Ex 6.1,10 - Chapter 6 Class 12 Application of Derivatives

Last updated at April 16, 2024 by Teachoo

Transcript

Ex 6.1, 10 A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall ?Let AB be the ladder
& OB be the wall & OA be the ground.
Given
Length of ladder is 5 m
AB = 5 cm
Let OA = 𝑥 cm & OB = 𝑦 cm
Given that
Bottom of ladder is pulled away from the wall at the rate of 2 cm/ s
i.e. 𝒅𝒚/𝒅𝒕 = 2 cm/sec
We need to calculate
at which rate height of ladder on the wall is decreasing when foot of the ladder is 4 m away from the wall
i.e. We need to calculate 𝒅𝒙/𝒅𝒕 when 𝒚 = 4 cm.
Since Wall OB is perpendicular to the ground OA
Using Pythagoras theorem
(OB)2 + (OA)2 = (AB)2
𝑦2 + 𝑥2 = (5)2
𝒚𝟐 + 𝒙𝟐 = 25
Differentiating w.r.t time
(𝑑(𝑦2 + 𝑥2))/𝑑𝑡 = (𝑑(25))/𝑑𝑡
(𝑑 (𝑦2))/𝑑𝑡 + (𝑑(𝑥2))/𝑑𝑡 = 0
(𝑑(𝑦2))/𝑑𝑡 × 𝑑𝑦/𝑑𝑦 + (𝑑(𝑥2))/𝑑𝑡 × 𝑑𝑥/𝑑𝑥 = 0
(𝑑(𝑦2))/𝑑𝑦 × 𝑑𝑦/𝑑𝑡 + (𝑑(𝑥2))/𝑑𝑥 × 𝑑𝑥/𝑑𝑡 = 0
2𝑦 × 𝒅𝒚/𝒅𝒕 + 2𝑥 × 𝑑𝑥/𝑑𝑡 = 0
2𝑦 × 2 + 2𝑥 × 𝑑𝑥/𝑑𝑡 = 0
("From (1): " 𝑑𝑦/𝑑𝑡= 2 m/s)
2𝑥 𝑑𝑥/𝑑𝑡 = –4𝑦
𝒅𝒙/𝒅𝒕 = (−𝟐𝒚)/𝒙
Putting 𝑦 = 4 cm
├ 𝑑𝑥/𝑑𝑡┤|_(𝑦 = 4) =(−2 × 4)/𝑥
├ 𝒅𝒙/𝒅𝒕┤|_(𝒚 = 𝟒) =(−𝟖)/𝒙
To find 𝑑𝑥/𝑑𝑡 , we need to find value of x for y = 4
Finding value of x
We know that
𝑥2 + 𝑦2 = 25
Putting 𝑦 =4
𝑥2 + 42 = 25
𝑥2 = 25 – 16
𝑥2 = 9
𝑥 = 3
Putting x = 3 in (2)
𝑑𝑥/𝑑𝑡=(−8)/𝑥
𝑑𝑥/𝑑𝑡 = (−8)/3
Since x is in cm & time is in sec
∴ 𝑑𝑥/𝑑𝑡 = (−𝟖)/𝟑 cm/sec
Hence, height of ladder on the wall is decreasing at rate of 𝟖/𝟑 cm/sec

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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