A particle moves along a path described by $y = 4 − x^2$ . At what point along the curve are $x$ and $y$ changing at the same rate? Why is the answer $\left(-\frac 1 2 , \frac{15}{4} \right)$? I have no idea how to approach this problem. Can someone guide me/explain it to me step by step? I have a final on this in less than 2 hours and I'm freaking out!
 A: Hint: $\frac{dy}{dx}=1$. And calculate this ratio which equals the derivative of your function!
A: You have
$$
y=4-x^2,
$$
then
$$
\frac{dy}{dx}=-2x.
$$
Since we are finding the point that $x$ and $y$ are changing at same rate, then
$$
\begin{align}
\frac{dy}{dt}&=\frac{dx}{dt}\\
\frac{dy}{dx}&=\frac{dt}{dt}\\
\frac{dy}{dx}&=1\\
-2x&=1\\
x&=-\frac12
\end{align}
$$
and
$$
\begin{align}
y&=4-x^2\\
y&=4-\frac14\\
&=\frac{15}4.
\end{align}
$$
A: If
$y = 4 - x^2$,
and $x = x(t)$ is a function of $t$, then by the chain rule
$\dfrac{dy}{dt} = \dfrac{dy}{dx} \dfrac{dx}{dt}; \tag{1}$
and since
$\dfrac{dy}{dx} = -2x, \tag{2}$
we have
$\dfrac{dy}{dt} = -2x\dfrac{dx}{dt}, \tag{3}$
so if
$\dfrac{dy}{dt} = \dfrac{dx}{dt} \ne 0, \tag{4}$
then
$x = -\dfrac{1}{2}; \tag{5}$
now 
$y = 4- x^2 = \dfrac{15}{4}, \tag{6}$
the given result.  And that's how it works!
Hope this helps!  Cheerio,
and as always,
Fiat Lux!!!
A: $\frac{dy}{dt}=\frac{dx}{dt}$
Therefore, By chain rule:
$$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{dy}{dx}=1$$
$\frac{dy}{dx}=-2x$
$\implies x=\frac{-1}{2};y=\frac{15}{4}$
