Posted by **Meg** on Friday, July 8, 2011 at 10:06am.

The foot of a 30-foot 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?

- Math Calculus -
**Reiny**, Friday, July 8, 2011 at 10:17am
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)

- Math Calculus -
**Meg**, Friday, July 8, 2011 at 10:59am
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

- Math Calculus -
**Reiny**, Friday, July 8, 2011 at 11:55am
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

- Math Calculus -
**Megan**, Friday, July 8, 2011 at 3:03pm
Thank you very much for your help :)

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