Velocity and Acceleration — Question 1

Question 1 – Solution

Strategy. Differentiate position componentwise; speed is the magnitude of velocity.

Step 1: Velocity and acceleration $$\mathbf v(t)=\mathbf r'(t)=\left\langle 2t-2,3t^2-3,4\right\rangle,$$ and $$\mathbf a(t)=\mathbf v'(t)=\left\langle 2,6t,0\right\rangle.$$ Therefore $$\boxed{\text{speed}=\|\mathbf v(t)\| =\sqrt{(2t-2)^2+(3t^2-3)^2+16}}.$$

Step 2: Values at $t=1$ $$\mathbf r(1)=\left\langle -1,-2,5\right\rangle,\quad \boxed{\mathbf v(1)=\left\langle 0,0,4\right\rangle},\quad \boxed{\mathbf a(1)=\left\langle 2,6,0\right\rangle}.$$ The speed is $\boxed{4}$.

Step 3: Tangent line Since $\mathbf v(1)\ne\mathbf 0$, it gives the tangent direction: $$\boxed{\mathbf L(s)=\left\langle -1,-2,5\right\rangle+s\left\langle 0,0,1\right\rangle}.$$ The simplified direction is parallel to velocity. At $s=0$ the line passes through the correct position.