# A pole of length 20 m is fixed at one end to the ground. The free end is raised such that its height rises at 4 m/s. What happens to the area under the pole.

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### 1 Answer

The pole of length 20 m is fixed at one end to the ground and the other end is free. This end is raised such that its height increases at the rate 4 m/s. Let the heightÂ at any moment of time be H.

A triangle is formed by the pole, the ground, and a line perpendicular to the ground and joining the free end to the ground.

Let the distance of the base of the perpendicular drawn from the free end to the ground from the fixed end of the pole be X.

As a right angled triangle is formed. `X^2 + H^2 = 20^2`

Take the implicit derivative of both the sides with respect to time.

`2*X*(dX)/(dt) + 2*H*(dH)/(dt) = 0`

H is changing at 4 m/s.

=> `X*(dX)/(dt) + H*4 = 0`

=> `(dX)/(dt) = -4*H/X`

=> `(dX)/(dt) = -4*H/sqrt(400 - H^2)`

The area of the triangle is `A = (1/2)*X*H`

`(dA)/(dt) = 2*sqrt(400 - H^2) + (1/2)*H*(-4*H/sqrt(400 - H^2))`

=> `(dA)/(dt) = 2*sqrt(400 - H^2) - (2*H^2)/sqrt(400 - H^2)`

The rate of change of the area is dependent on the value of H.

**The area of the triangle changes at a rate equal to `2*sqrt(400 - H^2) - (2*H^2)/sqrt(400 - H^2)` **