How fast do Cars approach each other

Let $x$ be the distance from car A to the intersection and $y$ be the distance from car B to the intersection. Let $z$ be the distance between the cars. By the Pythagorean theorem, $z=\sqrt{x^2+y^2}$ . We calculate the derivative of $z$ with respect to time, $\frac{\mathrm{d}z}{\mathrm{d}t}$ :

$\begin{aligned} \frac{\mathrm{d}z}{\mathrm{d}t} &= \frac{\mathrm{d}}{\mathrm{d}t}\left(x^2 + y^2\right)^{\frac{1}{2}} \\ &= \frac{1}{2}\left(x^2 + y^2\right)^{-\frac{1}{2}}\left(2x\frac{\mathrm{d} x}{\mathrm{d}t} + 2y\frac{\mathrm{d}y}{\mathrm{d}t}\right) \\ &= \frac{1}{2}\left(x^2 + y^2\right)^{-\frac{1}{2}}\left(2x\cdot50 + 2y\cdot60\right) && \color{#3D99F6} \frac{\mathrm{d} x}{\mathrm{d}t}=50,\frac{\mathrm{d} y}{\mathrm{d}t}=60 \\ &= \frac{1}{2}\left(.3^2 + .4^2\right)^{-\frac{1}{2}}\left(2\cdot .3\cdot50 + 2\cdot .4\cdot60\right) && \color{#3D99F6} \text{Evaluate at }(x,y) = (.3,.4) \\ &= \frac{1}{2}\frac{1}{.5}(30+48) && \color{#3D99F6} .3^2 + .4^2 = .5^2 \\ &= \boxed{78} \end{aligned}$

By Pythagorean's Theorem on the right triangle given, $x^2 + y^2 = z^2$ .

By implicit differentiation, $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 2x \frac{dz}{dt}$

Plugging in values $x = 0.3$ , $y = 0.4$ , $z = \sqrt{0.3^2 + 0.4^2} = 0.5$ , $\frac{dx}{dt} = 50$ , and $\frac{dz}{dt} = 60$ gives $2(0.3)(50) + 2(0.4)(60) = 2(0.5) \frac{dz}{dt}$ .

Solving this equation gives $\frac{dz}{dt} = \boxed{78}$ .

Let $C$ the intersection be the origin $C(0,0)$ , the distance from $C$ to car $A$ be $x$ and the distance from $C$ to car $B$ be $y$ . Then rate of change of $x$ or $\dfrac {d x}{dt} = - 50$ , since $x$ is reducing. Similarly, $\dfrac {dy}{dt} = -60$ . Let the distance between the two cars be $s(x,y)$ . Then we have $s(x,y) = \sqrt{x^2+y^2}$ and its rate of change is given by:

$\begin{aligned} \frac d{dt}s(x,y) & = \frac {\partial s}{\partial x} \cdot \frac {dx}{dt} + \frac {\partial s}{\partial y} \cdot \frac {dy}{dt} \\ & = \frac {x(-50)}{\sqrt{x^2+y^2}} + \frac {y(-60)}{\sqrt{x^2+y^2}} & \small \color{#3D99F6} \text{Putting }x=0.3 \text{ and }y=0.4 \\ & = - \frac {0.3(50)}{0.5} - \frac {0.4(60)}{0.5} \\ & = - 78 \end{aligned}$

Therefore, the absolute value of rate of change of $s$ at $x=0.3$ and $y=0.4$ is $\boxed{78}$ .