yes, label your right-angled triangle this way:
hypotenuse = h , (the rope)
vertical = 1 , (height of dock above bow)
horizontal = x , (path of the boat)
you are given dh/dt = 1 m/s
when x = 8, h^2 = 1^2 + 8^
h = √65
from x^2 + 1^2 = h^2
2x dx/dt = 2h dh/dt
then dx/dt = 2(8)(1)/(2√65)
= 8/√65 m/s or appr. 0.992 m/s
I need help with this problem. I think I figured out how to set it up but I keep getting stuck in the middle.
A boat is pulled into a dock by a rope attatched to the bow of the boat and passing through a pulley on the dock that is 1m higher than the bow of the boat. If the rope is pulled in at a rate of 1m/s, how fast is the boat approaching the dock when it is 8m from the dock?
You would set up a right triangle and use the pythagoreom therom to solve I think. I need to figure out dx/dt right? How would I do that?
3 answers
got my substitution backwards in the last two lines
should be:
dx/dt = 2√65(1)/16
= √65/8 or 1.0079 m/s
should be:
dx/dt = 2√65(1)/16
= √65/8 or 1.0079 m/s
thanx, that is what i thought i just had trouble sustituting all the numbers back in.
0 answers