## Mimi_x3 3 years ago Two straight roads intersect at right angles. Two men, A and B, are 100 km from the intersection, one on each road. They drive towards the intersection at 30km^-1 and 40 khmh^-1 respectively. Find the distance of each driver from the intersection as a function of t, the time in hours for which they are driving. Hence find their distance apart, d(t), at any time. For what value of t is the distance apart least?

1. Ishaan94

ah you posted it again

2. Mimi_x3

Yup, i was staring it for hours i didn't even know how to do it -_-

3. Mimi_x3

I tried working backwards but it didn't work. Or using the pythagorus theorem *sigh*

4. saifoo.khan

that looks soo scary.

5. Mimi_x3

LOL, I know its killing me =/

6. Ishaan94

|dw:1317388650593:dw|

7. Mimi_x3

I know, I already have the diagram, but how can I apply the pythagorus theorem ? I tried it nd I got huge numbers

8. Ishaan94

|dw:1317388745649:dw|

9. Mimi_x3

Like this : 3999.23 / dt = 2000a + 2000b / dt ?

10. IanT

|dw:1317388825437:dw| this is the graph of time against distance. Very badly drawn.

11. Ishaan94

$H^2 = (100 - 30t)^2 +(100-40t)^2$

12. Ishaan94

$$t$$ is in hours

13. Mimi_x3

Don't you find H as well ? That is 141.42 ?

14. SimonS

Let A be the location of the driver driving at 40 km/hr and B be the location of the driver driving at 30 km/hr A and B are measured by distance from the intersection. Then the distance you want is $d = \sqrt{A^2 + B^2}$ Now initially of course, i.e., when t = 0, A = B = 100 km and hence d= 100.sqrt(2) which is approximately 141 km. Now, what about for times t > 0. Well, A drives towards the intersection so if t is in hours: A(t) = 100 - 40t Making sense so far? If so. What is B(t) ?

15. IanT

from the graph, the slope is -0.333... The equation of the line is t = (-0.0333...)d + 3.333...

16. JamesJ

SimonS is trying to explain what Ishaan is saying and exactly right. So Mimi, what is B(t)?

17. JamesJ

and for the record what Ian is saying is unfortunately wrong.

18. Mimi_x3

B(t) is the distance B from the intersection ?

19. SimonS

Yes

20. JamesJ

The distance from the intersection of the driver driving at 30 km/hr

21. IanT

For my sake, can you point out the error. Reading the other posts, I realize I'm not helping with the problem, but can you show me where I went wrong?

22. JamesJ

@mimi, do you see where the formula for A(t) came from?

23. Mimi_x3

Yup, do i expand (100-30t)^2 ?

24. JamesJ

No.

25. JamesJ

Or not yet. So A(t) = 100 - 40t and hence B(t) = ... ?

26. Mimi_x3

B(t) = 100-30t

27. JamesJ

Right. Good.

28. Mimi_x3

Then the pythagoras theorem xD

29. JamesJ

so d(t)^2 = A(t)^2 + B(t)^2 = .... Now expand.

30. JamesJ

@Ian: you'll see that the equation we ultimately arrive at for d is not linear in t. Where you went wrong was writing down an equation without having really thought through the problem.

31. Mimi_x3

H^2 = 2500t^2 - 14000t + 20000 What do i do next ? differentiate it ?

32. phi

you want H(t)= sqrt(stuff) then differentiate, set to zero and solve for t

33. phi

And, if you are comfortable with the chain rule, it wasn't necessary to expand all that stuff. But you already did that work...

34. JamesJ

Yes. So now you follow the usual procedure to find the minimum of a function. By the way, H is at a minimum exactly when H^2 is at a minimum. So you can take the square root, but it's going to make the differentiation a little more difficult for no more insight. If I was doing this problem I would analyze H^2(t), the equation you have written down.

35. IanT

My graph is actually for t and the distance of A from the intersection. Not the final function. Am I correct, unhelpful as it may be?

36. JamesJ

ah, in which case, the graph was right.

37. Mimi_x3

Can I differentiate it keeping h^2 ?

38. phi

@JamesJ True, but the question asks for h(t)

39. JamesJ

Let f(t) = h^2(t). Differentiate this function.

40. Mimi_x3

f(t) ?

41. Ishaan94

won't make any difference differentiate with h^2 so it is easy

42. JamesJ

I'm just saying it's easier to differentiate the function f(t) = h^2(t). For the ultimate answer you need h(T) for whatever T gives the minimum. But it's easier to deal with the differentiation if you keep it in the h^2 form.

43. JamesJ

But if this causes anyone confusion, differentiate the function h(t)

44. Mimi_x3

h^2 = 5000t-1400 h = squareroot ( 5000t-1400) Is that right >

45. JamesJ

No. As you write above: H^2 = 2500t^2 - 14000t + 20000

46. Mimi_x3

Don't I differentiate it >

47. SimonS

Yes, but you made a few mistakes and your notation is wrong. (h^2) ' = 5000t - 14000

48. Mimi_x3

Okay, the its h' = squareroot (5000t-1400)

49. Mimi_x3

then*

50. JamesJ

No, no, definitely not. (h^2) ' does not equal (h')^2

51. phi

you are dropping a zero on 14000

52. JamesJ

Just set (h^2) ' = 0 and solve for t

53. JamesJ

i.e., (h^2) ' = 5000t - 14000 = 0 means t = .... Now for that t, h(t) = ....

54. Mimi_x3

Oh I got it 2.8 thank you (: