## anonymous one year ago I can't think of how to label my illustration for this question... I need someone to direct me in the right way. A baseball diamond is a square with sides of length 90 ft. A batter hits the ball and runs toward first base with a speed of 26 ft/s. At what rate is his distance from second base changing when he is halfway to first base?

1. anonymous

|dw:1434166696015:dw|

2. anonymous

I've misplaced #1 and 3 but it shouldn't make a difference

3. anonymous

@ganeshie8 thank god you're here! Can you help me on this?

4. ganeshie8

|dw:1434167956990:dw|

5. anonymous

Mhmm and how do I label the info given in the question

6. ganeshie8

|dw:1434168228668:dw|

7. ganeshie8

apply pythagorean theorem, differentiate and do the usual stuff

8. wolf1728

Just to make things easier, most people visualize a baseball diamond with home plate at the bottom, 1st base on the left, etc.

9. anonymous

So is this what I have to do... $90^2+(90-x)^2=y^2$ $2(90-x)=2yy'$ $y'=\frac{ 2(90-x) }{ 2y }$

10. Michele_Laino

hint: using the drawing of @ganeshie8 we can write: $y = \sqrt {{{90}^2} + {x^2}}$ furthermore, please keep in mind that: $x = x\left( t \right) = vt = 26t$

11. anonymous

Wouldn't it be $y=\sqrt{90^2+(90-x)^2}$

12. Michele_Laino

the right triangle is: |dw:1434170152802:dw|

13. anonymous

I'm a bit confused

14. Michele_Laino

when I write: $y\left( t \right) = \sqrt {{{90}^2} + x{{\left( t \right)}^2}}$ am I right?

15. Michele_Laino

@Tracy96

16. anonymous

yes

17. Michele_Laino

|dw:1434170615708:dw|

18. Michele_Laino

ok, now try to differentiate y(t) with respect to the time t, what do you get?

19. Michele_Laino

please keep in mind that you have to use the chain rule, since: $x = x\left( t \right) = vt = 26t$

20. anonymous

Just a quick question, why is it x(t) on one of the legs?

21. Michele_Laino

it is the distance between the current position of the player, from the first base

22. Michele_Laino

|dw:1434170805543:dw|

23. anonymous

Ok got it

24. Michele_Laino

ok! Now what is: $\frac{{dy}}{{dt}} = y' = ...$?

25. anonymous

Is it $y'=\frac{ 1 }{ 2 }\sqrt{(x(t))^2 + 90^2} \times2x(t)x'(t)$ ??

26. Michele_Laino

use this rule: $y = \sqrt x , \Rightarrow y' = \frac{1}{{2\sqrt x }}$

27. anonymous

$\frac{ 2x(t)x'(t) }{ 2\sqrt{x(t)^2+90^2} }$ like that?

28. Michele_Laino

that's right!

29. anonymous

Relief! haha ok now what?

30. Michele_Laino

now, as I wrote before, we have: $x = x\left( t \right) = vt = 26t$

31. Michele_Laino

so, what is: $x' = \frac{{dx}}{{dt}} = ...?$

32. anonymous

26 ?

33. Michele_Laino

perfect!

34. Michele_Laino

next, we have to substitute the expression of x and its derivative, into the above firtst derivative of y(t)

35. anonymous

Wait so why is x(t) = vt ? Is that coming from ---> distance = velocity * time

36. Michele_Laino

yes! I think that in your exercise, it is supposed that the motion of the player is an uniform motion

37. anonymous

Ohh gotcha! So we first find y' and then x' and then substitute x' into the equation we got from y'... am i following?

38. Michele_Laino

yes!

39. anonymous

Alrighty! So then $y'=\frac{ 52x(t) }{ 2\sqrt{x(t)^2 +90^2} }$ right?

40. Michele_Laino

ok! Now simplify: 52/2=26 and substitute x= vt

41. anonymous

Ohh so $y'=\frac{ 26(26t) }{ \sqrt{(26t)^2+90^2} }$

42. Michele_Laino

that's right!

43. anonymous

Great!

44. Michele_Laino

next we have to compute the time at which the player is at midpoint between the first base and the home plate

45. Michele_Laino

in other words we have to solve this equation: $45 = 26t$ what is t?

46. anonymous

t = 1.73

47. Michele_Laino

ok!

48. anonymous

or should i leave it as a fraction 45/26 ?

49. Michele_Laino

finally, replace t with 1.73 into your last formula for y' (t), and you will get your answer

50. Michele_Laino

better is the fraction form

51. anonymous

Right cuz then I can cancel things out and at the end i get 25/90

52. Michele_Laino

53. anonymous

Oh wait i made a mistake

54. anonymous

$\frac{ 25(45) }{ \sqrt{45^2 +90^2} }$

55. Michele_Laino

ok!

56. anonymous

Is that correct?

57. Michele_Laino

yes!

58. anonymous

Ahhh awesome!! I really bugged you didn't I?

59. Michele_Laino

It is all right! :)

60. anonymous

61. Michele_Laino

62. anonymous

Oh my goodness I love you so much! :D It's also asking to find at what rate the player's distance from third base is changing at the same moment... would I have to start all over?

63. Michele_Laino

yes! I think so!

64. anonymous

Oh no! :S

65. Michele_Laino

it is necessary, since we have to rewrite the function y(t), after that the procedure is the same as before

66. anonymous

|dw:1434173455402:dw| is this drawing right?

67. Michele_Laino

if we pick as origin of our x-axis the first base, then we have: |dw:1434173776450:dw| so: $y\left( t \right) = \sqrt {{{90}^2} + {{\left\{ {90 - x\left( t \right)} \right\}}^2}}$

68. anonymous

Ohhh so now I get what's happening!

69. anonymous

That's gonna be the only difference correct?

70. Michele_Laino

correct!

71. anonymous

Otherwise I just do the same thing as before

72. Michele_Laino

yes! the procedure is the same as before, only the function y(t) is different

73. anonymous

I honestly don't know how to thank you!! You're a lifesaver! Many thanks!

74. Michele_Laino

:)