Applications Of Derivatives Question 423
Question: A ladder 5 m in length is resting against vertical wall. The bottom of the ladder is pulled along the ground away from the wall at the rate of $ 1.5,m/\sec $ . The length of the highest point of the ladder when the foot of the ladder $ 4.0,m $ away from the wall decreases at the rate of
[Kurukshetra CEE 1996]
Options:
A) 2 m/sec
B) 3 m/sec
C) 2.5 m/sec
D) 1.5 m/sec
Show Answer
Answer:
Correct Answer: A
Solution:
According to fig. $ x^{2}+y^{2}=25 $ …..(i)
Differentiate (i) w.r.t. t, we get
$ 2x\frac{dx}{dt}+2y\frac{dy}{dt}=0 $ …………..(ii)
Here $ x=4 $ and $ \frac{dx}{dt}=1.5 $ From (i), $ 4^{2}+y^{2}=25\Rightarrow y=3 $
$ \therefore $ From (ii), $ 2(4)(1.5)+2(3)\frac{dy}{dt}=0 $ So, $ \frac{dy}{dt}=-2m/\sec $
Hence, length of the highest point decreases at the rate of 2m/sec.
BETA
