A 20m ladder is leaning against a wall. Top of ladder is sliding down at 3m per second and the bottom is sliding away from the wall at 4m per second. How high is the top of the ladder and far is the bottom of the ladder from the wall.
Thanks, Scott! I got the height as 16 and distance as 12m.
To solve this problem, we can use the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse (the longest side) is equal to the sum of the squares of the lengths of the other two sides.
Let's denote the height of the ladder as "h" and the distance of the bottom of the ladder from the wall as "d". Since we want to find the height and the distance, we can say that:
h^2 + d^2 = 20^2
We also know that the top of the ladder is sliding down at a rate of 3m/s and the bottom is sliding away from the wall at a rate of 4m/s. This means that the rates of change of h and d can be expressed as:
dh/dt = -3m/s
dd/dt = 4m/s
To find how fast the height is changing with respect to time (dh/dt), we can differentiate the equation h^2 + d^2 = 400 with respect to time (t):
2h * dh/dt + 2d * dd/dt = 0
Since we are looking for the rate of change of the height when h = 12 (half the length of the ladder), we know that d = sqrt(20^2 - h^2) = sqrt(20^2 - 12^2) = 16.
Plugging in the values:
2(12) * dh/dt + 2(16) * 4 = 0
24 * dh/dt + 128 = 0
Simplifying the equation:
dh/dt = -128 / 24 = -5.33m/s
Therefore, the top of the ladder is sliding down at a rate of approximately 5.33m/s.
To find the distance of the bottom of the ladder from the wall when h = 12, we can substitute into the original Pythagorean theorem equation:
h^2 + d^2 = 20^2
12^2 + d^2 = 400
144 + d^2 = 400
d^2 = 400 - 144
d^2 = 256
d = √256
d = 16
Therefore, the bottom of the ladder is 16 meters away from the wall.
To find the height of the top of the ladder and the distance of the bottom from the wall, we can use the concept of similar triangles.
Let's assume the height of the top of the ladder from the ground is h and the distance of the bottom of the ladder from the wall is d.
We are given that the top of the ladder is sliding down at a rate of 3m/s and the bottom is sliding away from the wall at a rate of 4m/s.
Using similar triangles, we can set up the following proportions:
h/20 = (h - 3t)/t --> Equation 1
d/20 = (d + 4t)/t --> Equation 2
Where t represents time in seconds.
To solve for h, we can rearrange Equation 1:
20(h - 3t) = ht
20h - 60t = ht
20h = ht + 60t
20h = t(h + 60)
h = (t(h + 60))/20
h = t(h/20 + 3)
Now, let's solve for d using Equation 2:
20(d + 4t) = dt
20d + 80t = dt
20d = dt - 80t
20d = t(d - 80)
d = (t(d - 80))/20
d = t(d/20 - 4)
From the obtained expressions for h and d, we can see that both depend on time (t). Therefore, we need to know the specific time at which we want to calculate the height and distance.
h = height up wall
d = distance from wall
by Pythagoras... h² + d² = 20²
differentiating
... 2 h dh/dt + 2 d dd/dt = 0
substituting
... (2 * h * -3) + (2 * d * 4) = 0
... 8 d = 6 h ... d = ¾ h
h² + (¾ h)² = 400
solve for h
substitute back to find d