A ladder 25 m long leans against a vertical all. If the lower end is being moved away from the wall at the rate 6m/sec, how fast is the height of the top decreasing when the lower end is 7m from the wall?
at the moment in question, the ladder top is 24m up the wall.
x^2+y^2 = 25^2
2x dx/dt + 2y dy/dt = 0
2(7)(6) + 2(24) dy/dt = 0
dy/dt = -4/7 m/s
start by drawing a diagram. you should notice that you have a triangle with the ladder as the hypoteneuse. you are given that the lower end is moving away from the wall at 6m/sec, so your dx/dt =6.
To find y when x=7:
x^2 +y^2 = 25^2
7^2 +y^2 = 625
49 + y^2 = 625
y^2 = 576
y = 24
Then, since we know that x is changing:
x^2 + y^2 = 25^2
2x(dx/dt) + 2y(dy/dt) = 0 divide by 2:
x(dx/dt) + y(dy/dt) = 0
plug in values that you know:
7(6) + 24(dy/dt) = 0
24(dy/dt) = -42
dy/dt = -1.75 m/sec
To determine how fast the height of the top of the ladder is decreasing, we need to find the derivative of the height with respect to time. Let's denote the height of the top of the ladder as "h" and the distance between the base of the ladder and the wall as "x".
We know that the ladder is 25 m long, so we have the relationship:
x^2 + h^2 = 25^2
We need to find dh/dt, the rate at which the height is changing with respect to time. To do this, we differentiate the above equation implicitly with respect to time:
2x(dx/dt) + 2h(dh/dt) = 0
Simplifying the equation, we get:
x(dx/dt) + h(dh/dt) = 0
We are given that dx/dt (the rate at which x is changing with respect to time) is 6 m/sec, and we need to find dh/dt (the rate at which h is changing with respect to time) when x = 7 m.
Substituting the given values, we have:
7(6) + h(dh/dt) = 0
Simplifying the equation further:
42 + h(dh/dt) = 0
Since we are looking for dh/dt, we can isolate it by moving the terms:
h(dh/dt) = -42
Dividing both sides by h:
dh/dt = -42/h
Now we just need to determine the value of h when x = 7 m. We can use the Pythagorean theorem to find it:
x^2 + h^2 = 25^2
7^2 + h^2 = 25^2
49 + h^2 = 625
h^2 = 625 - 49
h^2 = 576
h = 24
Now we can substitute this value into the equation for dh/dt:
dh/dt = -42/24 = -7/4
Therefore, when the lower end of the ladder is 7 m from the wall, the height of the top is decreasing at a rate of 7/4 m/sec.