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A person stands 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 meters per second. At a certain instant, the car is 24 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) -26 (B) -sqrt269 (C) -12 (D) -(169)/(12)

A person stands 1010 meters east of an intersection and watches a car driving towards the intersection from the north at 1313 meters per second.\newlineAt a certain instant, the car is 2424 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) −26-26\newline(B) −269 -\sqrt{269} \newline(C) −12-12\newline(D) −16912 -\frac{169}{12}

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Full solution

Q. A person stands 1010 meters east of an intersection and watches a car driving towards the intersection from the north at 1313 meters per second.\newlineAt a certain instant, the car is 2424 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) −26-26\newline(B) −269 -\sqrt{269} \newline(C) −12-12\newline(D) −16912 -\frac{169}{12}
  1. Draw Triangle: First, let's draw a right triangle where the car's distance from the intersection is one leg (2424 meters), and the person's distance from the intersection is the other leg (1010 meters). We're looking for the rate of change of the hypotenuse, which represents the distance between the car and the person.
  2. Calculate Initial Distance: Using the Pythagorean theorem, we calculate the initial distance between the car and the person.\newlineDistance2=242+102\text{Distance}^2 = 24^2 + 10^2\newlineDistance2=576+100\text{Distance}^2 = 576 + 100\newlineDistance2=676\text{Distance}^2 = 676\newlineDistance=676\text{Distance} = \sqrt{676}\newlineDistance=26 meters\text{Distance} = 26 \text{ meters}
  3. Find Rate of Change: Now, we need to find the rate of change of this distance. Since the car is moving towards the intersection, the distance between the car and the person is decreasing. We'll use related rates to find this rate of change.
  4. Use Related Rates: 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 1010 meters), and the distance between the car and the person as zz.
  5. Simplify Equation: The Pythagorean theorem in terms of xx, yy, and zz is: x2+y2=z2x^2 + y^2 = z^2 Differentiating both sides with respect to time tt, we get: 2x(dxdt)+2y(dydt)=2z(dzdt)2x\left(\frac{dx}{dt}\right) + 2y\left(\frac{dy}{dt}\right) = 2z\left(\frac{dz}{dt}\right)
  6. Apply Values: Since yy is constant, dydt\frac{dy}{dt} is 00, and we can simplify the equation to:\newline2xdxdt=2zdzdt2x\frac{dx}{dt} = 2z\frac{dz}{dt}
  7. Calculate Rate of Change: We know that dxdt\frac{dx}{dt} (the speed of the car) is −13-13 meters per second (negative because xx is decreasing), and we want to find dzdt\frac{dz}{dt} (the rate of change of the distance between the car and the person).
  8. Calculate Rate of Change: We know that dxdt\frac{dx}{dt} (the speed of the car) is −13-13 meters per second (negative because xx is decreasing), and we want to find dzdt\frac{dz}{dt} (the rate of change of the distance between the car and the person).Plugging in the values we have:\newline2×24×(−13)=2×26×(dzdt)2 \times 24 \times (-13) = 2 \times 26 \times (\frac{dz}{dt})\newline−624=52×(dzdt)-624 = 52 \times (\frac{dz}{dt})\newline(dzdt)=−62452(\frac{dz}{dt}) = \frac{-624}{52}\newline(dzdt)=−12(\frac{dz}{dt}) = -12 meters per second

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