#### To determine

**To find:** Find the validity of the model when top of the ladder approaches to the ground.

#### Answer

The model not valid when top of the ladder approaches to the ground.

#### Explanation

**Given:**

A ladder which is rests against a vertical wall and the on the horizontal ground.

Suddenly the bottom of the ladder slides away from the ground at a rate of 1 ft/s.

**Formula used:**

(1) Chain rule: dydx=dydu⋅dudx

(2) Pythagorean Theorem.

**Calculation:**

Let us assume that L be the length of the ladder and x be the horizontal distance from the wall to the lower end of the ladder, and y be the vertical distance from the ground to the top end of the ladder as shown in the Figure 1 given below.

Since the ladder slides away from the wall so the distance x become increases and y decreases.

Therefore x and y are function of the time t

Now from the Pythagorean Theorem.

x2+y2=L2

Differentiate with respect to the time t .

ddt[x2+y2]=ddt[L2]2xdxdt+2ydydt=0 [Qdydx=dydu⋅dudx]2xdxdt=−2ydydtxdxdt=−ydydt

On further simplification

dydt=−xydxdt

As top of the ladder approaches to the ground, that is y→0 .

Taking limit of dydt as y→0.

limy→0[dydt]=limy→0[−xy⋅dxdt]=−∞

Therefore, as y→0 the value of dydx=−∞ which does not make physical sense.

For example, the model predicts that for sufficiently small y,the top of the ladder start moving speed of greater than the speed of light, but according to the special theory of relativity ,nothing can move faster than light. Therefore the model is not appropriate for the sufficiently small value of y.

Due to the reason which is actually happen that the ladder leave the wall at some point in its decent.

Therefore, the model not valid when top of the ladder approaches to the ground.