Can I apply regular derivative instead of implicit describing rectangle diagonal?

Question

The sides of this rectangle increase in such a way that dz/dt=1 and dx/dt=3dy/dt. At the instant when x=4 and y=3, what is the value of dx/dt?

Rectangle with diagonal z

I solve it like this:

$\dfrac{dz}{dy}=(\sqrt{x^2+y^2})'=\dfrac{1}{2\sqrt{x^2+y^2}}2y$

$\dfrac{dz}{dy}=\dfrac{dz}{dt}*\dfrac{dt}{dy}$

$\dfrac{y}{\sqrt{x^2+y^2}}=1*\dfrac{dt}{dy}$, so $\dfrac{dy}{dt}=\dfrac{\sqrt{x^2+y^2}}{y}$

Given that $\dfrac{dx}{dt}=3\dfrac{dy}{dt}$, conclude $\dfrac{dx}{dt}=3\dfrac{\sqrt{x^2+y^2}}{y}$

Substituting $x=4, y=3$ the answer is 5.

The answer in textbook uses implicit derivative:

$z^2=x^2+y^2$, which implies $2z\dfrac{dz}{dt}=2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=\dfrac{dx}{dt}(2x+\dfrac{2}{3}y)$

Substituting values for x,y and $\dfrac{dz}{dt}=1$ we get that answer is 1.

Where did I get wrong? My suspicions are:

• Maybe it's incorrect to use $\dfrac{dz}{dy}$ when z depends on x too
• Working with z instead of $z^2$ is prohibited

Answer 1

The problem with your answer is that your first derivative only takes the derivative with respect to $y$. $x$ is essentially ignored. $\frac{dz}{dy}$ is not the same as $\frac{\partial z}{\partial y}$.

I would, indeed, start with $z^2 = x^2 + y^2$ because it is just a much simpler problem. Starting the way you did isn't problematic, it just becomes a much harder equation.

$$ z = \sqrt{x^2 + y^2} \\ d(z) = d\left(\sqrt{x^2 + y^2}\right) \\ dz = \frac{1}{2}\frac{1}{\sqrt{x^2 + y^2}} d(x^2 + y^2) \\ dz = \frac{1}{2}\frac{1}{\sqrt{x^2 + y^2}} (2xdx + 2ydy)\\ dz = \frac{x}{\sqrt{x^2 + y^2}}dx + \frac{y}{\sqrt{x^2 + y^2}}dy \\ $$ We can divide both sides by $dt$ to get

$$ \frac{dz}{dt} = \frac{x}{\sqrt{x^2 + y^2}}\frac{dx}{dt} + \frac{y}{\sqrt{x^2 + y^2}}\frac{dy}{dt} \\ $$ Then, putting in our substitutions, we get $$ 1 = \frac{x}{\sqrt{x^2 + y^2}}\frac{dx}{dt} + \frac{1}{3}\frac{y}{\sqrt{x^2 + y^2}}\frac{dx}{dt} \\ 1 = \frac{x + \frac{1}{3}y}{\sqrt{x^2 + y^2}}\frac{dx}{dt} \\ \frac{dx}{dt} = \frac{\sqrt{x^2 + y^2}}{x + \frac{1}{3}y} \\ \frac{dx}{dt} = \frac{\sqrt{4^2 + 3^2}}{4 + \frac{1}{3}3} \\ \frac{dx}{dt} = \frac{5}{5} = 1 $$

Also note that you could have used $\frac{dz}{dy}$ just fine as well, but you MUST apply the derivative with respect to $y$ as well for that to work. So, your first step would result in:

$$ \frac{dz}{dy} = \frac{1}{2}\frac{1}{\sqrt{x^2 + y^2}} (2x\frac{dx}{dy} + 2y\frac{dy}{dy})\\ $$ And the $\frac{dy}{dy}$ would reduce to $1$, but I wanted to be sure that you saw it.