Math Calculus
posted by Meg on .
The foot of a 30foot ladder is being pulled away from a vertical wall at a rate of 1 foot per minute. When the top of the ladder is 6 feet from the ground, at what rate is the top of the ladder moving down the wall?

Let the foot of the ladder be x ft from the wall, and let the top of the ladder by y ft above the ground
x^2 + y^2 = 30^2
2x dx/dt + 2y dy/dt = 0
x dx/dt + y dy/dt = 0
when y = 6
x^2 + 36 = 900
x = √864
given: dx/dt = 1 , x = √864 , y = 6
√864(1) + 6dy/dt = 0
dy/dt = √864/6 ft/min or appr. 4.9 ft/min
(the negative sign shows that y is decreasing or the ladder is moving down the wall) 
Reini  How did you ended up getting x dx/dt + y dy/dt=0?
my professor said something with
a^2 + b^2 = c^2
then differentiate it so 2a(da/dt) + 2b(db/dt) = 2c(dc/dt)
so a^2+b^2=c^2
a^2 + 6^2 = 30^2
so a=sqrt(864)
then I differentiate
let da/dt = 1
b = 6
c=30
2a(da/dt)+2b(db/dt)=2c(dc/dt)
2sqrt(864)(1)+2(6)(db/dt)=2(30)(0)
=24sqrt(6)+12db/dt = 0
but how do i sold for db/dt?
is this way correct?
thanks 
that is exactly what I did, except I used x and y instead of a and b
remember that your c is a constant, so its derivative is zero (I used the actual 30^2)
I had defined x as the distance along the ground, and we were told the base of the ladder moved out"at a rate of 1 foot per minute"
this is the same as saying dx/dt = 1
from your
24sqrt(6)+12db/dt = 0
12db/dt = 24√6
db/dt = 24√6/12 = 4.9 , the same answer as I had
so yes, you are correct 
Thank you very much for your help :)