Saved Bookmarks
| 1. |
Aladder 5 m long is leaning against a wall. The bottom of the ladder is pulledalong the ground, away from the wall, at the rate of 2 cm/s. How fast is itsheight on the wall decreasing when the foot of the ladder is 4 m away fromthe wall? |
|
Answer» We can draw a right angle triangle with the given details. Please refer to video to see the figure. From the figure, `x^2 +y^2 = 5^2` Differentiating it w.r.t. `t`, `=>2xdx/dt+2ydy/dt = 0->(1)` It is given that, `dx/dt = 2` cm/s `x = 4`m `= 400` cm `:.y = sqrt(5^2-4^2 ) = 3`m `= 300` cm So, putting these values in (1), `=>2(400)(2)+2(300)dy/dt = 0` `=>dy/dt = -1600/600 = -8/3` cm/s So, height is decreasing at the rate `8/3` cm/s. |
|