A 17 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=3.
at t = 0
x^2 + y^2 = 17^2
4^2 + y^2 = 17^2
y = sqrt(289 - 16) = sqrt (273)
2x dx/dt + 2 y dy/dt = 0
dy/dt = -(x/y) dx/dt
dy/dt = -[ 4/sqrt(273) ] *(3)
That is the speed at t = 0
to get it at t = 3, you need to do that at every infintesimal step. It may hit the ground before you hit 3 seconds :)
try
y = (289-x^2)^.5
dy/dx = {.5 [(289-x^2)^-.5 ](- 2x)}
dy/dt = {.5 [(289-x^2)^-.5 ](- 2x)} dx/dt
dy/dt = - [(289-x^2)^-.5 ] x dx/dt
dy/dt = -3(289-x^2)^-.5 ] x
now at t = 3, x = 4+9 = 13
so
dy/dt = -3(289-169)^-.5 ] 13
whew :) check my arithmetic
-3 *13 / 10.95 = - 3.56 ft/s
To find the velocity of the top of the ladder, we need to relate the motion of the ladder to that of its top. We can use the Pythagorean theorem to determine the relationship between the sides of the ladder.
Let's define three variables:
- Let x be the distance from the bottom of the ladder to the wall.
- Let y be the height of the ladder above the ground.
- Let t be the time in seconds.
According to the problem, the bottom of the ladder moves away from the wall at a rate of 3 ft/sec. Therefore, the rate of change of x with respect to time (dx/dt) is 3 ft/sec.
Using the Pythagorean theorem, we have x^2 + y^2 = 17^2. Differentiating this equation with respect to time, we can find a relationship between dx/dt, dy/dt, and x, y.
2x (dx/dt) + 2y (dy/dt) = 0
At t=0, x=4 and y=√(17^2 - 4^2) = 15. Therefore, we have:
2(4)(3) + 2(15) (dy/dt) = 0
Simplifying this equation, we get:
24 + 30 (dy/dt) = 0
30(dy/dt) = -24
(dy/dt) = -24/30
(dy/dt) = -4/5
So, the velocity of the top of the ladder at time t=3 seconds is -4/5 ft/sec.
In conclusion, the velocity of the top of the ladder at time t=3 is -4/5 ft/sec, indicating that the ladder is sliding down the wall.