d) -35/3 = change in area ft^2 per second

e) 2/15 = change in angle radians per second

2/15 rad/sec = 180/pi x 2/15 = 24/pi degrees= about

7.64 degrees per second

It might be useful to look at

a, b and c, especially c

25^2 - 7^2 = 625 - 49 = 576

sqr576 = height of the ladder = 24

l^2 = h^2 + b^2 where b=7, l=25, h=24

take the derivative

2ll' = 2hh' + 2bb'

ll' = hh' + bb', l' =0 the ladder is not changing its length

0 = 24h' + 7(2)

24h' = -14

h' = -14/24 = -7/12 feet per second = change in position of the top of the ladder to the ground

change 7 to 15 and 24 becomes 20 and h=-2(15)/20 =-3/2

change b' to 20 and b becomes 15, and h =-2(20)/15=-8/3

Area of the triangle= A = half base times height = bh/2

take the derivative with b=20, h=15, b'=2, h'=-8/2

A' = (1/2)(bh' +hb') = (1/2)(20(-8/3) +15(2)) =(1/2)(-160/3 + 90/3) = (1/2)(-70/3) = -70/6 = - 35/3 ft^2/sec

Let the angle between the ladder, l, and wall, h, = B

tanB =opposite side over adjacent side = b/h

b=20, h=15, h'=-8/3, b'=2,

secB = l/h = 25/15=5/3

tanB = b/h

take the derivative

(tanB)' = (b/h)'

sec^2B(B') = (hb'-bh')/h^2 = (15(2) - (20)(-8/3))/15^2

(5/3)^2 (B') = (30 +160/3)/225 =(90/3 + 160/3)/225 = (250/3)/(225) =10/27

B' = (10/27)(9/25) = 2/15

It's easier computations with

sinB=20/25 =4/5 cosB=3/5

l'=0

the ladder doesn't change length

sinB= b/l

take the derivative

(sinB)' =(b/l)'

cosB(B') = (lb'-bl')/l^2 = (25(2) -20(0))/25 = 25(2)/25^2 = 2/25

(3/5)B' = 2/25

l'=0, the ladder doesn't change length

B' = 2/25(5/3) = 2/15

change in the angle = B' = 2/15 rad/sec = about 7.64 degrees per second

check that answer

in one second b increases from 20 to 22

sinB increases from 4/5 to 22/25

B increases from 53.13 to 61.64

61.64-53.13= 8.51 which is close to 7.64, less than 1 degree apart. B' is the instantaneous change at a point, = 7.64

8.51 is the average change over an interval of 1 second