an 8ft 2 by 4 is leaning against a 10ft wall. THe lower end of the 2 by 4 is pulled away from the wall at the rate of 1ft/sec. How fast is the top of the 2 by 4 moving toward the ground (a) when it is 5 ft from the ground and (b) when it is 4 ft from the ground?
make a diagram showing a right angled triangle with hypotenuse of 8
let the base be x and the height be y
given : dx/dt = 1
find: dy/dt when x=5
a) when x=5 , 2+ y^2 = 64
y = √39
x^2 + y^2 = 64
2x dx/dt + 2y dy/dt = 0
dy/dt = -x dx/dt/y
= -5(1)/√39 = - .8 ft/sec
(the negative value of dy/dt shows y is getting smaller)
b) follow the same steps as above
Well, well, well, it seems we have a case of a wobbly 2 by 4 on our hands! Now, let's put our thinking clown caps on and figure this out.
(a) When the 2 by 4 is 5 ft from the ground, we can start by using some good old Pythagoras. We know that the length of the 2 by 4 is constant at 8 ft, the height of the wall is 10 ft, and we want to find the rate of change of the top of the 2 by 4. So, we plug in these numbers and solve for the rate:
(8^2) + (x^2) = (10^2), where x is the distance the top of the 2 by 4 is from the ground.
Solving that equation gives us x ≈ 6 ft. Since we know the lower end is being pulled away from the wall at the rate of 1 ft/sec, we can say that the top of the 2 by 4 is moving down at a rate of 1 ft/sec as well. From that, we can conclude that the top of the 2 by 4 is moving toward the ground at a rate of 1 ft/sec when it is 5 ft from the ground.
(b) Now, as the 2 by 4 moves closer to the ground, when it is 4 ft from the ground, we can go through the same process. Plugging in the numbers once again, we find that x ≈ 6.74 ft. Since the lower end is still being pulled away from the wall at the rate of 1 ft/sec, we can determine that the top of the 2 by 4 is moving toward the ground at a rate of 1 ft/sec when it is 4 ft from the ground as well.
So, in both cases, the top of the 2 by 4 is moving toward the ground at a rate of 1 ft/sec.
To solve this problem, we can use related rates and the concept of similar triangles.
Let's assume that the 8ft 2 by 4 forms a right triangle with the ground and the wall, where the 2 by 4 is the hypotenuse. Let's call the distance from the top of the 2 by 4 to the ground h and the distance from the top of the 2 by 4 to the wall x.
Given:
- The height of the wall is 10ft.
- The bottom of the 2 by 4 is pulled away from the wall at a rate of 1ft/sec.
We need to find:
- The rate at which the top of the 2 by 4 (h) is moving towards the ground.
To solve for this rate, we'll need to find a relation between x, h, and their rates.
From the similar triangles formed by the 2 by 4, the wall, and the ground, we have the following ratio:
x / h = 10 / 8
Differentiating both sides with respect to time (t), we get:
(dx / dt) / h = (10 / 8)(dh / dt)
Simplifying, we have:
dx / dt = (10 / 8)(dh / dt) * h
Now, let's solve the problem using the given information.
(a) When the top of the 2 by 4 is 5ft from the ground:
In this case, h = 5ft.
dx / dt = (10 / 8)(dh / dt) * h
dx / dt = (10 / 8)(-1) * 5
dx / dt = -25 / 8 ft/sec
Therefore, the top of the 2 by 4 is moving towards the ground at a rate of -25/8 ft/sec when it is 5 ft from the ground.
(b) When the top of the 2 by 4 is 4ft from the ground:
In this case, h = 4ft.
dx / dt = (10 / 8)(dh / dt) * h
dx / dt = (10 / 8)(-1) * 4
dx / dt = -5 ft/sec
Therefore, the top of the 2 by 4 is moving towards the ground at a rate of -5 ft/sec when it is 4 ft from the ground.
To solve this problem, we can use related rates, which involve finding the rate at which a quantity is changing with respect to another related quantity. In this case, we need to find the rate at which the top of the 2 by 4 is moving toward the ground.
Let's denote the height of the top end of the 2 by 4 above the ground as "y" and the distance from the bottom end of the 2 by 4 to the wall as "x".
Given:
- The rate at which the bottom end of the 2 by 4 is being pulled away from the wall is 1 ft/sec.
- The height of the wall is 10 ft.
We can form a right triangle using the 2 by 4 as the hypotenuse, with the height of the wall forming one leg and the distance from the bottom end to the wall forming the other leg.
Using the Pythagorean theorem, we have:
x^2 + y^2 = (8 ft)^2 = 64 ft^2
Taking the derivative of both sides with respect to time, we get:
2x(dx/dt) + 2y(dy/dt) = 0
Now we can solve for dx/dt, which represents the rate at which the top of the 2 by 4 is moving toward the ground.
(a) When the top of the 2 by 4 is 5 ft from the ground:
We need to find dx/dt when y = 5 ft. First, let's find x using the equation x^2 + y^2 = 64 ft^2:
x^2 + (5 ft)^2 = 64 ft^2
x^2 = 64 ft^2 - 25 ft^2
x^2 = 39 ft^2
x = sqrt(39) ft
Now let's substitute these values into the derived equation and solve for dx/dt:
2(sqrt(39))(dx/dt) + 2(5 ft)(-1 ft/sec) = 0
2(sqrt(39))(dx/dt) = 10 ft/sec
dx/dt = 10 ft/sec / (2(sqrt(39)))
dx/dt ≈ 0.80 ft/sec
Therefore, when the top of the 2 by 4 is 5 ft from the ground, it's moving toward the ground at a rate of approximately 0.80 ft/sec.
(b) When the top of the 2 by 4 is 4 ft from the ground:
Using the same process, let's find dx/dt when y = 4 ft:
x^2 + (4 ft)^2 = 64 ft^2
x^2 = 64 ft^2 - 16 ft^2
x^2 = 48 ft^2
x = sqrt(48) ft = 4(sqrt(3)) ft
Substituting these values into the derived equation, we get:
2(4(sqrt(3)))(dx/dt) + 2(4 ft)(-1 ft/sec) = 0
8(sqrt(3))(dx/dt) = 8 ft/sec
dx/dt = 8 ft/sec / (8(sqrt(3)))
dx/dt ≈ 0.33 ft/sec
Therefore, when the top of the 2 by 4 is 4 ft from the ground, it's moving toward the ground at a rate of approximately 0.33 ft/sec.