# Why does solving this equation with differential equations and related rates yield different results?

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*}

• 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 $\frac{y}{x}\ne\frac{\mathrm{d}y}{\mathrm{d}x}$. Sep 22, 2020 at 1:58
• so are you saying the derivative of $y=xtanθ$ with respect to x is not $tanθ$? Sep 22, 2020 at 2:40
• 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 $y$. Sep 22, 2020 at 7:34

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.