$$ Taking $h\to 0$ yields $$\boxed{\mathbf r'(t)=\left\langle 2t,3t^2-3,2\right\rangle}.$$

Step 2: Point and direction At $t=1$, $$\mathbf r(1)=\left\langle 0,-2,3\right\rangle,\qquad \mathbf r'(1)=\left\langle 2,0,2\right\rangle.$$ The derivative is nonzero, so it supplies a valid tangent direction.

Step 3: Tangent line Using a new line parameter $s$, $$\boxed{\mathbf L(s)=\left\langle 0,-2,3\right\rangle+s\left\langle 2,0,2\right\rangle}.$$ Substitution at $s=0$ confirms the point, and the line direction is parallel to $\mathbf r'(1)$.