or #1, use
r^2 = x^2 + y^2
x^2 + y^2 = 169
2x dx/dt + 2y dy/dt = 0 , I differentiated with respect to t (time)
or dy/dt = (-x/y)dx/dt
when x = 12, dx/dt = 5 ft/s
find dy/dt at that moment
12^2 + y^2 = 169
y = 5
dy/dt = (-12/5)(5) ft/s = -12 ft/sec
the negative shows it is descending.
for #2, use
tanØ = y/x
xtanØ = y
xsec^ Ø dØ/dt + tanØ dx/dt = dy/dt
so for our given case:
x = 12 , y = 5, r = 13
tanØ = 5/12 , secØ = 13/12, so sec^2 Ø = 169/144,
dx/dt = +5, dy/dt = -12 , dØ/dt = ????
sub in ...
12(169/144)dØ/dt + (5/12)(5) = -12
I got
dØ/dt = -1
Now remember that the derivatives of the above trig functions are only valid if Ø is in radians, so at that moment, the angle would be decreasing at 1 rad/sec
better check my arithmetic, but it sure came out nicely.
A 13 foot ladder is leaning against the house when it's base starts to slide away. By the time is 12 feet from the house the base is moving at the rate of 5 ft/sec.
1.How fast is the top of the ladder sliding down the wall at that moment?
2. At what rate is the angle theta between the ladder and the ground changing at that moment?
I am unsure of how to find the angle theta.
1 answer
1 answer