A ladder 13 m long leans against a wall. The foot of the ladder is pulled along the ground away from the wall, at the rate of 1.5m/sec. How fast the angle θ between the ladder and the ground is is changing when the foot of the ladder is 12 m away from the wall.

Asked by Aaryan | 1 year ago |  66

##### Solution :-

Given as a ladder 13 m long leans against a wall.

The foot of the ladder is pulled along the ground away from the wall, at the rate of 1.5m/sec

As to find how fast is the angle θ between the ladder and the ground is changing when the foot of the ladder is 12 m away from the wall

Suppose AC be the position of the ladder initially, then AC = 13m.

DE be the position of the ladder after being pulled at the rate of 1.5m/sec, then DE = 13m as shown in the below figure.

Therefore it is given that foot of the ladder is pulled along the ground away from the wall, at the rate of 1.5 m/sec

So, $$\dfrac{dx}{dt}$$ = 1.5 m/sec ...(i)

Considering the ΔABC, it is fight angled triangle, therefore on applying pythagoras theorem

And in same for triangle secθ =$$\dfrac{13}{12}$$ ...(iii)

Differetiating equation (ii) with respect to time

Substitute the value of x,y,h and $$\dfrac{dx}{dt}$$,

Value of h is always constant as the ladder is not increasing in size, thus the equation becomes

And consider the same triangle tanθ =$$\dfrac{y}{x}$$

Differentiate the above equation with respect to time

Substitute the value of secθ, x,y,h and $$\dfrac{dx}{dt}$$, the equation becomes

Thus the angle θ between the ladder and the ground is changing at the rate of 0.3 rad/sec (because angle cannot be negative) when the foot of the ladder is 12 m away from the wall.

Answered by Sakshi | 1 year ago

### Related Questions

#### Given the sum of the perimeters of a square and a circle, show that the sum of their areas is least when one side

Given the sum of the perimeters of a square and a circle, show that the sum of their areas is least when one side of the square is equal to diameter of the circle.

#### A wire of length 20 m is to be cut into two pieces. One of the pieces will be bent into shape

A wire of length 20 m is to be cut into two pieces. One of the pieces will be bent into shape of a square and the other into shape of an equilateral triangle. Where the wire should be cut so that the sum of the areas of the square and triangle is minimum?

#### A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square

A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the lengths of the two pieces so that the combined area of the circle and the square is minimum?