**Example:**
A particle is moving around a circle of radius 5
around the origin. As it passes through the point $(3,4)$, its $x$
position is changing at a rate $\dfrac{dx}{dt}\Big|_{{x=3},\, {y=4}} = 2$.
How fast is $y$
changing at that instant?
**Solution:** We know that the equation for the circle is $$x^2 + y^2 =
25.$$ To find a relationship between the **rates of change** of $x$ and $y$ with respect to time, we can implicitely differentiate the equation above with respect to $t$.
$$ 2 x \frac{dx}{dt} + 2 y \frac{dy}{dt} = 0.$$
We know that when the particle is passing $(3,4)$, then its velocity is $\displaystyle{\frac{dx}{dt}\Big|_{{x=3},\, {y=4}} = 2}$, so we can solve for
$\dfrac{dy}{dt}\Big|_{{x=3},\, {y=4}}$. Since
$$ 2(3)(2) + 2(4)\frac{dy}{dt}\Big|_{{x=3},\, {y=4}} = 0,$$ we have
$\dfrac{dy}{dt}\Big|_{{x=3},\, {y=4}}=
-\frac{12}{8}=-\frac{3}{2}.$ |