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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. At a certain instant, the car is 8 meters from the intersection. What is the rate of change of the distance between the car and the person at that instant (in meters per second)? Choose 1 answer: (A) -sqrt65 (B) -(8)/(17) (C) -2.125 (D) -17

A person stands 1515 meters east of an intersection and watches a car driving towards the intersection from the north at 11 meter per second.\newlineAt a certain instant, the car is 88 meters from the intersection.\newlineWhat is the rate of change of the distance between the car and the person at that instant (in meters per second)?\newlineChoose 11 answer:\newline(A) −65 -\sqrt{65} \newline(B) −817 -\frac{8}{17} \newline(C) −2-2.125125\newline(D) −17-17

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Full solution

Q. A person stands 1515 meters east of an intersection and watches a car driving towards the intersection from the north at 11 meter per second.\newlineAt a certain instant, the car is 88 meters from the intersection.\newlineWhat is the rate of change of the distance between the car and the person at that instant (in meters per second)?\newlineChoose 11 answer:\newline(A) −65 -\sqrt{65} \newline(B) −817 -\frac{8}{17} \newline(C) −2-2.125125\newline(D) −17-17
  1. Calculate Hypotenize: Now, we need to use the Pythagorean theorem to find the hypotenuse (the distance between the car and the person).\newlineHypotenuse2=82+152\text{Hypotenuse}^2 = 8^2 + 15^2\newlineHypotenuse2=64+225\text{Hypotenuse}^2 = 64 + 225\newlineHypotenuse2=289\text{Hypotenuse}^2 = 289\newlineHypotenuse=289\text{Hypotenuse} = \sqrt{289}\newlineHypotenuse=17\text{Hypotenuse} = 17 meters.
  2. 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 11 meter per second.
  3. Apply Chain Rule: Let's denote the distance between the car and the intersection as xx, the distance between the person and the intersection as yy (which is constant at 1515 meters), and the distance between the car and the person as zz. We have dzdt=(dxdz)∗(dzdt)\frac{dz}{dt} = \left(\frac{dx}{dz}\right) * \left(\frac{dz}{dt}\right). Since yy is constant, dydt=0\frac{dy}{dt} = 0, and we only need to consider dxdt\frac{dx}{dt} and dzdt\frac{dz}{dt}.
  4. Substitute Values: Using the chain rule, we differentiate both sides of the equation z2=x2+y2z^2 = x^2 + y^2 with respect to time tt. \newline2zdzdt=2xdxdt+2ydydt2z\frac{dz}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}\newlineSince dydt=0\frac{dy}{dt} = 0, the equation simplifies to 2zdzdt=2xdxdt2z\frac{dz}{dt} = 2x\frac{dx}{dt}
  5. Calculate Rate of Change: Now we plug in the values we know: x=8x = 8 meters, dxdt=−1\frac{dx}{dt} = -1 meter/second (since the distance is decreasing), and z=17z = 17 meters.\newline2(17)dzdt=2(8)(−1)2(17)\frac{dz}{dt} = 2(8)(-1)\newline34dzdt=−1634\frac{dz}{dt} = -16\newlinedzdt=−1634\frac{dz}{dt} = \frac{-16}{34}\newlinedzdt=−817\frac{dz}{dt} = \frac{-8}{17} meters per second.
  6. Calculate Rate of Change: Now we plug in the values we know: x=8x = 8 meters, dxdt=−1\frac{dx}{dt} = -1 meter/second (since the distance is decreasing), and z=17z = 17 meters.\newline2(17)dzdt=2(8)(−1)2(17)\frac{dz}{dt} = 2(8)(-1)\newline34dzdt=−1634\frac{dz}{dt} = -16\newlinedzdt=−1634\frac{dz}{dt} = \frac{-16}{34}\newlinedzdt=−817\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 −817\frac{-8}{17} meters per second.\newlineThe correct answer is (B) −(8)(17)\frac{-(8)}{(17)}.

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