A ladder 25 feet long that was leaning against a vertical wall begins to slip. Its top slides down the wall while its bottom moves along the lever ground at a constant speed of 4 ft/sec. How fast is the top of the ladder moving when it is 20 feet above the ground?
Just about every Calculus book I have ever seen uses this question as an introductory example of "rate of change" problem.
At a time of t seconds, let the foot of the ladder be x ft from the wall, and the top of the ladder by y ft above the ground>
then,
x^2 + y^2= 25^2
2x dx/dt + 2y dy/dt = 0
or
dy/dt = -2x dx/dt/2y = -x dx/dt/y
given: dx/dt = 4 ft/sec
find : dy/dt when y = 20
when y = 20
x^2 + 20^2 = 25^2
x^2 = 225
x = 15
dy/dt = -15(4)/20 ft/sec
= -3 ft/sec
(the negative tells me that y is decreasing)
To solve this problem, we can use related rates. Let's denote the distance of the top of the ladder from the ground as x.
First, let's identify the given information:
- The ladder is 25 feet long.
- The bottom of the ladder is moving along the ground at a constant speed of 4 ft/sec.
To find how fast the top of the ladder is moving, we need to find dx/dt (the rate at which x is changing) when x = 20 ft.
To relate the variables x and t, we'll use the Pythagorean theorem:
x^2 + y^2 = 25^2
Differentiating both sides with respect to t, we get:
2x(dx/dt) + 2y(dy/dt) = 0
Since we know dy/dt is 4 ft/sec (constant speed of the bottom of the ladder), we can substitute this value:
2x(dx/dt) + 2y(4) = 0
Simplifying the equation at x=20 ft:
2(20)(dx/dt) + 2y(4) = 0
40(dx/dt) + 8y = 0
To find y, we can use the Pythagorean theorem:
x^2 + y^2 = 25^2
(20)^2 + y^2 = 25^2
400 + y^2 = 625
y^2 = 625 - 400
y^2 = 225
y = 15 ft
Substituting y = 15 ft into the equation:
40(dx/dt) + 8(15) = 0
40(dx/dt) + 120 = 0
40(dx/dt) = -120
dx/dt = -120/40
dx/dt = -3 ft/sec
Therefore, the top of the ladder is moving at a rate of -3 ft/sec when it is 20 feet above the ground. The negative sign indicates it is moving downward.
To solve this problem, we need to use related rates. Related rates involve finding how the change in one quantity affects the change in another quantity. In this case, we want to find how the height of the ladder changes with respect to time.
Let's define some variables:
Let x be the distance from the bottom of the ladder to the wall (along the horizontal ground).
Let y be the distance from the top of the ladder to the ground (along the vertical wall).
We are given that the ladder is 25 feet long, so we have the relation: x^2 + y^2 = 25^2. [Equation 1]
We are also given that the bottom of the ladder is moving along the ground at a constant speed of 4 ft/sec. This means dx/dt = 4. [Equation 2]
We want to find dy/dt, the rate at which the top of the ladder is moving when it is 20 feet above the ground.
To find dy/dt, we need to differentiate both sides of Equation 1 with respect to time t.
Differentiating x^2 + y^2 = 25^2 implicitly:
2x(dx/dt) + 2y(dy/dt) = 0.
Substituting the given value of dx/dt = 4 and the value of x when y = 20 into the equation, we can solve for dy/dt, which is the rate at which the top of the ladder is moving when it is 20 feet above the ground.
Hope this helps!