How would I solve the following problem?
A particle is moving on the circular orbit $x^2+y^2=41$. As it passes through the point $(5,4)$ its y coordinate is decreasing at the rate of $2$ units per second. At what rate is the x coordinate changing?
I know the distance formula is $x^2+y^2=1$
so I did $\frac{d}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}$
which is $x\frac{dx}{dt}+y\frac{dy}{dt}=0$
$x\frac{dx}{dt}=-y\frac{dy}{dt}$
$\frac{dx}{dt}=\frac{-y}{x}(-2)$
$\frac{dx}{dt}=\frac{-4}{5}(-2)$
but would this correct.
Best Answer
The problem was done correctly, except for the worrying reference to the "distance formula" $x^2+y^2=1$.
We are given that $x^2+y^2=41$. We are also given some information about $\frac{dy}{dt}$, and are asked about $\frac{dx}{dt}$. This is a related rates problem, easier than most. We are given a rate (rate of change, derivative), and are as ked about another derivative, the derivative of a related quantity. Normally, we solve such a problem by finding a formula that captures the relationship between the quantities. Then we differentiate. But in this case the relationship is given to us explicitly.
Start from $x^2+y^2=41$. Differentiate both sides with respect to $t$. We get $$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0,$$ which looks somewhat better as $$x\frac{dx}{dt}+y\frac{dy}{dt}=0.\tag{$1$}$$ Now freeze the situation at the instant when $x=5$ and $y=4$. At that instant, we have $\frac{dy}{dt}=-2$. So at that instant $$5\frac{dx}{dt}+(4)(-2)=0,$$ which says that at that instant $\dfrac{dx}{dt}=\dfrac{8}{5}$.
We substituted particular values and then solved. But your style (solve and then substitute) works just as well. From $(1)$ we get $$\frac{dx}{dt}=-\frac{y}{x}\frac{dy}{dt}.$$ Set $x=5$, $y=4$, and $\frac{dy}{dt}=-2$.