Question

Need to know the formula to use and how to work the implicit differentiation afterwards! A particle on the x-axis is moving to the right at 2 units per second. A second particle is moving down the y-axis at the rate of 3 units per second. At a certain instant the first particle is at the point (5,0) and the second is at the point (0,7). How rapidly is the angle between the x-axis and the line joining the two particles changing at that instant? Are the particles moving towards or away from each other at that instant?

Answer 1

so hmm check the picture below

thus $\bf tan(\theta)=\cfrac{y}{x}\impliedby \textit{using the quotient rule} \\\\\\ sec^2(\theta)\cfrac{d\theta}{dt}=\cfrac{\frac{dy}{dt}x-y\frac{dx}{dt}}{x^2}\implies \cfrac{1}{cos^2(\theta)}\cdot \cfrac{d\theta}{dt}=\cfrac{\frac{dy}{dt}x-y\frac{dx}{dt}}{x^2} \\\\\\ \cfrac{d\theta}{dt}=cos^2(\theta)\cdot \cfrac{\frac{dy}{dt}x-y\frac{dx}{dt}}{x^2}\qquad \begin{cases} \cfrac{dy}{dt}=3\\ \cfrac{dx}{dt}=2\\ x=5\\ y=7\\ [cos(\theta)]^2=\left( \frac{5}{\sqrt{74}} \right)^2=\frac{25}{74} \end{cases}$

$\bf \cfrac{d\theta}{dt}=\cfrac{\frac{25}{74}(3\cdot 5-7\cdot 2)}{5^2}\implies \cfrac{d\theta}{dt}=\cfrac{\frac{25}{74}(1)}{25}\implies \cfrac{d\theta}{dt}=\cfrac{\frac{25}{74}}{\frac{25}{1}} \\\\\\ \cfrac{d\theta}{dt}=\cfrac{25}{74}\cdot \cfrac{1}{25}\implies \cfrac{d\theta}{dt}=\cfrac{1}{74}$

now, are the particles moving away or towards them? well, the dθ/dt is postive, that means, the angle is opening, not closing, so, they're moving away from each other