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enter image description here Sorry for the figure being so large I was unsure how to shrink it.

The question asked is about a beam anchored at angle $\theta$ to two perpendicular axes (at points a and b). The beam slides along them at a constant speed $-V$ in the $x$ direction and $V_b$ in the $y$ direction. The goal is to solve for $V_b$ in terms of $\theta$ and $V$. I solved this equation in two different ways, the first using related rates and the second using a differential equation, and don't understand why they yield different results, and I was hoping someone could shed some light on it for me, as I think I violated some mathematical rule when solving with related rates.

Solve attempt 1 using related rates:

\begin{align*} \frac{\mathrm{d}x}{\mathrm{d}t} &= -v\\ \frac{\mathrm{d}y}{\mathrm{d}t} &= v_b\\ y &=x\tan\theta\\ \frac{\mathrm{d}y}{\mathrm{d}x} &=\tan\theta\\ \frac{\mathrm{d}y}{\mathrm{d}t} &=\frac{\mathrm{d}y}{\mathrm{d}x}\frac{\mathrm{d}x}{\mathrm{d}t}\\ \frac{\mathrm{d}y}{\mathrm{d}t} &=-v\tan\theta\\ v_b &=-v\tan\theta \end{align*}

Solve attempt 2 using differential equations (where $L$ is the beam)

\begin{align*} x^2+y^2 &=L^2\\ 2x\frac{\mathrm{d}x}{\mathrm{d}t}+2y\frac{\mathrm{d}y}{\mathrm{d}t} &=0\\ -2xv+2yv_b &=0\\ v_b &=\frac{x}{y}v\\ v_b &=\frac{v}{\tan\theta} \end{align*}

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  • $\begingroup$ In attempt 1, your third line is $y=x\tan\theta$ but your fourth line is $\frac{\mathrm{d}y}{\mathrm{d}x}=\tan\theta$. Note that $\frac{y}{x}\ne\frac{\mathrm{d}y}{\mathrm{d}x}$. $\endgroup$
    – user338955
    Sep 22, 2020 at 1:58
  • $\begingroup$ so are you saying the derivative of $y=xtanθ$ with respect to x is not $tanθ$? $\endgroup$
    – figbar
    Sep 22, 2020 at 2:40
  • $\begingroup$ If you claim that $\frac{\mathrm{d}y}{\mathrm{d}x}=\tan\theta$, then you are claiming that $\tan\theta$ is a constant. But it isn't; it depends on $x$ and $y$. $\endgroup$
    – user338955
    Sep 22, 2020 at 7:34

1 Answer 1

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In your attempt 1, you cannot deduce $\frac{dy}{dx}=\tan\theta$ from the equation $y=x\tan\theta$, because $\theta$ is also a function of $x$. You should instead deduce $\frac{dy}{dx}$ from the equation $$x^2+y^2=L^2.\tag1$$ Implicit differentiation of (1) wrt $x$ gives $$2x + 2y\frac{dy}{dx}=0,$$ which yields $\frac{dy}{dx}=-\frac xy$. When you plug this value of $\frac{dy}{dx}$ into $$ \frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt},$$ you will end up with the same result as in attempt 2.

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