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A 39-meter ladder is sliding down a vertical wall so the distance between the bottom of the ladder and the wall is increasing at 10 meters per minute. At a certain instant, the bottom of the ladder is 36 meters from the wall. What is the rate of change of the distance between the top of the ladder and the ground at that instant (in meters per minute)? Choose 1 answer: (A) -12 (B) -(25)/(6) (C) -24 (D) -10

A 3939-meter ladder is sliding down a vertical wall so the distance between the bottom of the ladder and the wall is increasing at 1010 meters per minute.\newlineAt a certain instant, the bottom of the ladder is 36 \mathbf{3 6} meters from the wall.\newlineWhat is the rate of change of the distance between the top of the ladder and the ground at that instant (in meters per minute)?\newlineChoose 11 answer:\newline(A) −12-12\newline(B) −256 -\frac{25}{6} \newline(C) −24-24\newline(D) −10-10

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Q. A 3939-meter ladder is sliding down a vertical wall so the distance between the bottom of the ladder and the wall is increasing at 1010 meters per minute.\newlineAt a certain instant, the bottom of the ladder is 36 \mathbf{3 6} meters from the wall.\newlineWhat is the rate of change of the distance between the top of the ladder and the ground at that instant (in meters per minute)?\newlineChoose 11 answer:\newline(A) −12-12\newline(B) −256 -\frac{25}{6} \newline(C) −24-24\newline(D) −10-10
  1. Triangle Description: We're dealing with a right triangle where the ladder is the hypotenuse, the distance from the wall is one leg, and the height of the ladder above the ground is the other leg.
  2. Variable Assignment: Let's call the distance from the wall to the bottom of the ladder xx, and the height of the ladder above the ground yy. We know that xx is changing at 1010 meters per minute.
  3. Pythagoras' Theorem: Using Pythagoras' theorem, we have x2+y2=392x^2 + y^2 = 39^2, since the ladder is 3939 meters long.
  4. Differentiation: Differentiate both sides of the equation with respect to time tt to find the rate of change of yy with respect to time. So we get 2x(dxdt)+2y(dydt)=02x\left(\frac{dx}{dt}\right) + 2y\left(\frac{dy}{dt}\right) = 0.
  5. Initial Values: We know dxdt=10\frac{dx}{dt} = 10 meters per minute (since xx is increasing at 1010 meters per minute), and at the instant we're considering, x=36x = 36 meters.
  6. Calculate yy: We need to find yy at the instant x=36x = 36 meters. Using Pythagoras' theorem again, we have 362+y2=39236^2 + y^2 = 39^2.
  7. Substitute Values: Calculate y2=392−362=1521−1296=225y^2 = 39^2 - 36^2 = 1521 - 1296 = 225.
  8. Solve for dydt\frac{dy}{dt}: Take the square root of y2y^2 to find yy. So y=225=15y = \sqrt{225} = 15 meters.
  9. Isolate dydt\frac{dy}{dt}: Now we can substitute xx, dxdt\frac{dx}{dt}, and yy into the differentiated equation: 2⋅36⋅10+2⋅15⋅(dydt)=02\cdot36\cdot10 + 2\cdot15\cdot\left(\frac{dy}{dt}\right) = 0.
  10. Final Result: Solve for dydt\frac{dy}{dt}: 720+30⋅dydt=0720 + 30\cdot\frac{dy}{dt} = 0.
  11. Final Result: Solve for dydt\frac{dy}{dt}: 720+30(dydt)=0720 + 30\left(\frac{dy}{dt}\right) = 0.Isolate dydt\frac{dy}{dt}: 30(dydt)=−72030\left(\frac{dy}{dt}\right) = -720.
  12. Final Result: Solve for dydt\frac{dy}{dt}: 720+30(dydt)=0720 + 30\left(\frac{dy}{dt}\right) = 0.Isolate dydt\frac{dy}{dt}: 30(dydt)=−72030\left(\frac{dy}{dt}\right) = -720.Divide both sides by 3030 to find dydt\frac{dy}{dt}: dydt=−72030=−24\frac{dy}{dt} = \frac{-720}{30} = -24 meters per minute.

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