A 15 ft ladder leans against a wall. The bottom of the ladder is 3 ft from the wall at time t=0 and slides away from the wall at a rate of 3ft/sec.
Find the velocity of the top of the ladder at time t=3.
a building is 3ft. from a 8ft. fence that surround propert a worker wants to wash a window in the building 13ft. from the ground. he will need what size ft. ladder
To find the velocity of the top of the ladder at time t=3, we can use the concept of related rates.
Let's denote:
- x(t) as the horizontal distance between the bottom of the ladder and the wall at time t
- y(t) as the vertical distance between the top of the ladder and the ground at time t
Given that the bottom of the ladder is sliding away from the wall at a rate of 3 ft/sec, we have dx/dt = 3. The ladder has a fixed length of 15 ft, so x and y are related by the Pythagorean theorem:
x^2 + y^2 = 15^2
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 dy/dt at t=3, let's substitute x=3 and dx/dt=3 into the equation above:
2(3)(3) + 2y(dy/dt) = 0
18 + 2y(dy/dt) = 0
To find dy/dt, we need to solve for y(dy/dt):
2y(dy/dt) = -18
Divide both sides by 2y:
dy/dt = -18 / (2y)
Now, let's substitute x=3 and y into the equation x^2 + y^2 = 15^2 to solve for y:
3^2 + y^2 = 15^2
9 + y^2 = 225
y^2 = 216
y = √216
Now substitute y=√216 into the equation for dy/dt:
dy/dt = -18 / (2√216)
Simplifying the expression:
dy/dt = -9 / √216
Finally, substitute t=3 into the equation we just derived:
dy/dt = -9 / √216 at t = 3
Hence, the velocity of the top of the ladder at time t=3 is -9 / √216 ft/sec.
To find the velocity of the top of the ladder at time t=3, we need to find the rate at which the top of the ladder is moving away from the wall.
Let's assume that the distance from the bottom of the ladder to the wall at time t is x. We know that the bottom of the ladder is initially 3 ft from the wall, so x = 3 ft.
We also know that the ladder slides away from the wall at a rate of 3 ft/sec. So the rate of change of x with respect to time t is dx/dt = 3 ft/sec.
Now, let's use the Pythagorean theorem to relate the distance from the top of the ladder to the wall to the distance from the bottom of the ladder to the wall. The length of the ladder is always 15 ft, so according to the Pythagorean theorem, we have:
x^2 + (15^2) = h^2
Where h is the distance from the top of the ladder to the wall.
To find dh/dt, the rate of change of h with respect to time t, we need to differentiate this equation with respect to t.
Differentiating both sides of the equation, we get:
2x(dx/dt) = 2h(dh/dt)
Substituting the values we know:
2(3)(3) = 2h(dh/dt)
18 = 2h(dh/dt)
Simplifying, we have:
9 = h(dh/dt)
We are asked to find dh/dt when t = 3. Using the equation we obtained above, we can substitute h = 15 into the equation and solve for dh/dt:
9 = 15(dh/dt)
dh/dt = 9/15 = 0.6 ft/sec
Therefore, the velocity of the top of the ladder at time t=3 is 0.6 ft/sec.
Let the foot of the ladder be x feet from the wall
let the height of the ladder be y feet
then x^2 + y^2 = 225
differentiate with respect to t, this is a "related rate problem"
2x dx/dt + 2y dy/dt = 0 (1)
when t=3 sec,
x = 9 feet (you knew the rate)
then 81+y^2=225
y=12 feet
sub x=9, y=12, dx/dt = 3 into (1) and solve for dy/dt
I got dy/dt = -9/4 ft/sec
(the negative value shows that y was decreasing at 9/4 ft/sec at that instant)