since x^2+y^2 = 36
x dx/dt + y dy/dt = 0
since we want dy/dt = -dx/dt,
(x-y) dx/dt = 0
true when x=y, as might be expected.
Or, when x = 6/√2
All that stuff about 4m and .5 m/s is just noise, unless there are other parts to the problem.
the top of a ladder 6 m long is resting against a vertical wall on a level pavement, when the ladder begins to slide outwards. At the moment when the foot of the ladder is 4 m from the wall, it is sliding away from the wall @ 0.5m/sec. How far is the foot from the wall when it and the top are moving at the same rate>
2 answers
I guess this is a calculus problem.
When I drew this out, I got a right triangle
x = length the foot is from the wall = 4
6 = length of ladder = hypothenuse
h = unknown height of wall
dx/dt = .5 m/sec
use pythag theorem to find h:
x^2 + h^2 = 6^2
4^2+h^2 = 6^2
h =4.47
Now find the rate of h with respect to time ( how fast the ladder is sliding down) so I guess differentiate with respect to time:
d/dt [ x^2 + h^2 ] = d/dt(36)
This is where the chain rule applies:
2x*dx/dt + 2h * dh/dt = 0 (derivative of constant is 0)
so now plug in everything:
(2*6*.5) + (2*4.47 * dh/dt) = 0
6 + 8.94*dh/dt = 0
8.94*dh/dt = -6
dh/dt = -.671 m/s (notice that it's negative, because its sliding down!)
I'm not sure what to do problem here, I might be completely wrong here…
When I drew this out, I got a right triangle
x = length the foot is from the wall = 4
6 = length of ladder = hypothenuse
h = unknown height of wall
dx/dt = .5 m/sec
use pythag theorem to find h:
x^2 + h^2 = 6^2
4^2+h^2 = 6^2
h =4.47
Now find the rate of h with respect to time ( how fast the ladder is sliding down) so I guess differentiate with respect to time:
d/dt [ x^2 + h^2 ] = d/dt(36)
This is where the chain rule applies:
2x*dx/dt + 2h * dh/dt = 0 (derivative of constant is 0)
so now plug in everything:
(2*6*.5) + (2*4.47 * dh/dt) = 0
6 + 8.94*dh/dt = 0
8.94*dh/dt = -6
dh/dt = -.671 m/s (notice that it's negative, because its sliding down!)
I'm not sure what to do problem here, I might be completely wrong here…