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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) (20)/(3) (B) 20 (C) (147)/(20) (D) 7

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) 203 \frac{20}{3} \newline(B) 2020\newline(C) 14720 \frac{147}{20} \newline(D) 77

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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) 203 \frac{20}{3} \newline(B) 2020\newline(C) 14720 \frac{147}{20} \newline(D) 77
  1. Set Up Problem: We got a 2929-meter ladder against a wall. The top is sliding down at 77 meters per minute. When the bottom is 2121 meters out, we gotta find how fast it's moving away from the wall.
  2. Identify Variables: Let's call the distance from the wall to the bottom of the ladder xx, and the distance from the ground to the top of the ladder yy. We're given yy is decreasing at 77 meters per minute, so dydt=−7\frac{dy}{dt} = -7 meters per minute.
  3. Apply Pythagoras Theorem: We got a right triangle with the ladder, the wall, and the ground. So, by Pythagoras, we got x2+y2=292x^2 + y^2 = 29^2.
  4. Differentiate with Respect to Time: Differentiate both sides with respect to time tt to find dxdt\frac{dx}{dt}. So, 2x(dxdt)+2y(dydt)=02x\left(\frac{dx}{dt}\right) + 2y\left(\frac{dy}{dt}\right) = 0.
  5. Calculate Height of Ladder: Plug in the values we know: x=21x = 21, dydt=−7\frac{dy}{dt} = -7. We get 2⋅21⋅dxdt+2⋅y⋅(−7)=02\cdot21\cdot\frac{dx}{dt} + 2\cdot y\cdot(-7) = 0.
  6. Substitute Values: We need yy, the height of the ladder on the wall when the bottom is 2121 meters out. So, y2=292−212y^2 = 29^2 - 21^2. That's y2=841−441y^2 = 841 - 441.
  7. Calculate Height: Calculate y2y^2: y2=400y^2 = 400. So, y=20y = 20 meters (since yy is a distance, it's positive).
  8. Solve for dx/dt: Now we got yy, so plug it into the differentiated equation: 2×21×dxdt+2×20×(−7)=02 \times 21 \times \frac{dx}{dt} + 2 \times 20 \times (-7) = 0.
  9. Simplify Equation: Simplify the equation: 42(dxdt)−280=042\left(\frac{dx}{dt}\right) - 280 = 0.
  10. Find dxdt\frac{dx}{dt}: Solve for dxdt\frac{dx}{dt}: 42(dxdt)=28042\left(\frac{dx}{dt}\right) = 280. So, (dxdt)=28042\left(\frac{dx}{dt}\right) = \frac{280}{42}.
  11. Find dxdt\frac{dx}{dt}: Solve for dxdt\frac{dx}{dt}: 42(dxdt)=28042\left(\frac{dx}{dt}\right) = 280. So, dxdt\frac{dx}{dt} = 28042\frac{280}{42}.Calculate dxdt\frac{dx}{dt}: dxdt\frac{dx}{dt} = 28042=203\frac{280}{42} = \frac{20}{3} meters per minute.

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