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A 29-meter ladder is sliding down a vertical wall so the distance between the top of the ladder and the ground is decreasing at 7 meters per minute. At a certain instant, the bottom of the ladder is 21 meters from the wall. What is the rate of change of the distance between the bottom of the ladder and the wall at that instant (in meters per minute)? Choose 1 answer: (A) 7 (B) (147)/(20) (C) 20 (D) (20)/(3)

A 2929-meter ladder is sliding down a vertical wall so the distance between the top of the ladder and the ground is decreasing at 77 meters per minute.\newlineAt a certain instant, the bottom of the ladder is 2121 meters from the wall.\newlineWhat is the rate of change of the distance between the bottom of the ladder and the wall at that instant (in meters per minute)?\newlineChoose 11 answer:\newline(A) 77\newline(B) 14720 \frac{147}{20} \newline(C) 2020\newline(D) 203 \frac{20}{3}

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Q. A 2929-meter ladder is sliding down a vertical wall so the distance between the top of the ladder and the ground is decreasing at 77 meters per minute.\newlineAt a certain instant, the bottom of the ladder is 2121 meters from the wall.\newlineWhat is the rate of change of the distance between the bottom of the ladder and the wall at that instant (in meters per minute)?\newlineChoose 11 answer:\newline(A) 77\newline(B) 14720 \frac{147}{20} \newline(C) 2020\newline(D) 203 \frac{20}{3}
  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. Pythagorean Theorem: Let xx be the distance from the bottom of the ladder to the wall, and yy be the height of the ladder above the ground. We have x2+y2=292x^2 + y^2 = 29^2 because of the Pythagorean theorem.
  3. Differentiation: Differentiate both sides with respect to time tt to find the rates of change. We get 2xdxdt+2ydydt=02x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 since the length of the ladder doesn't change.
  4. Given Rates: We know dydt=−7\frac{dy}{dt} = -7 meters per minute (since the top of the ladder is sliding down), and we need to find dxdt\frac{dx}{dt} when x=21x = 21 meters.
  5. Equation Simplification: Plug in the values we know: 2×21×dxdt+2×y×(−7)=02\times21\times\frac{dx}{dt} + 2\times y\times(-7) = 0.
  6. Calculate y: We need to find yy when x=21x = 21. Using the Pythagorean theorem, y2=292−212y^2 = 29^2 - 21^2. Calculate y2=841−441y^2 = 841 - 441.
  7. Solve for dxdt\frac{dx}{dt}: y2=400y^2 = 400, so y=20y = 20 meters (since yy is positive as it's a distance).
  8. Final Calculation: Now we can solve for dxdt\frac{dx}{dt}: 2×21×(dxdt)−2×20×7=02 \times 21 \times \left(\frac{dx}{dt}\right) - 2 \times 20 \times 7 = 0.
  9. Final Calculation: Now we can solve for dxdt\frac{dx}{dt}: 2×21×dxdt−2×20×7=02\times 21\times\frac{dx}{dt} - 2\times 20\times 7 = 0.Simplify the equation: 42×dxdt−280=042\times\frac{dx}{dt} - 280 = 0.
  10. Final Calculation: Now we can solve for dxdt\frac{dx}{dt}: 2×21×dxdt−2×20×7=02\times 21\times\frac{dx}{dt} - 2\times 20\times 7 = 0.Simplify the equation: 42×dxdt−280=042\times\frac{dx}{dt} - 280 = 0.Add 280280 to both sides: 42×dxdt=28042\times\frac{dx}{dt} = 280.
  11. Final Calculation: Now we can solve for dxdt\frac{dx}{dt}: 2×21×dxdt−2×20×7=02\times 21\times\frac{dx}{dt} - 2\times 20\times 7 = 0.Simplify the equation: 42×dxdt−280=042\times\frac{dx}{dt} - 280 = 0.Add 280280 to both sides: 42×dxdt=28042\times\frac{dx}{dt} = 280.Divide by 4242 to solve for dxdt\frac{dx}{dt}: dxdt=28042\frac{dx}{dt} = \frac{280}{42}.
  12. Final Calculation: Now we can solve for dxdt\frac{dx}{dt}: 2×21×(dxdt)−2×20×7=02\times 21\times\left(\frac{dx}{dt}\right) - 2\times 20\times 7 = 0.Simplify the equation: 42×(dxdt)−280=042\times\left(\frac{dx}{dt}\right) - 280 = 0.Add 280280 to both sides: 42×(dxdt)=28042\times\left(\frac{dx}{dt}\right) = 280.Divide by 4242 to solve for dxdt\frac{dx}{dt}: dxdt=28042\frac{dx}{dt} = \frac{280}{42}.Calculate dxdt\frac{dx}{dt}: dxdt=203\frac{dx}{dt} = \frac{20}{3} meters per minute.

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