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)
a^2 + 6^2 = 30^2
then I differentiate
let da/dt = 1
b = 6
=24sqrt(6)+12db/dt = 0
but how do i sold for db/dt?
is this way correct?
- 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
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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