Some Applications of the Chain Rule

Suppose a robot has been programmed to move in the $\mathrm{xy}$-plane so at time $\mathrm{t}$ its $\mathrm{x}$ coordinate will be $\sin (t)$ and its $\mathrm{y}$ coordinate will be $t^{2}$. Both $x$ and $y$ are functions of the independent parameter t, $x(t)=\sin (t)$ and $y(t)=t^{2}$, and the path of the robot (Fig. 4) can be found by plotting $(x, y)=(x(t), y(t))$ for lots of values of $t$.

$\begin{array}{c|c|c|c} \mathrm{t} & \mathrm{x}(\mathrm{t})=\sin (\mathrm{t}) & \mathrm{y} \mathrm{t})=\mathrm{t}^{2} & \text { plot point at} \\ \hline 0 & 0 & 0 & (0,0) \\ .5 & .48 & .25 & (.48, .25) \\ 1.0 & .84 & 1 & (.84,1) \\ 1.5 & 1.00 & 2.25 & (1,2.25) \\ 2.0 & .91 & 4 & (.91,4) \end{array}$ 

Typically we know $\mathrm{x}(\mathrm{t})$ and $\mathrm{y}(\mathrm{t})$ and need to find $\mathrm{dy} / \mathrm{dx}$, the slope of the tangent line to the graph of $(\mathrm{x}(\mathrm{t}), \mathrm{y}(\mathrm{t}))$. The Chain Rule says that $\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{\mathrm{dy}}{\mathrm{dx}} \cdot \frac{\mathrm{d} \mathrm{x}}{\mathrm{dt}}$, so, algebraically solving for $\frac{\mathrm{dy}}{\mathrm{dx}}$, we get $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy} / \mathrm{dt}}{\mathrm{dx} / \mathrm{dt}}$.

If we can calculate $\mathrm{dy} / \mathrm{dt}$ and $\mathrm{dx} / \mathrm{dt}$, the derivatives of $\mathrm{y}$ and $\mathrm{x}$ with respect to the parameter $\mathrm{t}$, then we can determine $\mathrm{dy/dx}$, the rate of change of $\mathrm{y}$ with respect to $\mathrm{x}$.

Example 8: Find the slope of the tangent line to the graph of $(x, y)=\left(\sin (t), t^{2}\right)$ when $t=2 ?$
Solution: $\mathrm{dx} / \mathrm{dt}=\cos (\mathrm{t})$ and $\mathrm{dy} / \mathrm{dt}=2 \mathrm{t} .$ When $\mathrm{t}=2$, the object is at the point $\left(\sin (2), \mathbf{2}^{2}\right) \approx(.91,4)$ and the slope of the tangent line to the graph is $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy} / \mathrm{dt}}{\mathrm{dx} / \mathrm{dt}}=\frac{2 \mathrm{t}}{\cos (\mathrm{t})}=\frac{2 \cdot 2}{\cos (\mathbf{2})} \approx \frac{4}{-.42} \approx-9.61$.

Practice 5: Graph $(x, y)=(3 \cos (t), 2 \sin (t))$ and find the slope of the tangent line when $t=\pi/2$.

When we calculated $\frac{d y}{d x}$, the slope of the tangent line to the graph of $(\mathrm{x}(\mathrm{t}), \mathrm{y}(\mathrm{t}))$, we used the derivatives $\frac{d x}{d t}$ and $\frac{d y}{d t}$, and each of these derivatives also has a geometric meaning:

$\frac{d x}{d t}$ measures the rate of change of $\mathrm{x}(\mathrm{t})$ with respect to $\mathrm{t}$ - it tells us whether the $\mathrm{x}$-coordinate is increasing or decreasing as the t-variable increases.

$\frac{d y}{d t}$ measures the rate of change of $\mathrm{y}(\mathrm{t})$ with respect to $\mathrm{t}$.

Example 9: For the parametric graph in Fig. 5, tell whether $\frac{d x}{d t}, \frac{d y}{d t}$ and $\frac{d y}{d x}$ is positive or negative when $\mathrm{t=2}$.

Solution: As we move through the point $\mathrm{B}$ (where $\mathrm{t}=2$ ) in the direction of increasing values of $t$, we are moving to the left so $x(t)$ is decreasing and $\frac{d x}{d t}$ is negative.

Similarly, the values of $\mathrm{y}(\mathrm{t})$ are increasing so $\frac{d y}{d t}$ is positive. Finally, the slope of the tangent line, $\frac{d y}{d x}$, is negative.

(As check on the sign of $\frac{d y}{d x}$ we can also use the result $\frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{\text { positive }}{\text { negative}}$  = $\text{negative}$.)

Practice 6: For the parametric graph in the previous example, tell whether $\frac{d x}{d t}, \frac{d y}{d t}$ and $\frac{d y}{d x}$ is positive or negative when $\mathrm{t}=1$ and when $\mathrm{t}=3$.