Expert: Ahmed Salami - 11/17/2009

Question
i am trying to understand this problem:

1- A ladder 25 feet long is leaning against the wall of aa house. The base of the ladder is pulle away from the wall at a rate of 2 feet per second. How fast is the top moving down the wall when the base of the ladder is (a) 7 feet, (b) 15 feet, and (c) 24 feet from the wall?

2- Consider the right formed by the moving ladder, the side of the house, and the ground in Exercise 1. When  the base is 7 feet from the wall, find the rate at which the ares of the triangle is changing

1. Let x=the distance the base of the ladder is from the bldg, and y=the height.  Thus, x^2 + y^2 = 25^2 = 625
differentiating wrt 't': 2x x' + 2y y'=0 ==> y'=-(x/y) x'
y=sqrt(625-x^2)...y'=-(x/sqrt(625-x^2)) x'
(a) x=7 ==> x'=-(7/24) 2 ft/sec = -7/12 ft/sec
(b) x=15 ==> x'=-(15/20) 2 ft/sec = -3/2 ft/sec
(c) x=24 ==> x'=-(24/7) 2 ft/sec = -48/7 ft/sec

2. Area, A=xy/2...differentiating wrt 't' gives
A'=xy'/2 + x'y/2 when x=7, y=24, x'=2, y'=-7/12
so A'=7(-7/12)/2 + 2(24)/2 = 527/24 sq. ft./sec

i understand everything up until it says A'=xy'/2 + x'y/2

my question is: Why is the x multiplied by the derivative of y and vice versa? in the first part of the problem x is multiplied by x' so why does it change in part 2?  

Answer
Hi Lauren,
In the first part, there were no xy terms and it was straightforward to differentiate. We differentiate the term and add x'(i.e dx/dt) or y' as appropriate.
In the second part, there is an xy term which means we have to differentiate using the product rule.
By the product rule,
P = uv
P' = uv' + u'v
Its a standard rule of differentiation. We simply take one of the variables as constant and differentiate the other and do vice versa for the second term and add.
Lets say, as an example, that we wanted to find P' for
P = x²y³
P' = x²(y³)' + (x²)'y³
  = x².3y².y' + 2x.x'.y³
  = 3x²y²y' + 2xy³x'

I hope i have helped. You can always get back to me.

Regards