Question

A 13-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate of 8 feet per second, at what rate is the area of the triangle formed by the wall, the ground, and the ladder changing, in square feet per second, at the instant the bottom of the ladder is 12 feet from the wall?

Answer 1

`(da)/(dt)=-95.2m/s^2`

Explanation:

From the question we are told that:

Slant height
`h=13ft`

Velocity
`v=(dx)/(dt)=8`

Distance
`d=12ft`

Generally the equation for area is mathematically given by

`A=0.5*x*sqrt(169 - x^2)`

Differentiating

`(da)/(dx)=(169-2 x^2)/((2 sqrt(169-x^2)))`

Multiplying through by dx/dt

`(da)/(dx)*(dx)/(dt)=(169-2 x^2)/((2 √(169-x^2)))*(dx)/(dt)`

`(da)/(dt)=(169-2x^2)/((2 √(169-x^2)))*(dx)/(dt)`

`(da)/(dt)=\frac{169-(2*12^(2))}{2*\sqrt{(169-12^(2))}}*8`

`(da)/(dt)=-95.2m/s^2`