Multiple Solutions? Suppose ${x^2}+{y^2} = 45$ and $x=2y$ ... find $\frac{dy}{dt}$

Question

Suppose ${x^2}+{y^2} = 45$ and $x=2y$ for positive values of $x$ and $y$ find $\frac{dy}{dt}$ when $\frac{dx}{dt}=2$

I am new to implicit differntiation and related rates. When I attempted to solve this problem, I was met with multiple solutions. It is my first time working with two seperate equations and was wondering why two solutions is possible and what they would mean in context. Thank you.

Solution 1:

$$\frac{d}{dt}[x=2y]$$ $$\implies\frac{dx}{dt}=2\frac{dy}{dt}$$ $$\implies2=2\frac{dy}{dt}$$ $$\implies\frac{dy}{dt}=1$$

Solution 2: $$\frac{d}{dt}[{x^2}+{y^2} = 45]$$ $$\implies2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$$ $$\implies2(2y)(2)+2y\frac{dy}{dt}=0$$ $$\implies4+\frac{dy}{dt}=0$$ $$\implies\frac{dy}{dt}=-4$$ Solution 3: (admittedly the same as sol. 2) $${x^2}+{y^2} = 45 \quad \Bbb{and} \quad x=2y$$ $$\implies{(2y)^2}+{y^2} = 45$$ $$\implies {5y^2} = 45$$ $$\implies {y} = 3 \Bbb {\quad [ignore\, \pm\, because\, of\, original\, question]}$$ $$\Bbb{and}\quad{x=6}$$ $$\Bbb{we\,recall} \quad 2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$$ $$\implies 2(6)(2)+2(3)\frac{dy}{dt}=0$$ $$\implies \frac{dy}{dt}=-\frac{24}{6}=-4$$

Answer 1

The context is elementary; it is a context where one can perfectly understand the situation before approaching less traditional situations; in other words, we can practice there like you did. It's elementary geometry: tangents at the intersection points of a circle and a straight line; the tangents are perpendicular to the radii at the two intersection points $\color{blue}{B=(6,3)}$ and $B'=(-6,-3 )$. You ignore $B'$. The idea is to parameterize the circle with a functional relation (implicit function), here $\color{green}{\begin{cases} x=2t \\ y=\varphi(t)\end{cases}}$, at least locally (hence the small rectangle drawn around $\color{blue}B$ in the illustration.)

Hence the $\frac{dx}{dt}=2$, which first allows you by parameterizing the line to obtain $4t^2+t^2=45\iff t=\mp3$ and the points $\color{blue}B$ and $B'$. Note that for the line the parameterization is $\begin{cases} x=2t \\ y=t\end{cases}$, but this is obviously no longer the case for the circle parameterization and that is why I introduced a function $\color{green}{\varphi}$.

You finally get in your "solution 2" $\frac{dy}{dt}=-4$, hence $\color{green}{\begin{cases} \frac{dx}{dt}=2 \\ \frac{dy}{dt}=\varphi^{'}(3)=-4\end{cases}}$

Which gives you a direction vector of the tangent $T$ at $\color{blue}B$, $\color{red}{u=(2,-4)}$.

Answer 2

This is not a typical implicit differentiation problem. If that was the case, we would be discussing if the equations can define $x$ as a function of $y$ or vice-versa. In this case, the equations and the positivity condition dictate that $x=3$ and $y=6$. So, if you want to imagine that $x,y$ are functions of a new variable $t$, you can only conclude that they are constant functions and that the derivatives with respect to $t$ are both zero.