## eva12 Group Title A boat is pulled into a dock by means of a rope attached to a pulley on the dock. The rope is attached to the front of the boat, which is 7 feet below the level of the pulley. If the rope is pulled through the pulley at a rate of 10 ft/min, at what rate will the boat be approaching the dock when 100 ft of rope is out? one year ago one year ago

1. baldymcgee6 Group Title

|dw:1350689236706:dw|

2. eva12 Group Title

how do you get the answer or solve it?

3. baldymcgee6 Group Title

You must find a way to relate rate of the pully rope to the rate of the boat using the right angle triangle

4. eva12 Group Title

i do not understand this problem or how to approach it to solve it

5. eva12 Group Title

can someone help me

6. baldymcgee6 Group Title

|dw:1350690450622:dw|

7. baldymcgee6 Group Title

you can express the triangle as 7^2 + x^2 =r^2

8. eva12 Group Title

then what will x be and would r will stand for rate rigth

9. baldymcgee6 Group Title

$\frac{ dr }{ dt } = -10ft/\min$$\frac{ dx }{ dt }|_{r=100}$

10. baldymcgee6 Group Title

I'm just calling r the length of ROPE

11. eva12 Group Title

so the equation 7^2+x^2=r^2 will be dr/dt=2x+2r

12. baldymcgee6 Group Title

when we differentiate 7^2 + x^2 =r^2, we will get: 49 + 2x* dx/dt = 2r* dr/dt

13. eva12 Group Title

then i pug in 100 for r and 10 for x in the equation

14. baldymcgee6 Group Title

no, dr/dt = -10 ft/min.. we don't know what x is

15. eva12 Group Title

49 +2x=2(100)= will = x is 75.5

16. baldymcgee6 Group Title

no, we don't solve for x as x is always changing. that is essentially what the question is asking.

17. eva12 Group Title

so then how do i approach then

18. baldymcgee6 Group Title

okay... so here's what we do, we have to equations that will help us solve this problem, namely: 7^2 + x^2 =r^2 49 + 2x* dx/dt = 2r* dr/dt are you with me so far?

19. eva12 Group Title

do i solve for 1 equation then the 2 one

20. baldymcgee6 Group Title

First lets deal with this equation: 49 + 2x* dx/dt = 2r* dr/dt what are we looking for, well the question asks, "at what rate will the boat be approaching the dock?" The rate at which the boat is approaching the dock is represented by dx/dt (the rate of change of x), so we need to re arrange this equation for dx/dt

21. eva12 Group Title

dx/dt=2r*dr/dt-49/2x

22. baldymcgee6 Group Title

good.. what do we know, we know what dr/dt is, and we know what that r=100 at the point of interest. but we don't know what x is at this given instance, so what we need to do is express x in terms of r in the other other equation: 7^2 + x^2 =r^2 x = sqrt(r^2-49) now substitute that into our change of rates equation, and we know that r =100, and dr/dt= -10: dx/dt = (2(100)*(-10)-49)/2(sqrt(100^2-49) = your answer

23. eva12 Group Title

-102198.6893

24. baldymcgee6 Group Title

no, it shouldn't be that big

25. baldymcgee6 Group Title

you must have typed it in wrong in your calculator

26. eva12 Group Title

-10.27

27. eva12 Group Title

thanks

28. baldymcgee6 Group Title

you're very welcome