A10-meter ladder is sliding down a vertical wall so the distance between the top of the ladder and the ground is decreasing at 3 meters per minute. at a certain instant, the bottom of the ladder is 6 meters from the wall. what is the rate of change of the area formed by the ladder at that instant (in square meters per minute)?

The rate of change of the area formed by the ladder at that instant is 7 square metres per minute.

Step-by-step explanation:

Let x be the length of wall. And y be the distance of ladder from the wall.

Then,  the decrement in the distance between the top of the ladder and the ground is,

Use Pythagoras theorem with base y and height x, and length as hypotaneous as,

Differentiate the above equation with respect to time as,

The area is given as,

Differentiate to obtain the change in area as,

Here, y= 6 and x =8 by Pythagoras theorem. Substitute the values as,

Thus, the rate of change of the area formed by the ladder at that instant is 7 square metres per minute.

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Hence the rate of change in area increases by 7 square meters.

Step-by-step explanation:

Let  'y' denotes the length of the wall.

and 'x' denotes the length of the ladder from the wall.

denotes the change of the top of the ladder and the ground.

now using Pythagorean theorem with base x and height y and hypotenuse is 10 ( since the length of the ladder is 10 units).

Hence,

On differentiating with respect to time t we have:

Hence,

The area is given by:

now the change in area is given by:

here x=6 and y=8 (by using pythagorean theorem)

we get

Hence the rate of change in area increases by 7 square meters.

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