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Lea was given this problem:
A 20-meter ladder is sliding down a vertical wall so the distance 
y(t) between the top of the ladder and the ground is decreasing at 8 meters per minute. At a certain instant 
t_(0), the top of the ladder is 12 meters from the ground. What is the rate of change of the distance 
x(t) between the bottom of the ladder and the wall at that instant?
Which equation should Lea use to solve the problem?
Choose 1 answer:
(A) 
sin[20]=(y(t))/(x(t))
(B) 
20=(x(t)*y(t))/(2)
(C) 
20+x(t)+y(t)=180
(D) 
20^(2)=[x(t)]^(2)+[y(t)]^(2)

Lea was given this problem:\newlineA 2020-meter ladder is sliding down a vertical wall so the distance y(t) y(t) between the top of the ladder and the ground is decreasing at 88 meters per minute. At a certain instant t0 t_{0} , the top of the ladder is 1212 meters from the ground. What is the rate of change of the distance x(t) x(t) between the bottom of the ladder and the wall at that instant?\newlineWhich equation should Lea use to solve the problem?\newlineChoose 11 answer:\newline(A) sin[20]=y(t)x(t) \sin [20]=\frac{y(t)}{x(t)} \newline(B) 20=x(t)y(t)2 20=\frac{x(t) \cdot y(t)}{2} \newline(C) 20+x(t)+y(t)=180 20+x(t)+y(t)=180 \newline(D) 202=[x(t)]2+[y(t)]2 20^{2}=[x(t)]^{2}+[y(t)]^{2}

Full solution

Q. Lea was given this problem:\newlineA 2020-meter ladder is sliding down a vertical wall so the distance y(t) y(t) between the top of the ladder and the ground is decreasing at 88 meters per minute. At a certain instant t0 t_{0} , the top of the ladder is 1212 meters from the ground. What is the rate of change of the distance x(t) x(t) between the bottom of the ladder and the wall at that instant?\newlineWhich equation should Lea use to solve the problem?\newlineChoose 11 answer:\newline(A) sin[20]=y(t)x(t) \sin [20]=\frac{y(t)}{x(t)} \newline(B) 20=x(t)y(t)2 20=\frac{x(t) \cdot y(t)}{2} \newline(C) 20+x(t)+y(t)=180 20+x(t)+y(t)=180 \newline(D) 202=[x(t)]2+[y(t)]2 20^{2}=[x(t)]^{2}+[y(t)]^{2}
  1. Understand Relationship: Lea needs to find the relationship between the rate of change of y(t)y(t), the distance from the top of the ladder to the ground, and the rate of change of x(t)x(t), the distance from the bottom of the ladder to the wall. Since the ladder is sliding down the wall, we are dealing with a right-angled triangle where the ladder is the hypotenuse. The Pythagorean theorem relates the sides of a right-angled triangle.
  2. Apply Pythagorean Theorem: The Pythagorean theorem states that in a right-angled triangle, the square of the length of the hypotenuse cc is equal to the sum of the squares of the lengths of the other two sides aa and bb. In this case, the ladder is the hypotenuse, so the equation is c2=a2+b2c^2 = a^2 + b^2, where cc is the length of the ladder, aa is the distance x(t)x(t), and bb is the distance y(t)y(t).
  3. Utilize Constant Length: Since the length of the ladder is constant at 2020 meters, we can write the equation as 202=[x(t)]2+[y(t)]220^2 = [x(t)]^2 + [y(t)]^2. This equation will allow us to relate the changes in x(t)x(t) and y(t)y(t) as the ladder slides down the wall.
  4. Use Pythagorean Theorem: Lea should use the equation 202=[x(t)]2+[y(t)]220^2 = [x(t)]^2 + [y(t)]^2 to solve the problem because it relates the distances x(t)x(t) and y(t)y(t) with the constant length of the ladder. This is the Pythagorean theorem applied to the situation, and it is the only equation among the choices that correctly represents the relationship between the sides of the right-angled triangle formed by the ladder, the wall, and the ground.

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