A ladder is 25 feet long and leaning against a vertical wall at 3 ft/sec. Suppose we wish to determine how fast the top of the ladder is sliding down the wall when the bottom is 15 ft from the wall
Thank you!
To determine how fast the top of the ladder is sliding down the wall when the bottom is 15 ft from the wall, we can use related rates.
Let's define the following variables:
- Let x represent the distance from the bottom of the ladder to the wall.
- Let y represent the distance from the top of the ladder to the ground.
- Let z represent the length of the ladder.
According to the problem, we have the following information:
- The ladder is 25 feet long, so z = 25 ft.
- The ladder is leaning against the vertical wall, so x + y = z.
To find dy/dt, the rate at which the top of the ladder is sliding down the wall, we need to differentiate the equation x + y = z with respect to time t. We have:
d(x + y)/dt = dz/dt.
Since we are given dx/dt = 3 ft/sec, we can substitute it into the equation:
d(15 + y)/dt = d(25)/dt.
Differentiating both sides with respect to time:
0 + dy/dt = 0.
Simplifying the equation:
dy/dt = 0.
Therefore, the rate at which the top of the ladder is sliding down the wall when the bottom is 15 ft from the wall is 0 ft/sec.
To determine how fast the top of the ladder is sliding down the wall, we can use related rates and the Pythagorean theorem.
Let's identify our known and unknown variables:
Known variables:
- Length of the ladder (l) = 25 ft
- Rate at which the ladder is leaning against the wall (dw/dt) = 3 ft/sec
- Distance between the bottom of the ladder and the wall (x) = 15 ft
Unknown variable:
- Rate at which the top of the ladder is sliding down the wall (dy/dt) = ?
Using the Pythagorean theorem, we can write:
x² + y² = l²
Differentiating both sides with respect to time (t), we get:
2x(dx/dt) + 2y(dy/dt) = 0
Rearranging this equation, we have:
dy/dt = -x(dx/dt) / y
Now, let's substitute the known values into the equation:
dy/dt = -15(3) / y
Since y represents the height of the ladder above the ground, we can calculate the value of y using the Pythagorean theorem. When x = 15 (distance from the wall), we can solve for y:
y² = 25² - 15²
y² = 625 - 225
y² = 400
y = 20 ft
Now, we can substitute y = 20 ft into the equation to find dy/dt:
dy/dt = -15(3) / 20
dy/dt = -45 / 20
dy/dt = -9/4 ft/sec
Therefore, the top of the ladder is sliding down the wall at a rate of 9/4 ft/sec when the bottom is 15 ft from the wall.
Fay/Grace --
You need to show some work on what you've posted. Otherwise, the math tutors will think you just want to copy the answers down.
use right angle formula
c^2=a^2+b^2
then take the derivative..with respect to time..
2c dc/dt=2ada/dt + 2b db/dt
you know dc/dt, da/dt, solve for db/dt.