I am reviewing some things in calculus, I have these two sample problems with their answers, that I tried to solve another way and got different results. I'd like to understand why what I made yielded different results.
For the first problem, I thought that we have $\frac{dx}{dt}=\frac{dy}{dt}$ and this meant that we need to differentiate implicitly with respect to $x$ and then with respect to $y$:
$$\frac{dy}{dt} = 2x $$
$$1=2x \frac{dx}{dt}$$
Now we have:
$$2x=\frac{1}{2x} \implies x=\pm \frac{1}{2}$$
Where we get two points:
$$\left(\frac{1}{2},\frac{1}{4}\right)\quad \left(-\frac{1}{2},\frac{1}{4}\right)$$
Instead of just one, as given in the answer.
In the second one, I tried to answer in the same way, I thought that we needed to find $\frac{dy}{dt}=4 \frac{dx}{dt}$. Differentiating implicitly, we get:
$$\frac{dy}{dt}= 2x-6$$
$$1= 2x\frac{dx}{dt}-6\frac{dx}{dt}$$
Now we have:
$$2x-6=\frac{4}{2x-6} \implies x \in \{2,4\} $$
And again, we have two points:
$$\left(2,-8\right)\quad \left(4,-8\right)$$
Which is completely off from the point found in the answer. I am a bit confused with the following: Why the way I did fail here? Also, as I have found more solutions than the answer in the first example, I suspect it fails there too.
In both problems, you are given $x$ is a function of $t$ and $y$ is a function of $x$. Writing $x = x(t)$ and $y = y(x(t))$ may help you conceptualise this.
Then, to differentiate $y$ with respect to $t$, we use the chain rule: $$\frac{dy}{dt} = \frac{dy}{dx}\frac{dx}{dt}.$$
For the first problem, we have $$\frac{dy}{dt} = \frac{dy}{dx}\frac{dx}{dt} = 2x\frac{dx}{dt}.$$ If $\frac{dy}{dt} = \frac{dx}{dt}$, then $x = \frac{1}{2}$ which implies $y = (\frac{1}{2})^2 = \frac{1}{4}$.
You can see that the solution of the second problem given in the book uses the same method.