Answer:
(i) 7.2 feet per minute.
(ii) No, the rate would be different.
(iii) The rate would be always positive.
(iv) the resultant change would be constant.
(v) 0 feet per min
Explanation:
Let the length of ladder is l, x be the height of the top of the ladder from the ground and y be the length of the bottom of the ladder from the wall,
By making the diagram of this situation,
Applying Pythagoras theorem,
![l^2 = x^2 + y^2-----(1)](https://tex.z-dn.net/?f=l%5E2%20%3D%20x%5E2%20%2B%20y%5E2-----%281%29)
Differentiating with respect to t ( time ),
( l = 26 feet = constant )
![\implies 2y\frac{dy}{dt} = -2x\frac{dx}{dt}](https://tex.z-dn.net/?f=%5Cimplies%202y%5Cfrac%7Bdy%7D%7Bdt%7D%20%3D%20-2x%5Cfrac%7Bdx%7D%7Bdt%7D)
![\implies \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}](https://tex.z-dn.net/?f=%5Cimplies%20%5Cfrac%7Bdy%7D%7Bdt%7D%3D-%5Cfrac%7Bx%7D%7By%7D%5Cfrac%7Bdx%7D%7Bdt%7D)
We have,
![y = 10, \frac{dx}{dt}= -3\text{ feet per min}](https://tex.z-dn.net/?f=y%20%3D%2010%2C%20%5Cfrac%7Bdx%7D%7Bdt%7D%3D%20-3%5Ctext%7B%20feet%20per%20min%7D)
![\frac{dy}{dt}=\frac{3x}{10}-----(X)](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdt%7D%3D%5Cfrac%7B3x%7D%7B10%7D-----%28X%29)
(i) From equation (1),
![26^2 = x^2 + 10^2](https://tex.z-dn.net/?f=26%5E2%20%3D%20x%5E2%20%2B%2010%5E2)
![676=x^2 + 100](https://tex.z-dn.net/?f=676%3Dx%5E2%20%2B%20100)
![576 = x^2](https://tex.z-dn.net/?f=576%20%3D%20x%5E2)
![\implies x = 24\text{ feet}](https://tex.z-dn.net/?f=%5Cimplies%20x%20%3D%2024%5Ctext%7B%20feet%7D)
From equation (X),
![\frac{dy}{dt}=\frac{3\times 24}{10}=7.2\text{ feet per min}](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdt%7D%3D%5Cfrac%7B3%5Ctimes%2024%7D%7B10%7D%3D7.2%5Ctext%7B%20feet%20per%20min%7D)
(ii) From equation (X),
![\frac{dy}{dt}\propto x](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdt%7D%5Cpropto%20x)
Thus, for different value of x the value of
would be different.
(iii) Since, distance = Positive number,
So, the value of y will always a positive number.
Thus, from equation (X),
The rate would always be a positive.
(iv) The length of the ladder is constant, so, the resultant change would be constant.
i.e. x = increases ⇒ y = decreases
y = decreases ⇒ y = increases
(v) if ladder hit the ground x = 0,
So, from equation (X),
![\frac{dy}{dt}=0\text{ feet per min}](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdt%7D%3D0%5Ctext%7B%20feet%20per%20min%7D)