- #1
karush
Gold Member
MHB
- 3,240
- 5
9 A Ladder 13ft long is leaning against the side of a building.
If the foot of the ladder is pulled away from the building at a constant rate of 8in per second how fast is the area of triangle formed by the ladder, the building and the ground changing (in feet squared per second) at the instant when the top of the ladder is 12 feet above the ground
https://www.physicsforums.com/attachments/1520
$$A=\frac{1}{2}BH=\frac{1}{2}xy$$
$$\frac{d}{dt}A=\frac{d}{dt} \left(\frac{1}{2} xy \right)
\Rightarrow \frac{dA}{dt}=\frac{1}{2}x\frac{dy}{dt}+\frac{1}{2}y\frac{dx}{dt}$$
since $$\frac{dy}{dt}\text{ at y }= 12\text { is }\frac{-5 ft}{12 ft}\cdot \frac {2ft}{3 sec} = -\frac {5 ft}{18 sec}$$
then substituting
$$\frac{dA}{dt} = \frac{119}{36}\frac{ft^2}{sec}$$
hopefully
If the foot of the ladder is pulled away from the building at a constant rate of 8in per second how fast is the area of triangle formed by the ladder, the building and the ground changing (in feet squared per second) at the instant when the top of the ladder is 12 feet above the ground
https://www.physicsforums.com/attachments/1520
$$A=\frac{1}{2}BH=\frac{1}{2}xy$$
$$\frac{d}{dt}A=\frac{d}{dt} \left(\frac{1}{2} xy \right)
\Rightarrow \frac{dA}{dt}=\frac{1}{2}x\frac{dy}{dt}+\frac{1}{2}y\frac{dx}{dt}$$
since $$\frac{dy}{dt}\text{ at y }= 12\text { is }\frac{-5 ft}{12 ft}\cdot \frac {2ft}{3 sec} = -\frac {5 ft}{18 sec}$$
then substituting
$$\frac{dA}{dt} = \frac{119}{36}\frac{ft^2}{sec}$$
hopefully