A person stands $30$ meters east of an intersection and watches a car driving away from the intersection to the north at $17$ meters per second. $\newline$ At a certain instant, the car is $16$ meters from the intersection. $\newline$ What is the rate of change of the distance between the car and the person at that instant (in meters per second)? $\newline$ Choose $1$ answer: $\newline$ (A) $\sqrt{1189}$ $\newline$ (B) $34$ $\newline$ (C) $36$ . $125$ $\newline$ (D) $8$

Full solution

Q. A person stands

30

meters east of an intersection and watches a car driving away from the intersection to the north at

17

meters per second.

\newline

At a certain instant, the car is

16

meters from the intersection.

\newline

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

\newline

Choose

1

answer:

\newline

(A)

\sqrt{1189}

\newline

(B)

34

\newline

(C)

36

125

\newline

(D)

8

Calculate Initial Distance: First, let's find the initial distance between the car and the person using the Pythagorean theorem. $\newline$ Distance $^2 = 30^2 + 16^2$ $\newline$ Distance $^2 = 900 + 256$ $\newline$ Distance $^2 = 1156$ $\newline$ Distance $= \sqrt{1156}$ $\newline$ Distance $= 34$ meters.
Find Rate of Change: Now, we need to find the rate of change of this distance. The car is moving away at $17 \, \text{m/s}$ , which is the rate of change of the leg of the triangle parallel to the car's path. $\newline$ We'll use the derivative of the Pythagorean theorem to find the rate of change of the hypotenuse. $\newline$ Let's call the distance between the car and the person ' $z$ ', the distance of the car from the intersection ' $y$ ', and the distance of the person from the intersection ' $x$ '. $\newline$ So, $z^2 = x^2 + y^2$ $\newline$ Differentiating both sides with respect to ' $t$ ', we get: $\newline$ $2z\frac{dz}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}$ $\newline$ Since $x$ is constant ( $30 \, \text{m}$ ), $\frac{dx}{dt} = 0$ , so we can simplify to: $\newline$ $z$ $0$
Derivative Calculation: Plugging in the values we have: $\newline$ $2 \times 34\frac{dz}{dt} = 2 \times 16 \times 17$ $\newline$ $68\frac{dz}{dt} = 544$ $\newline$ $\frac{dz}{dt} = \frac{544}{68}$ $\newline$ $\frac{dz}{dt} = 8 \, \text{m/s}.$