MA005: Calculus I

If we know the position of an object at every time, then we can determine its speed. The formula for speed comes from the distance formula and looks a lot like it, but with derivatives.

Proof: The speed of an object is the limit, as $\Delta \mathrm{t} \rightarrow 0$, of $\frac{\text { change in position }}{\text { change in time }}$. (Fig. 6 )

$\begin{gathered} \frac{\text { change in position }}{\text { change in time }}=\frac{\sqrt{(\Delta x)^{2}+(\Delta y)^{2}}}{\Delta t}=\sqrt{\frac{(\Delta x)^{2}+(\Delta y)^{2}}{(\Delta t)^{2}}} \\ \qquad \qquad \qquad \qquad =\sqrt{\left(\frac{\Delta x}{\Delta t}\right)^{2}+\left(\frac{\Delta y}{\Delta t}\right)^{2}} \\ \quad \qquad \qquad \qquad \qquad \qquad \qquad \rightarrow \sqrt{\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}} \qquad \text { as } \Delta t \rightarrow 0 \end{gathered}$

Exercise 10: Find the speed of the object whose location at time $\mathrm{t}$ is $(\mathrm{x}, \mathrm{y})=\left(\sin (\mathrm{t}), \mathrm{t}^{2}\right)$ when $\mathrm{t}=0$ and $\mathrm{t}=1$.

Solution: $\mathrm{dx} / \mathrm{dt}=\cos (\mathrm{t})$ and $\mathrm{dy} / \mathrm{dt}=2 \mathrm{t}$ so speed $=\sqrt{(\cos (\mathrm{t}))^{2}+(2 \mathrm{t})^{2}}=\sqrt{\cos ^{2}(\mathrm{t})+4 \mathrm{t}^{2}}$.

Practice 7: Show that an object whose location at time $t$ is $(x, y)=(3 \sin (t), 3 \cos (t))$ has a constant speed. (This object is moving on a circular path).

Practice 8: Is the object whose location at time $\mathrm{t}$ is $(\mathrm{x}, \mathrm{y})=(3 \cos (\mathrm{t}), 2 \sin (\mathrm{t}))$ travelling faster at the top of the ellipse ( at $\mathrm{t}=\pi / 2$ ) or at the right edge of the ellipse (at $\mathrm{t}=0$)?