Two cars are driving away from an intersection in perpendicular directions. $\newline$ The first car's velocity is $7$ meters per second and the second car's velocity is $3$ meters per second. $\newline$ At a certain instant, the first car is $5$ meters from the intersection and the second car is $12$ meters from the intersection. $\newline$ What is the rate of change of the distance between the cars at that instant (in meters per second)? $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{99}{13}$ $\newline$ (B) $\mathbf{1 3}$ $\newline$ (C) $\sqrt{58}$ $\newline$ (D) $\frac{71}{13}$

Full solution

Q. Two cars are driving away from an intersection in perpendicular directions.

\newline

The first car's velocity is

7

meters per second and the second car's velocity is

3

meters per second.

\newline

At a certain instant, the first car is

5

meters from the intersection and the second car is

12

meters from the intersection.

\newline

What is the rate of change of the distance between the cars at that instant (in meters per second)?

\newline

Choose

1

answer:

\newline

(A)

\frac{99}{13}

\newline

(B)

\mathbf{1 3}

\newline

(C)

\sqrt{58}

\newline

(D)

\frac{71}{13}

Identify Variables: Now we'll find the rate of change of the distance between the cars using derivatives. Let's call the distance between the cars $s$ , the distance of the first car from the intersection $x$ , and the distance of the second car from the intersection $y$ . According to the Pythagorean theorem, $s^2 = x^2 + y^2$ . Differentiating both sides with respect to time $t$ , we get: $2s\frac{ds}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}$ Now we plug in the velocities of the cars for $\frac{dx}{dt}$ and $\frac{dy}{dt}$ , and the current distance $s$ to find $\frac{ds}{dt}$ . $x$ $0$ $x$ $1$ $x$ $2$ $x$ $3$ $x$ $4$ meters per second