a ladder 10m long is leaning against a vertical wall with its other end on the ground. the top end of the ladder is sliding down the wall , when the top end is 6m from the ground it is sliding down at 2m/sec. how fast is the bottom moving away from the wall at this instant?
-3/2
x^2+y^2 = 10^2
so, when x=6, y=8
2x dx/dt + 2y dy/dt = 0
Now just plug in your values to find dx/dt
To find the rate at which the bottom of the ladder is moving away from the wall, we can use related rates. Let's denote the height of the wall as "x" and the distance between the bottom of the ladder and the wall as "y."
Given:
- The ladder is 10m long.
- The top end of the ladder is sliding down the wall at a rate of 2m/sec.
- When the top end is 6m from the ground, we need to find how fast the bottom is moving away from the wall.
We can use the Pythagorean theorem to relate the three distances involved:
x^2 + y^2 = 10^2
Differentiating the equation with respect to time (t), we get:
2x(dx/dt) + 2y(dy/dt) = 0
Since we are interested in finding dy/dt when y = 6m, we can substitute the given values:
2x(dx/dt) + 2(6)(dy/dt) = 0
Knowing that dx/dt is given as -2m/s (since the top end of the ladder is sliding down), we can solve for dy/dt:
2x(-2) + 2(6)(dy/dt) = 0
-4x + 12(dy/dt) = 0
12(dy/dt) = 4x
dy/dt = (4/12)x
dy/dt = (1/3)x
Now, we need to determine x when y = 6m. Using the Pythagorean theorem equation:
x^2 + 6^2 = 10^2
x^2 + 36 = 100
x^2 = 100 - 36
x^2 = 64
x = √64
x = 8m
Finally, we substitute x = 8m into the equation for dy/dt:
dy/dt = (1/3)(8) = 8/3 m/s
Therefore, the bottom of the ladder is moving away from the wall at a speed of 8/3 m/s when the top end is 6m from the ground.
To solve this problem, we can use the concept of related rates. We need to find the rate at which the bottom end of the ladder is moving away from the wall (dx/dt), given that the top end of the ladder is sliding down the wall at a rate of 2m/sec (dy/dt) when it is 6m from the ground.
Let's define the variables:
dy/dt = -2m/sec (negative because the ladder is sliding down)
y = distance of the top end of the ladder from the ground
x = distance of the bottom end of the ladder from the wall
We need to find dx/dt.
Using the Pythagorean theorem, we have:
x^2 + y^2 = 10^2 (since the ladder is 10m long)
Differentiating both sides of the equation with respect to time t, we get:
2x(dx/dt) + 2y(dy/dt) = 0
Since we are interested in finding dx/dt when y = 6m, we substitute the values into the equation:
2x(dx/dt) + 2(6)(-2) = 0
2x(dx/dt) - 24 = 0
Now, rearranging and solving for dx/dt:
2x(dx/dt) = 24
dx/dt = 24 / (2x)
dx/dt = 12 / x
To find dx/dt, we need to determine the value of x. We can find x by using the Pythagorean theorem with the given values of y and the ladder length:
x^2 + 6^2 = 10^2
x^2 + 36 = 100
x^2 = 64
x = 8m
Substituting x = 8 into the equation for dx/dt:
dx/dt = 12 / 8 = 1.5m/sec
Therefore, the bottom end of the ladder is moving away from the wall at a rate of 1.5m/sec at the instant when the top end is 6m from the ground and sliding down at a rate of 2m/sec.