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A 10-meter ladder is sliding down a vertical wall so the distance between the top of the ladder and the ground is decreasing at 3 meters per minute. At a certain instant, the bottom of the ladder is 6 meters from the wall. What is the rate of change of the area formed by the ladder at that instant (in square meters per minute)? Choose 1 answer: (A) 12 (B) -(7)/(2) (C) 7 (D) -6

A 1010-meter ladder is sliding down a vertical wall so the distance between the top of the ladder and the ground is decreasing at 33 meters per minute.\newlineAt a certain instant, the bottom of the ladder is 66 meters from the wall.\newlineWhat is the rate of change of the area formed by the ladder at that instant (in square meters per minute)?\newlineChoose 11 answer:\newline(A) 1212\newline(B) −72 -\frac{7}{2} \newline(C) 77\newline(D) −6-6

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Q. A 1010-meter ladder is sliding down a vertical wall so the distance between the top of the ladder and the ground is decreasing at 33 meters per minute.\newlineAt a certain instant, the bottom of the ladder is 66 meters from the wall.\newlineWhat is the rate of change of the area formed by the ladder at that instant (in square meters per minute)?\newlineChoose 11 answer:\newline(A) 1212\newline(B) −72 -\frac{7}{2} \newline(C) 77\newline(D) −6-6
  1. Triangle Area Formula: The ladder forms a right triangle with the wall and the ground. The area of a right triangle is (12)×base×height(\frac{1}{2}) \times \text{base} \times \text{height}.
  2. Variables and Area Calculation: Let xx be the distance from the bottom of the ladder to the wall, and yy be the distance from the top of the ladder to the ground. The area A=(12)×x×yA = (\frac{1}{2}) \times x \times y.
  3. Pythagorean Theorem Application: Given that the ladder is 1010 meters long, by the Pythagorean theorem, we have x2+y2=102x^2 + y^2 = 10^2.
  4. Differentiation for Rates of Change: Differentiate both sides with respect to time tt to find the rates of change: 2xdxdt+2ydydt=02x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0.
  5. Given Rate of Change: We know that dydt=−3\frac{dy}{dt} = -3 meters per minute (since the top is going down, it's negative), and we need to find dxdt\frac{dx}{dt}.
  6. Substitution for Variables: Substitute x=6x = 6 meters and y=(102−62)=(100−36)=64=8y = \sqrt{(10^2 - 6^2)} = \sqrt{(100 - 36)} = \sqrt{64} = 8 meters into the differentiated equation.
  7. Equation Simplification: Now we have 2×6×(dxdt)+2×8×(−3)=02\times 6\times \left(\frac{dx}{dt}\right) + 2\times 8\times (-3) = 0.
  8. Solving for dxdt\frac{dx}{dt}: Solve for dxdt\frac{dx}{dt}: 12⋅(dxdt)−48=012\cdot\left(\frac{dx}{dt}\right) - 48 = 0.
  9. Final Rate Calculation: Add 4848 to both sides: 12(dxdt)=4812\left(\frac{dx}{dt}\right) = 48.
  10. Area Rate of Change Formula: Divide by 1212: dxdt=4812=4\frac{dx}{dt} = \frac{48}{12} = 4 meters per minute.
  11. Substitution for Area Rate: Now we find the rate of change of the area dAdt=12⋅(x⋅dydt+y⋅dxdt)\frac{dA}{dt} = \frac{1}{2} \cdot (x\cdot\frac{dy}{dt} + y\cdot\frac{dx}{dt}).
  12. Area Rate Calculation: Substitute x=6x = 6, y=8y = 8, dydt=−3\frac{dy}{dt} = -3, and dxdt=4\frac{dx}{dt} = 4 into the equation for dAdt\frac{dA}{dt}.
  13. Area Rate Calculation: Substitute x=6x = 6, y=8y = 8, dydt=−3\frac{dy}{dt} = -3, and dxdt=4\frac{dx}{dt} = 4 into the equation for dAdt\frac{dA}{dt}.Calculate dAdt=12×(6×(−3)+8×4)\frac{dA}{dt} = \frac{1}{2} \times (6 \times (-3) + 8 \times 4).
  14. Area Rate Calculation: Substitute x=6x = 6, y=8y = 8, dydt=−3\frac{dy}{dt} = -3, and dxdt=4\frac{dx}{dt} = 4 into the equation for dAdt\frac{dA}{dt}.Calculate dAdt=12×(6×(−3)+8×4)\frac{dA}{dt} = \frac{1}{2} \times (6 \times (-3) + 8 \times 4).dAdt=12×(−18+32)\frac{dA}{dt} = \frac{1}{2} \times (-18 + 32).
  15. Area Rate Calculation: Substitute x=6x = 6, y=8y = 8, dydt=−3\frac{dy}{dt} = -3, and dxdt=4\frac{dx}{dt} = 4 into the equation for dAdt\frac{dA}{dt}.Calculate dAdt=12×(6×(−3)+8×4)\frac{dA}{dt} = \frac{1}{2} \times (6\times(-3) + 8\times4).dAdt=12×(−18+32)\frac{dA}{dt} = \frac{1}{2} \times (-18 + 32).dAdt=12×14=7\frac{dA}{dt} = \frac{1}{2} \times 14 = 7 square meters per minute.

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