A ladder 10 m long rests on horizontal ground and leans against a
vertical wall. The foot of the ladder is pulled away from the wall at 2
m/s. How fast is the top sliding down the wall when the foot of the
ladder is 6m from the wall?
Let x represent the length of the ladder =10meters+5meters=x =x=10m+5m =15m :the length of the ladder is 10m
Well, let me put on my "funny clown" hat and try to solve this for you.
Okay, so we have a ladder leaning against a wall. It's like the ladder is trying to have a meeting with the wall, but can't quite reach it. Poor ladder.
Anyway, we want to figure out how fast the top of the ladder is sliding down the wall. It's like the ladder is saying, "I'm down, but not out!"
So, let's use a little math to solve this puzzle. We know that the foot of the ladder is being pulled away from the wall at a rate of 2 m/s. That's like the wall is teasing the ladder, going, "Come on, I bet you can't reach me!"
Now, when the foot of the ladder is 6 meters away from the wall, we want to know how fast the top is sliding down. It's like the ladder is saying, "I'm not gonna let this setback bring me down!"
To solve this, we can use a little trigonometry. We'll call the distance from the top of the ladder to the wall "y" and the distance from the foot of the ladder to the wall "x".
Now, we know that the ladder is 10 meters long, so we have a right triangle. The ladder is the hypotenuse, and x and y are the legs.
Using the Pythagorean theorem, we have x^2 + y^2 = 10^2.
Now, let's take the derivative of both sides of this equation with respect to time. That's like the ladder saying, "Hey, I'm not just sitting here, I'm moving!"
So, we have 2x(dx/dt) + 2y(dy/dt) = 0.
Now, we want to find dy/dt when x = 6. The foot of the ladder is 6 meters away from the wall. It's like the ladder is saying, "I'm getting closer, one step at a time!"
Plugging in x = 6, we get 2(6)(2) + 2y(dy/dt) = 0.
Simplifying this, we have 12 + 2y(dy/dt) = 0.
Now, we can solve for dy/dt. It's like the ladder is saying, "I need to know how fast I'm sliding down!"
Rearranging the equation, we get 2y(dy/dt) = -12.
Dividing both sides by 2y, we have dy/dt = -6/y.
So, when the foot of the ladder is 6 meters away from the wall, the top is sliding down at a rate of -6/y m/s.
Now, we need to find y. It's like the ladder is saying, "I'm not just sliding down, I'm going down in style!"
Using the Pythagorean theorem again, we have 6^2 + y^2 = 10^2.
Simplifying this, we have 36 + y^2 = 100.
Subtracting 36 from both sides, we get y^2 = 64.
Taking the square root of both sides, we have y = 8.
So, when the foot of the ladder is 6 meters from the wall, the top is sliding down at a rate of -6/8 m/s.
And there you have it! The top of the ladder is sliding down at a rate of 3/4 m/s when the foot is 6 meters away from the wall. Just remember, even when the world is pulling you away, you can still make progress. Keep climbing!
To find the rate at which the top of the ladder is sliding down the wall, we can use related rates. Let's denote the distance between the top of the ladder and the ground as y and the distance between the foot of the ladder and the wall as x.
Given:
- dx/dt = 2 m/s (Rate at which the foot of the ladder is pulled away from the wall)
- x = 6 m (Distance between the foot of the ladder and the wall)
We are asked to find dy/dt (Rate at which the top of the ladder is sliding down the wall).
By applying the Pythagorean theorem, we have the equation: x^2 + y^2 = 10^2.
Differentiating both sides of the equation with respect to time t, we get:
2x(dx/dt) + 2y(dy/dt) = 0
Substituting the given values, we can solve for dy/dt:
2(6)(2) + 2y(dy/dt) = 0
12 + 2y(dy/dt) = 0
2y(dy/dt) = -12
dy/dt = -12 / (2y)
To find the value of y, we can use the Pythagorean theorem:
6^2 + y^2 = 10^2
36 + y^2 = 100
y^2 = 100 - 36
y^2 = 64
y = √64
y = 8
Substituting the value of y into the equation for dy/dt:
dy/dt = -12 / (2*8)
dy/dt = -12 / 16
dy/dt = -0.75 m/s
Therefore, the top of the ladder is sliding down the wall at a rate of 0.75 m/s.
To find the rate at which the top of the ladder is sliding down the wall, we can use related rates. Let's denote the distance from the foot of the ladder to the wall as x, and the distance from the top of the ladder to the ground as y.
We are given that dx/dt = 2 m/s, which represents the rate at which x is changing. We need to find dy/dt, the rate at which y is changing when x = 6.
Using the Pythagorean theorem, we can relate x, y, and the length of the ladder (10 m):
x^2 + y^2 = 10^2
Differentiating both sides with respect to time t, we get:
2x (dx/dt) + 2y (dy/dt) = 0
Now, we substitute the given values:
2(6)(2) + 2y (dy/dt) = 0
12 + 2y (dy/dt) = 0
2y (dy/dt) = -12
(dy/dt) = -12 / (2y)
(dy/dt) = -6 / y
To find the value of y when x = 6, we can use the Pythagorean theorem:
6^2 + y^2 = 10^2
36 + y^2 = 100
y^2 = 64
y = 8 (ignoring the negative value since y represents a length)
Now, substituting y = 8 into the equation for dy/dt:
(dy/dt) = -6 / (8)
(dy/dt) = -3 / 4
Therefore, the top of the ladder is sliding down the wall at a rate of -3/4 m/s when the foot of the ladder is 6 m from the wall. The negative sign indicates that the top of the ladder is sliding downward.
x^2 + y^2 = 100 , where x is the base, y the height
2x dx/dt + 2y dy/dt = 0
given: dx/dt = 2 , and x = 6
then 36+y^2 = 100
y^2 = 64
y = 8
in
2x dx/dt + 2y dy/dt = 0
2(6)(2) + 2(8)dy/dt = 0
dy/dt = -24/16 = -3/2
it is sliding down at 3/2 m/sec