A 18 ft ladder leans against a wall. The bottom of the ladder is 4 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=1.
The velocity of ladder at time t=1 is __________ ft/sec.
look at the triangle.
c^2=b^2+h^2 b= base, h=height
take derivitave
d c^2/dt=0=2bdb/dt + 2h dh/dt
b=4, db/dt=3ft/sec h=sqrt(18^2-4^2)
find dh/dt
To find the velocity of the top of the ladder at time t=1, we first need to find an expression for the length of the ladder as a function of time.
Let's call the length of the ladder "L" and the distance between the bottom of the ladder and the wall "x". According to the problem, the bottom of the ladder is initially 4 ft from the wall at time t=0, and it slides away from the wall at a rate of 3 ft/sec, so we can express x as a function of time:
x(t) = 4 + 3t
Using the Pythagorean theorem, we know that the square of the length of the ladder is equal to the sum of the squares of the distances between the top and bottom of the ladder and the wall:
L^2 = x^2 + 18^2
Substituting x(t) into the equation, we get:
L^2 = (4 + 3t)^2 + 18^2
L^2 = 16 + 24t + 9t^2 + 324
L^2 = 9t^2 + 24t + 340
To find the velocity of the top of the ladder at time t=1, we need to find the derivative of L with respect to time (t):
dL/dt = 18t + 24
Now, let's find the velocity at time t=1 by substituting t=1 into the derivative:
V = dL/dt (t=1) = 18(1) + 24
V = 18 + 24
V = 42 ft/sec
Therefore, the velocity of the top of the ladder at time t=1 is 42 ft/sec.