related rate calculus problem?

Question

a boat is being pulled toward a dock by means of a rope attached to the front tip of the bow. initially there are 30 ft. of rope out and the rope is taught and being reeled in by a circular device the top of which is 10ft. higher than the point where the rope is attached to the boat. this circular device has a radius of 1ft. and turns at the rate of one revolution every pi seconds. how fast is the boat moving along the water when there are 15ft. of rope out?

Answer 1

Look closely for errors, but here you go:

x distance=sqrt(rope out^2 -y distance^2)

y-distance=10;

rope out= 30-2*t where t is time

so

x=sqrt(30-2*t)^2 -100);

dx/dt=(1/2)((30-2t)^2-100)^-1/2 * 2(30-2*t)*(-2)

the t value when rope out=15 is 7.5

so the answer is -6/25 ft/second

Answer 2

Radius of 1 ft @ rate of 1 rev/pi seconds --> 2pi*R/pi-seconds = 2R/second is the linear rate dh/dt, where h is the hypotenuse.

Then the horizontal component is:
h*cos(arcsin(10ft/h))

Take the derivative

Answer 3

[a] floor subject = circumference * top = 2? r * h top = 10cm + zero.1cm * t h = 10 + zero.1t A = 10? * (10 + zero.1t) = 100? + ? t dA/dt = fee of difference of subject with relation to time dA/dt = (100? + ? t)' = ? cm^two/sec [b] quantity = subject of base * top = ? r^two * (10 + zero.1t) V = 25? (10 + zero.1t) V = 250? + two.5? t dV/dt = two.5? cm^three/sec Hope this is helping!