Lesson A ladder foot slides on the ground

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A ladder foot slides on the ground

Problem 1

A 13 foot ladder leans against a wall. The foot of a ladder begins to slide away from the wall
at the rate of 1 foot per minute. When the foot is 5 ft from the wall, at what rate is the top of the ladder is falling?

Solution

Let x be horizontal distance from the wall and y be vertical coordinate.

Then from Pythagoras

    x^2 + y^2 = 13^2.    (1)


Here x = x(t) and y = y(t) are functions of time, t.


Differentiate equation  (1)  over t.  You will get

    2x*x'(t) + 2y*y'(t) = 0,

hence

    y't = - (x*x'(t))/y.


Evaluate it at the given values  x = 5 ft,  x'(t) = 1 ft/minute,  y =  =  =  = 12.


You will get  y'(t) = -  = - 5/12 ft/minute.    


ANSWER.  The top of the ladder moves vertically down at the rate of 5/12 ft per minute.

Problem 2

A 10 m ladder is leaning against a vertical wall with its foot on the same horizontal plane
as the base of the wall. If the lower end of the ladder is moving away from the wall horizontally at 1 m/sec,
how fast is the top of the ladder descending when the lower end is 4 meters from the wall?

Solution

Let x be horizontal distance from the wall and y be vertical coordinate.

Then from Pythagoras

    x^2 + y^2 = 10^2.    (1)


Here x = x(t) and y = y(t) are functions of time, t.


Differentiate equation  (1)  over t.  You will get

    2x*x'(t) + 2y*y'(t) = 0,

hence

    y't = - (x*x'(t))/y.


Evaluate it at the given values  x = 4 m,  x'(t) = 1 m/s,  y =  =  = .


You will get  y'(t) = -  = - 0.436 m/s  (rounded).    ANSWER

Problem 3

A 6 m ladder is placed against a wall. If the bottom of the ladder is pushed in towards the wall at 0.4 m/s,
how fast is the top of the ladder moving when the ladder reaches 4.5 m above the ground?

Solution

Imagine a coordinate plane with coordinate axes x and y (x horizontal, y vertical).


Imagine that the bottom of the ladder slides along the horizontal x-axis from some negative x-values to zero 
(from left to right).


Imagine that the top of the ladder slides along the vertical y-axis.


The length of the ladder is  d = .     (1)


Here x and y are, actually, functions of the time  x = x(t), y = y(t).


You are given that the derivative of the function x(t) over time t is 


    x'(t) = 0.4 m/s.


Square both sides of the formula (1)


    d^2 = x^2(t) + y^2(t).    (2)


Since the length of the ladder d is a constant,  d^2 is a constant, too,


Therefore, the derivative of d^2 over time is zero. 

From the other side, this derivative, from (2), is  2x*x' + 2y*y'.


Thus we have this equation


    2x*x' + 2y*y' = 0,  or,  canceling the factor 2,


    x*x' + y*y' = 0.    (3)


When y = 4.5 m above the ground, you have a right angled triangle with the hypotenuse 6 m and vertical leg of 4.5 m.

So, its horizontal leg is  x = -  = -3.969 meters.


Now you substitute these data  x= -3.969 m,  y= 4.5 m,  x' = 0.4 m/s  into the formula (3).


You get then


    y' = - (x*x')/y = -  = 0.353 m/s.    ANSWER

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    - OVERVIEW of my lessons on Calculus word problems

Use this file/link ALGEBRA-II - YOUR ONLINE TEXTBOOK to navigate over all topics and lessons of the online textbook ALGEBRA-II.

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