A ladder 10 meters long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2/ 3 meter per second.

Question

(a) What is the velocity of the top of the ladder when the base is given below? (Round your answers to three decimal places.) (m/sec)

-2 meters away from the wall

-6 meters away from the wall

-8 meters away from the wall

 

(b) Consider the triangle formed by the side of the house, ladder, and the ground. Find the rate at which the area of the triangle is changing when the base of the ladder is 6 meters from the wall.

 (m²/sec)

(c) Find the rate at which the angle between the ladder and the wall of the house is changing when the base of the ladder is 6 meters from the wall.

 (rad/sec)

Answer 1

this is a "related rates" problem, a subtopic of derivatives in textbooks or online

h^2= a^2 + b^2

h= ladder length

b= base = distance from bottom of the ladder to the wall

a = distance from top of the ladder to the bottom of the wall

take the derivative

2hh' = 2aa' + 2bb'

divide by 2

hh' = aa' +bb'

h'=0

b'=2/3

b=2

by the Pythagorean Theorem

a= sqr(10^2-2^2) = sqr96= 4sqr6

plug into the equation

10(0)=0 = a'4sqr6 + 2(2/3)

a' = -(4/3)/4sqr6 = -(1/3)/sqr6

= -1/3sqr6 = -sqr6/18

= about -0.1361 m/sec

= rate of speed of the top of the ladder down the wall

for b=6 or b= 8

replace 2 by 6 or 8 and recalculate the above

a=sqr(100-36) = sqr64 = 8

10(0) = 0 = a'8 +6(2/3)

a' =- 4/8 = -1/2 m/sec

Area = A =ba/2

take the derivative of Area with respect to time

A'(6) =(1/2)(ba'+ab') = (1/2)6(-1/2) + (1/2)8(2/3)

=-6/4+8/3

= -3/2 + 8/3

= (16-9)/6 =7/6 m/sec

= about 1.1667 m^2/sec

the angle problem is a little more difficult

the angle between ladder & ground when b=6, call it T

= tan^-1(a/b) =tan^-1(8/6)= .9273 radians

= about 53.13 degrees

or = sin^-1(8/10) = .9273 radians

sinT = a/h = ah^-1 = 0.8

(sinT)' = T'cosT = -ah'/h^2 + a'/h

T'cos.9273= .6T'= 0-1/20

T'= -.05/.6 = -08333...radians per second

which looks about right

as about 0.08 radians per second looks like a close guess

if you look at the angle change as b changes by 1 meter

angle rates are the most difficult related rates problems