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?
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 :)
To find the rate at which the top of the ladder is moving down the wall, we can use the concept of related rates.
Let's denote the distance from the top of the ladder to the ground as "y" and the distance from the foot of the ladder to the wall as "x." We are given that x is changing at a rate of 1 foot per minute.
Since the ladder forms a right triangle with the wall and the ground, we can use the Pythagorean theorem to relate x and y. The Pythagorean theorem states that in a right triangle, the square of the hypotenuse (the ladder) is equal to the sum of the squares of the other two sides.
Therefore, we have the equation: x^2 + y^2 = 30^2.
To determine the rate at which the top of the ladder is moving down the wall (dy/dt), we need to differentiate both sides of the equation with respect to time t.
Differentiating x^2 + y^2 = 30^2 with respect to t, we get:
2x(dx/dt) + 2y(dy/dt) = 0.
Since dx/dt represents the rate at which x is changing (1 foot per minute), we can substitute dx/dt as 1.
2x(1) + 2y(dy/dt) = 0.
Simplifying the equation, we have:
2x + 2y(dy/dt) = 0.
Since we are interested in finding the rate at which the top of the ladder is moving down the wall (dy/dt) when x = 6 feet, we substitute x = 6 into the equation:
2(6) + 2y(dy/dt) = 0.
12 + 2y(dy/dt) = 0.
Solving for dy/dt, we get:
2y(dy/dt) = -12.
dy/dt = -12 / (2y).
We now need to find the value of y when x = 6 feet. To do this, we can use the relationship between x, y, and the ladder length: x^2 + y^2 = 30^2.
Substituting x = 6 into the equation:
6^2 + y^2 = 30^2.
36 + y^2 = 900.
y^2 = 864.
y = √864 = 29.39 (approximately).
Now we can substitute this value into the equation for dy/dt:
dy/dt = -12 / (2 * 29.39).
The rate at which the top of the ladder is moving down the wall is approximately -0.205 feet per minute.
Therefore, the top of the ladder is moving down the wall at a rate of approximately 0.205 feet per minute.