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I am using James Stewarts Early Transcendentals Calculus, and Section 9.3 (which is where this problem comes from) doesn't seem to have anything remotely similar to the problem I am facing. No examples, nothing

I hate to just dump a homework problem off, but I really don't know where to begin. I thought "find the tangent line" at first, but it doesn't mention tangent anywhere, and I only thought of that because of previous math classes where I had to find the equation of a tangent line.

Reprinted problem below:

Find an equation of the curve that passes through the point
$(0, 6)$ and whose slope at $(x, y)$ is $\frac{x}{y}$.

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    $\begingroup$ We have $\frac{dy}{dx}=\frac{x}{y}$. This is a separable differential equation. $\endgroup$ – André Nicolas Jun 17 '16 at 6:49
  • $\begingroup$ Thank you! this is correct. Not sure why its x/y instead of y/x $\endgroup$ – dmscs Jun 17 '16 at 6:56
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    $\begingroup$ Passing through the point $(0,6)$ with slope $\frac{6}{0}$ wouldn't make much sense anyway, so it better be slope $\frac{x}{y}$. $\endgroup$ – Christoph Jun 17 '16 at 7:07
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    $\begingroup$ It is true that the slope of any differentiable function is the limit of $\frac{\Delta y}{\Delta x}$ as $\Delta x$ goes to $0$. However, in this case, they're saying that at any point along the graph with coordinates $(x, y)$, the slope is also $\frac xy$. This means that for any point $(x, y)$ on the graph, if you move lust a little bit (say, a distance $h$, which is very small) to the right, then the value of the function at $x+h$ is roughly $y + h\frac{x}{y}$. In other words, if you have a run of $h$, you get a rise of $h\cdot \frac xy$. $\endgroup$ – Arthur Jun 17 '16 at 7:16
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The slope at $(x,y)$ is another way of saying the derivative of the curve, so the question can be read as find the equation of a line that satisfies

\begin{equation} \frac{dy}{dx} = \frac{x}{y}, \end{equation}

and that passes through the point $(x,y) = (0,6)$.

Hint: this is a first order ODE that can be solved and will have one constant of integration that can be set such that the solution passes through the desired point.

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  • $\begingroup$ Why is it x/y? It seems like rise/run means it should be y/x $\endgroup$ – dmscs Jun 17 '16 at 6:54
  • $\begingroup$ @dmscs If you want to say that rise/run is the slope then you must interpret this as rise/run $ = dy/dx$. And the question states that the slope is equal to $x/y$. Also rise/run isn't really the same as $y/x$, it's more like rise/run $= \Delta y/ \Delta x$. $\endgroup$ – okrzysik Jun 17 '16 at 7:00
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It is variable separable differential equation.Integrating, $$ \frac{dy}{dx} = \frac{x}{y}, \, y {dy}-x {dx} = 0,\, y^2/2 -x^2/2 = c$$ Plug in BC: $$ 6^2 /2-0 = c = 18,\, y^2-x^2= 36. $$ There is no calculus law about y/x, any relation that agrees with the differentials seen as connecting separate terms is ok.

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Assume that $y$ is never $0.$ We have $$y'=x/y\implies yy'=x\implies (y^2)'=2 y y'=2 x.$$ So $y^2$ is an anti-derivative of the function $g(x)=2 x.$ So $y^2=x^2+K$ for some constant $K.$ When $x=0$ we have $36=6^2=y^2=x^2+K=K.$ So $y=\sqrt {x^2+36}.\quad$(We take the positive square root because $y>0$ when $x=0.$)

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You can use WolframAlpha to check your result: (link)

It recognizes the differential equation is separable: $$ y'(x) \, y(x) = x $$ And you can solve this by integrating both sides. $$ y \, \frac{dy}{dx} = x \iff \\ \int\limits_{y(x_0)}^{y(x)} y(\xi) \, dy = \int\limits_{x_0}^x \xi \, d\xi \iff \\ \frac{1}{2} \left( y(x)^2 - y(x_0)^2 \right) = x - x_0 $$

In our case $(x_0, y(x_0)) = (0,6)$, so we get: $$ \frac{1}{2} \left(y(x)^2 - 36 \right) = x \iff \\ y(x) = \sqrt{x+ 36} $$

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