## ta123 Group Title At noon, ship A is 150 km west of ship B. ship A is sailing east at 35 km/h and ship B is sailing north at 25km/h. How fast is the distance between the ships changing at 4:00PM? 2 years ago 2 years ago

1. ta123 Group Title

Can someone explain how to do this problem step by step, please

2. eseidl Group Title

We need a function describing distance between the ships as a function of the time. The initial distance between the ships is 150km, which occurs at what I will call time t=0. I will call the East direction my positive x and North my positive y. Lets refer everything from the point of view ship A which we will call at rest. Then ship B is approaching A with a horizontal component and moving away from A with a vertical component. Calling d=distance between the ships, we have: $d=\sqrt{(150-35t)^2+(25t)^2}$

3. ta123 Group Title

is there any way you could draw it?

4. eseidl Group Title

The draw function on the site is tough

5. ta123 Group Title

also what would plug into the t variable, time, to get rid of the t

6. ta123 Group Title

so you can't draw the figure?

7. eseidl Group Title

Now that we have our function we need to know how fast it changes with time. t=0 is noon here and t=4 would be 4:00pm. We find the derivative of this function (with respect to time) and then plug in t=4 to find out how fast the distance between the ships is changing at 4:00pm.

8. eseidl Group Title

i'll try to draw it...it's going to look brutal though hehe

9. eseidl Group Title

|dw:1318178380389:dw|

10. ta123 Group Title

so we have to take the derivative of d=((150-35t)^2+(25t)^2)^(1/2), then plug in 4 for t, right?

11. eseidl Group Title

The rectangle on the left is ship A and the rectangle on the right is ship B. d=distance between them (which has contributions from the horizontal distance,x and the vertical distance, y) x=(150-35t) y=25t

12. eseidl Group Title

Yep

13. eseidl Group Title

let me know what you get

14. eseidl Group Title

I get a rate of change of about +21.4km/h at time t=4 (4:00pm). i.e., the distance between the ships is increasing by this amount at 4:00pm

15. ta123 Group Title

well I got 1.379, can you show your step on how you got that number

16. ta123 Group Title

maybe I didn't take the derivative right because those squares inside the square root was kinda confusing on how to take the derivative of those properly.

17. eseidl Group Title

Yeah, expand them first, then take derivative.

18. eseidl Group Title

You should get:$\frac{d/dt(150-35t)^2+(25t)^2)}{2\sqrt{(150-35t)^2+(25t)^2)}}$

19. eseidl Group Title

After you expand and take the derivative of the numerator you have you function of how the distance changes with time. Plug in t=4 to find the value at 4:00pm

20. ta123 Group Title

wait did you use natural log to expand it?

21. eseidl Group Title

No. Just expand as usual and you get a polynomial on top. Take the derivative of this polynomial.

22. ta123 Group Title

is that rule or something because I haven't seen that before except its look similar to natural log

23. ta123 Group Title

like for example ln x=1/x

24. eseidl Group Title

The numerator should be:$d/dt(22500-2(150)(35)t+1225t^2+625t^2)$

25. eseidl Group Title

okay...going back to the distance function$d=\sqrt{w}$where w=all the stuff under the root sign Take the derivative with respect to t:$=\frac{d/dt(w)}{2\sqrt{w}}$

26. eseidl Group Title

I'm just using the chain rule here

27. eseidl Group Title

and the normal rule for taking derivatives of roots

28. ta123 Group Title

if your using the chain for the numerator should you do outside of derivative first which wil be 2(150-35t). Then inside which would be 2(150-35t)(35)

29. eseidl Group Title

You dont need the chain rule for the numerator (although you can use it). expand it first then take derivative.

30. eseidl Group Title

consider the function$y=\sqrt{x^3}$To take the derivative you use substitution rule: let u=x^3 $y=\sqrt{u}$Now take derivative wrt x:$dy/dx=(dy/du)(du/dx)=(\frac{1}{2\sqrt{u}})(3x^2)$No get rid of u, using u=x^3:$dy/dx=\frac{3x^2}{2\sqrt{x^3}}$Your problem is analogous to this.

31. eseidl Group Title

gotta go. good luck. hope I earned a medal ;)

32. ta123 Group Title

lastly did you use (a-b)^(2)=a^2-2ab+b^2 to expand the numerator?

33. ta123 Group Title

34. ta123 Group Title

when you comeback could you please show the format of your final polynomial function after taking the derivative of the numerator? mine was -10500+3700t/2((150-35t)^(2)+(25t)^(2))^(1/2), then I plug in 4 and got 200.99

35. eseidl Group Title

$\frac{d/dt(22500-10500t+1225t^2+625t^2)}{2\sqrt{(150-35t)^2+(25t)^2}}$

36. eseidl Group Title

This gives:$\frac{(0-10500+3700t)}{2\sqrt{(150-35t)^2+(25t)^2}}$

37. eseidl Group Title

Now plug in t=4:$\frac{14800-10500}{2\sqrt{(10)^2+(100)^2}}$Giving,$\frac{4300}{200.99}\approx21.4$

38. ta123 Group Title

Thanks!

39. eseidl Group Title

np :)

40. ta123 Group Title

hey esidl could you help me with one more problem?

41. eseidl Group Title

sure

42. ta123 Group Title

A boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is 1 m higher than the bow of the boat. If the rope is pulled in at a rate 1 m/s, how fast is the boat approaching the dock when it is 8 m from dock?

43. eseidl Group Title

Ok. We have another triangle here. The hypotenuse is shrinking at a rate of 1 m/s. The height of the triangle is constant at 1m. We want to know how fast the length, x, of the triangle is shrinking when x=8m.

44. ta123 Group Title

ok

45. eseidl Group Title

$c^2=x^2+y^2$But y=constant=1. So,$c=\sqrt{x^2-1}$Differentiating:$dc/dt=\frac{(2x)(dx/dt)}{\sqrt{x^2-1}}$Okay, we have a lot of stuf here now. Keeps your knowns and unknowns in mind. dc/dt is known, it is the rate the hypotenuse is shrinking i.e., 1m/s. We want dx/dt. What about x? Well, we want dx/dt when x=8 so this is known. We have:$1=\frac{8(dx/dt)}{\sqrt{63}}$So,$dx/dt=\sqrt{63}/8$

46. eseidl Group Title

oooops I made a boo boo. Forgot the factor of 2 in the denominator when I differentiated. But the answer take this into account, just a typo

47. ta123 Group Title

its ok, so its suppose dx/dt=squrt(63)/16 on the denominator? Also could you draw a figure again.

48. eseidl Group Title

no the answer is still dx/dt=sqrt63/8 I just didn't type the 2 in the denominator

49. ta123 Group Title

2X8=18

50. ta123 Group Title

oops! I meant 16

51. eseidl Group Title

|dw:1318192994858:dw|

52. eseidl Group Title

rectangle is the boat, circle is the pulley, 1m above the level of the boat. distance is the horizontal line

53. eseidl Group Title

again, dc/dt should have been:$dc/dt=\frac{2x(dx/dt)}{2\sqrt{x^2-1}}$cancel the 2 on top with the one on the bottom. everything else works out. answer is still the one given.

54. ta123 Group Title

ok thanks so much

55. eseidl Group Title

no worries...hope is correct (should be!)

56. ta123 Group Title

when you take the derivative of inside of square root should be 2x on the denominator

57. ta123 Group Title

unless that 2 on denominator was suppose to represent the 1/2 and the numerator represented taking the derivative of the inside of the squrt(x^2-1) in the denominator which would be 2X

58. eseidl Group Title

I made an error above; c^2=x^2+y^2=x^2+1 I put c^2=x^2-1 for some unknown reason lol answer should be sqrt(65)/8

59. ta123 Group Title

ok, Thanks again