Question

A ladder 20 ft long leans against a vertical wall. If the top end slides downwards at the rate of 2ft per second, what will be the rate at which the slope of the ladder changes?

(a) -19/54

(b) -21/54

(c) -23/54

(d) -25/54

I have been asked this question in exam.

My doubt stems from Application of Derivative topic in section Application of Derivatives of Mathematics – Class 12

Answer 1

Correct answer is (d) -25/54

Easiest explanation: Let the height on the wall be x and laser touches the ground at distance y from the wall. The length of the ladder is 20ft.

By Pythagoras theorem:

x^2 + y^2 = 400

Differentiating with respect to t:

2x(dx/dt) + 2y(dy/dt) = 0

Dividing throughout by 2:

x (dx/dt) + y (dy/dt) = 0

Now, dx/dt = -2 ft /s. negative because downwards

x(-2) + y (dy/dt) = 0  ………..(1)

When lower end is 12 ft from wall, let us find x:

x^2 + 12^2 = 400

x^2 = 400 – 144= 256

x = 16

x(2) + y (dy/dt) = 0 from (1)

16(-2) + 12 (dy/dt) = 0

-32 + 12(dy/dt) = 0

dy/dt = (32/12) = (8/3)

Thus, lower end moves on a horizontal floor when it is 12 ft from the wall at the rate of 8/3 ft/s

Now, assume that the ladder makes an angle θ with the horizontal plane at time t.

If, m be the slope of the ladder at time t, then,

 m = tanθ = x/y

Thus, dm/dt = d/dt(x/y) = [y(dx/dt) – x(dx/dt)]/y^2

Therefore, the rate of change of slope of ladder is,

[dm/dt]y = 2 = [12*(-2) – 16*(8/3)]/(12)^2

Now, putting the value of x = 16, when y = 12 and dx/dt = -2, dy/dt = 8/3

We get, [dm/dt]y = 12 = [12(-2) – 16(8/3)]/(12)^2 = -25/54