Hello, I am confused as to how they got 5y in this problem when they multiplied dx/dy by dy/dt in the fourth line.
I am also confused as to what dx/dt and dx/dy and dy/dt mean.
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$\begingroup$It would have been more obvious if that had inserted a line after line 3 which read:
$$\frac{dx}{dy}=y $$
Do you see why? (just differentiate line 3 w.r.t y).
They told you $$\frac{dy}{dt}=5$$ so line 5 is just putting the values in for each term.
If you look back into the history of math, there is a fascinating distinction of notation between Lagrange and Leibnitz. Lagrange wrote $$y=f(x)$$and $$y'=f'(x)$$Leibnitz came up with the alternate notation of $$y=f(x)$$ so $$\frac{dy}{dx}=d\frac{f(y)}{dx}$$Leibnitz' notation allows a lot more 'clever' use of fraction-like quantities, which usually work out - but can lead to problems if things aren't quite right (but you must be careful!).
See
$\endgroup$ 7 $\begingroup$$\frac{dy}{dt}$ and $\frac{dx}{dt}$ are just another way to denote derivatives of functions. $\frac{dy}{dt}$ is equivalent to $y'(t)$. Likewise, $\frac{dx}{dt}$ means the same thing as $x'(t)$. This notation is called Leibniz's notation because it was Leibniz who introduced it.
If you differentiate both sides of the equation $y^2-1=2x$ with respect to $t$ (since $x$ and $y$ are functions of $t$), you will get this:
$$2y\frac{dy}{dt}=2\frac{dx}{dt}\implies \frac{dx}{dt}=y\frac{dy}{dt}$$
$\frac{dy}{dt}$ is given. You also know that $x=12$. This means that at some point $t_0$, $x$ is equal to $12$. But since $y$ is a function of the same quantity, what would it be equal to at the same point $t_0$ when $x$ is $12$? Well, use the relationship between $y$ and $x$ that you're given—$y(t)=\sqrt{2x(t)+1}$:$$ y(t_0)=\sqrt{2x(t_0)+1}\\ y(t_0)=\sqrt{2\cdot 12+1}\\ y(t_0)=5\\y=5 $$
Just plug all the things that you now know into the expression that we came up with a few lines above:
$$\frac{dx}{dt}=5\cdot 5 = 25$$
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