A 26 foot ladder is lowered down a vertical wall at a rate of 3 feet per minute. The base of the ladder is sliding away from the wall. A. At what rate is the ladder sliding away from the wall when the base of the ladder is 10 feet from the wall? Would you expect the same rate when the ladder is any distance from the wall? Would you expect your answer to be positive when the base is any distance from the wall? Is there a physical reason why the rate is positive? At what rate is the ladder moving away from the wall when the ladder hits the ground?

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,

[tex]l^2 = x^2 + y^2-----(1)[/tex]

Differentiating with respect to t ( time ),

[tex]0=2x\frac{dx}{dt} + 2y\frac{dy}{dt}[/tex] ( l = 26 feet = constant )

[tex]\implies 2y\frac{dy}{dt} = -2x\frac{dx}{dt}[/tex]

[tex]\implies \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}[/tex]

We have,

[tex]y = 10, \frac{dx}{dt}= -3\text{ feet per min}[/tex]

[tex]\frac{dy}{dt}=\frac{3x}{10}-----(X)[/tex]

(i) From equation (1),

[tex]26^2 = x^2 + 10^2[/tex]

[tex]676=x^2 + 100[/tex]

[tex]576 = x^2[/tex]

[tex]\implies x = 24\text{ feet}[/tex]

From equation (X),

[tex]\frac{dy}{dt}=\frac{3\times 24}{10}=7.2\text{ feet per min}[/tex]

(ii) From equation (X),

[tex]\frac{dy}{dt}\propto x[/tex]

Thus, for different value of x the value of [tex]\frac{dy}{dt}[/tex] 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),

[tex]\frac{dy}{dt}=0\text{ feet per min}[/tex]