## mathcalculus 2 years ago Help with related rate problem please! (DISTANCE)

1. mathcalculus

2. Coolsector

hint -> you want dy/dt so: dy/dt = (dy/dx) * (dx/dt)

3. mathcalculus

which formula do i use?

4. Coolsector

you have dx/dt at this point and you have the point itself. that should help you

5. mathcalculus

right.

6. mathcalculus

got that.

7. Coolsector

dy/dt = (dy/dx) * (dx/dt)

8. Coolsector

dy/dx means derivative of y with respect to x dx/dt is given at this point

9. mathcalculus

i know.

10. Coolsector

so?

11. mathcalculus

but what formula should i use, for example there are problems for volume, circle, triangle, etc.

12. mathcalculus

distance formula?

13. Coolsector

ok so distance from the origin is s = sqrt(x^2 + y^2) so you want ds/dt = (ds/dy) * (dy/dt) + (ds/dx) * (dx/dt)

14. mathcalculus

what does s symbolize?

15. mathcalculus

right.

16. Coolsector

s - distance

17. Coolsector

so that is it. now you have it all

18. mathcalculus

thanks let me try

19. Coolsector

you can, as well, to express y in terms of x in the distance formula and then find ds/dt = (ds/dx) * (dx/dt) it will be much more simple

20. mathcalculus

right the whole 5*sqrt(2x+2) is throwing me off though,

21. mathcalculus

im not sure if i should find the derivative of that function with respect to time.

22. mathcalculus

and the whole distance formula too.

23. Coolsector

ok so if we do it the way i said at the end : "you can, as well, to express y in terms of x in the distance formula and then find ds/dt = (ds/dx) * (dx/dt) it will be much more simple" then you dont have to do much

24. Coolsector

cause s=sqrt(x^2 + y^2) = sqrt(x^2+50x+50) now ds/dt = (ds/dx) * (dx/dt)

25. Coolsector

you just need to find ds/dx

26. mathcalculus

i did this: i found s=10.0499 then i found the derivative which i got 5/sqrt(2x+2) *dx/dt

27. mathcalculus

i got ds/dt as 10 but i know it's wrong.

28. Coolsector

you dont need to find s. you need to find ds/dx

29. mathcalculus

ds/dx=10

30. mathcalculus

31. Coolsector

since we express y in terms of x we dont need to worry about it anymore.

32. mathcalculus

ok

33. Coolsector

ds/dx = (x + 25)/sqrt(x^2+50x+50) at the point x=1 it is ds/dx = 26 / sqrt(101) so ds/dt = (26 / sqrt(101))* 4 i might done some mistake though

34. mathcalculus

yeah im not sure what's happening.

35. Coolsector

why? s = sqrt(x^2+y^2) = sqrt(x^2+50x+50) so we want ds/dt ds/dt = ds/dx * dx/dt ds/dx = (x + 25)/sqrt(x^2+50x+50) ds/dx at this point = 26/sqrt(101) so ds/dt at this point = 26 * 4 /sqrt(101)

36. Coolsector

plugged into the derivative x=1 and dx/dt = 4

37. Coolsector

understand what i did ?

38. mathcalculus

sqrt(x^2+50x+50) ?

39. Coolsector

this is the distance.. s=sqrt(x^2 + y^2) = sqrt(x^2 + 50x + 50)

40. mathcalculus

i know, how'd you come up with this sqrt(x^2+50x+50)

41. mathcalculus

x^2+50x+50)

42. Coolsector

plugged y=5sqrt(2x+2)

43. mathcalculus

10=10

44. mathcalculus

50x+50?

45. mathcalculus

i plugged in y and x

46. Coolsector

what ?

47. Coolsector

y=5sqrt(2x+2) y^2 = 50x+50

48. mathcalculus

plugged y=5sqrt(2x+2). i plugged in y and x. 10=10

49. Coolsector

so now s = sqrt(x^2 + y^2) but since y^2 = 50x+50 s = sqrt(x^2+50x+50)

50. Coolsector

not the value of y at the point. plug y as a function of x

51. mathcalculus

okay thanks alot.

52. mathcalculus

i got it

53. Coolsector

are you sure ?

54. Coolsector

s = sqrt(x^2+y^2) = sqrt(x^2+50x+50) so we want ds/dt ds/dt = ds/dx * dx/dt when we calculate ds/dx we calculate is using s as a function of x. not plugging numerical values yet. ds/dx = (x + 25)/sqrt(x^2+50x+50) now plug x=1 ds/dx at this point = 26/sqrt(101) so ds/dt at this point = 26 * 4 /sqrt(101)