Finding rate of change

Chapter 6 Class 12 Application of Derivatives (Term 1)
Concept wise

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 = π₯ cm & OB = π¦ cm Given that Bottom of ladder is pulled away from the wall at the rate of 2 cm/ s i.e. ππ/ππ = 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 ππ/ππ when π = 4 cm. Since Wall OB is perpendicular to the ground OA Using Pythagoras theorem (OB)2 + (OA)2 = (AB)2 π¦2 + π₯2 = (5)2 ππ + ππ = 25 Differentiating w.r.t time (π(π¦2 + π₯2))/ππ‘ = (π(25))/ππ‘ (π (π¦2))/ππ‘ + (π(π₯2))/ππ‘ = 0 (π(π¦2))/ππ‘ Γ ππ¦/ππ¦ + (π(π₯2))/ππ‘ Γ ππ₯/ππ₯ = 0 (π(π¦2))/ππ¦ Γ ππ¦/ππ‘ + (π(π₯2))/ππ₯ Γ ππ₯/ππ‘ = 0 2π¦ Γ ππ/ππ + 2π₯ Γ ππ₯/ππ‘ = 0 2π¦ Γ 2 + 2π₯ Γ ππ₯/ππ‘ = 0 ("From (1): " ππ¦/ππ‘= 2 m/s) 2π₯ ππ₯/ππ‘ = β4π¦ ππ/ππ = (βππ)/π Putting π¦ = 4 cm β ππ₯/ππ‘β€|_(π¦ = 4) =(β2 Γ 4)/π₯ β ππ/ππβ€|_(π = π) =(βπ)/π To find ππ₯/ππ‘ , we need to find value of x for y = 4 Finding value of x We know that π₯2 + π¦2 = 25 Putting π¦ =4 π₯2 + 42 = 25 π₯2 = 25 β 16 π₯2 = 9 π₯ = 3 Putting x = 3 in (2) ππ₯/ππ‘=(β8)/π₯ ππ₯/ππ‘ = (β8)/3 Since x is in cm & time is in sec β΄ ππ₯/ππ‘ = (βπ)/π cm/sec Hence, height of ladder on the wall is decreasing at rate of π/π cm/sec