Question 1174226: A ladder 4m long at a construction site is resting against a wall. The bottom of the ladder is slipping away from the wall. Find the estimate of the instantaneous rate of change of the Height H of the top of the ladder with respect to the Distance D of the bottom of the ladder from the wall when the bottom of the ladder is 2.5m away from the wall. Use h = 0.01 as the central interval.
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Absolutely! Let's solve this problem step-by-step.
**1. Set up the Problem**
* **Ladder Length:** 4 meters (constant)
* **Distance from Wall (D):** 2.5 meters
* **Height on Wall (H):** We need to find this.
* **Rate of Change:** We want to find dH/dD (how the height changes with respect to the distance).
**2. Use the Pythagorean Theorem**
* We have a right triangle formed by the wall, the ground, and the ladder.
* The Pythagorean theorem states: D² + H² = 4² (where 4 is the ladder length)
* D² + H² = 16
**3. Express Height (H) in Terms of Distance (D)**
* H² = 16 - D²
* H = √(16 - D²)
**4. Find the Instantaneous Rate of Change (dH/dD)**
* We'll use the given central difference method to estimate the derivative.
* The formula for the central difference is:
* dH/dD ≈ [H(D + h) - H(D - h)] / (2h)
* Where:
* D = 2.5 meters
* h = 0.01 meters
**5. Calculate H(D + h) and H(D - h)**
* H(D + h) = H(2.5 + 0.01) = H(2.51) = √(16 - 2.51²) ≈ √(16 - 6.3001) ≈ √9.6999 ≈ 3.11446
* H(D - h) = H(2.5 - 0.01) = H(2.49) = √(16 - 2.49²) ≈ √(16 - 6.2001) ≈ √9.7999 ≈ 3.13048
**6. Substitute into the Central Difference Formula**
* dH/dD ≈ (3.11446 - 3.13048) / (2 * 0.01)
* dH/dD ≈ (-0.01602) / 0.02
* dH/dD ≈ -0.801
**Result**
* The estimated instantaneous rate of change of the height of the ladder with respect to the distance of the bottom of the ladder from the wall is approximately -0.801 meters per meter.
**Therefore, when the bottom of the ladder is 2.5 meters from the wall, the top of the ladder is sliding down the wall at a rate of approximately 0.801 meters for every meter the bottom slides away.**
Answer by ikleyn(52779)  (Show Source):
You can put this solution on YOUR website! .
A ladder 4m long at a construction site is resting against a wall. The bottom of the ladder is slipping away
from the wall. Find the estimate of the instantaneous rate of change of the Height H of the top of the ladder
with respect to the Distance D of the bottom of the ladder from the wall
when the bottom of the ladder is 2.5m away from the wall. Use h = 0.01 as the central interval.
~~~~~~~~~~~~~~~~~~~~~~~~~
First, the problem in the post is posed INCORRECTLY.
To be correct, the problem must provide the rate of the horizontal move of the base of the ladder
from the wall, as the input data. Without this input data, the problem CAN NOT BE SOLVED.
Second, this instruction "Use h = 0.01 as the central interval" is IRRELEVANT to the problem.
By knowing the mathematical level of those who write to this forum, I am 837% sure
that this "instruction" came mistakenly from the other problem.
Third, the "solution" by @CPhill is non-sensical collection/soup of words.
It does not solve the problem, and even can not be a subject for discussions.
So, my advise to you is to ignore his post, for the sake of safety of your mind.
Now I am ready to start my solution.
I will assume, that the rate of the horizontal move of the base of the ladder
from the wall is 0.5 meters per second.
S O L U T I O N
Let H be the height of the ladder (in meters) when it leans against the wall.
Let D be the distance of the ladder base from the wall (in meters).
Write the Pythagorean equation
H^2 + D^2 = 4^2 = 16, (1)
where '4' is the ladder's length, in meters.
When the ladder moves, we consider H and D as functions of time; H = H(t), D = D(t).
But the length of the ladder remains with no change.
Differentiate equation (1) with respect to time, keeping in mind that the right side is a constant.
You will get
-D*D'
2H*H' + 2D*D' = 0, or H*H' + D*D' = 0, or H' = -------. (2)
H
Calculate H: H =  = 3.1225 m (approximately).
Recall that D' is given D' = 0.5 m/s.
Substitute the values into formula (2) and get
H' =  = -0.40032 m/s.
ANSWER. Thus the ladder's upper point moves down along the vertical wall
with the rate of 0.4 m/s at the assigned time moment.
Solved correctly, with complete explanations.
For this class of problems, it is a standard methodology of their solution.
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