A person stands $15$ meters east of an intersection and watches a car driving towards the intersection from the north at $1$ meter per second. $\newline$ At a certain instant, the car is $8$ 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) $-\sqrt{65}$ $\newline$ (B) $-\frac{8}{17}$ $\newline$ (C) $-2$ . $125$ $\newline$ (D) $-17$

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Rate of travel: word problems

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Q. A person stands

15

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

1

meter per second.

\newline

At a certain instant, the car is

8

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{65}

\newline

(B)

-\frac{8}{17}

\newline

(C)

-2

125

\newline

(D)

-17

Calculate Hypotenize: Now, we need to use the Pythagorean theorem to find the hypotenuse (the distance between the car and the person). $\newline$ $\text{Hypotenuse}^2 = 8^2 + 15^2$ $\newline$ $\text{Hypotenuse}^2 = 64 + 225$ $\newline$ $\text{Hypotenuse}^2 = 289$ $\newline$ $\text{Hypotenuse} = \sqrt{289}$ $\newline$ $\text{Hypotenuse} = 17$ meters.
Find Rate of Change: Next, we'll use related rates to find the rate of change of the distance between the car and the person. Since the car is moving towards the intersection, the distance between the car and the intersection is decreasing at a rate of $1$ meter per second.
Apply Chain Rule: Let's denote the distance between the car and the intersection as $x$ , the distance between the person and the intersection as $y$ (which is constant at $15$ meters), and the distance between the car and the person as $z$ . We have $\frac{dz}{dt} = \left(\frac{dx}{dz}\right) * \left(\frac{dz}{dt}\right)$ . Since $y$ is constant, $\frac{dy}{dt} = 0$ , and we only need to consider $\frac{dx}{dt}$ and $\frac{dz}{dt}$ .
Substitute Values: Using the chain rule, we differentiate both sides of the equation $z^2 = x^2 + y^2$ with respect to time $t$ . $\newline$ $2z\frac{dz}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}$ $\newline$ Since $\frac{dy}{dt} = 0$ , the equation simplifies to $2z\frac{dz}{dt} = 2x\frac{dx}{dt}$
Calculate Rate of Change: Now we plug in the values we know: $x = 8$ meters, $\frac{dx}{dt} = -1$ meter/second (since the distance is decreasing), and $z = 17$ meters. $\newline$ $2(17)\frac{dz}{dt} = 2(8)(-1)$ $\newline$ $34\frac{dz}{dt} = -16$ $\newline$ $\frac{dz}{dt} = \frac{-16}{34}$ $\newline$ $\frac{dz}{dt} = \frac{-8}{17}$ meters per second.
Calculate Rate of Change: Now we plug in the values we know: $x = 8$ meters, $\frac{dx}{dt} = -1$ meter/second (since the distance is decreasing), and $z = 17$ meters. $\newline$ $2(17)\frac{dz}{dt} = 2(8)(-1)$ $\newline$ $34\frac{dz}{dt} = -16$ $\newline$ $\frac{dz}{dt} = \frac{-16}{34}$ $\newline$ $\frac{dz}{dt} = \frac{-8}{17}$ meters per second.So, the rate of change of the distance between the car and the person at that instant is $\frac{-8}{17}$ meters per second. $\newline$ The correct answer is (B) $\frac{-(8)}{(17)}$ .