This is the standard question in just about every textbook when introducing rates of change.
Did you make your diagram?
I see a right angled triangle.
At a time of t seconds, let the foot of the ladder be x ft from the wall, and let the top of the ladder be y ft above ground.
Clearly,
x^2 + y^2 = 18^2
differentiate with respect to t
2x dx/dt + 2y dy/dt = 0
at t = 2, x = 7,
49 + y^2 = 18^2
y = √275 or 5√11
sub in all we know,
2(7)(2) + 2(5√11) (dy/dt) = 0
dy/dt = -28/(10√11) ft/s
= appr .844 ft/sec
check my arithmetic
A 18 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 2ft/sec.
Find the velocity of the top of the ladder at time t=2
2 answers
Dude.... You're so smart.... Wow.
3 answers