Mimi_x3
  • Mimi_x3
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
ah you posted it again
Mimi_x3
  • Mimi_x3
Yup, i was staring it for hours i didn't even know how to do it -_-
Mimi_x3
  • Mimi_x3
I tried working backwards but it didn't work. Or using the pythagorus theorem *sigh*

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saifoo.khan
  • saifoo.khan
that looks soo scary.
Mimi_x3
  • Mimi_x3
LOL, I know its killing me =/
anonymous
  • anonymous
|dw:1317388650593:dw|
Mimi_x3
  • 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
anonymous
  • anonymous
|dw:1317388745649:dw|
Mimi_x3
  • Mimi_x3
Like this : 3999.23 / dt = 2000a + 2000b / dt ?
anonymous
  • anonymous
|dw:1317388825437:dw| this is the graph of time against distance. Very badly drawn.
anonymous
  • anonymous
\[H^2 = (100 - 30t)^2 +(100-40t)^2\]
anonymous
  • anonymous
\(t\) is in hours
Mimi_x3
  • Mimi_x3
Don't you find H as well ? That is 141.42 ?
anonymous
  • anonymous
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) ?
anonymous
  • anonymous
from the graph, the slope is -0.333... The equation of the line is t = (-0.0333...)d + 3.333...
JamesJ
  • JamesJ
SimonS is trying to explain what Ishaan is saying and exactly right. So Mimi, what is B(t)?
JamesJ
  • JamesJ
and for the record what Ian is saying is unfortunately wrong.
Mimi_x3
  • Mimi_x3
B(t) is the distance B from the intersection ?
anonymous
  • anonymous
Yes
JamesJ
  • JamesJ
The distance from the intersection of the driver driving at 30 km/hr
anonymous
  • anonymous
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?
JamesJ
  • JamesJ
@mimi, do you see where the formula for A(t) came from?
Mimi_x3
  • Mimi_x3
Yup, do i expand (100-30t)^2 ?
JamesJ
  • JamesJ
No.
JamesJ
  • JamesJ
Or not yet. So A(t) = 100 - 40t and hence B(t) = ... ?
Mimi_x3
  • Mimi_x3
B(t) = 100-30t
JamesJ
  • JamesJ
Right. Good.
Mimi_x3
  • Mimi_x3
Then the pythagoras theorem xD
JamesJ
  • JamesJ
so d(t)^2 = A(t)^2 + B(t)^2 = .... Now expand.
JamesJ
  • 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.
Mimi_x3
  • Mimi_x3
H^2 = 2500t^2 - 14000t + 20000 What do i do next ? differentiate it ?
phi
  • phi
you want H(t)= sqrt(stuff) then differentiate, set to zero and solve for t
phi
  • 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...
JamesJ
  • 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.
anonymous
  • anonymous
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?
JamesJ
  • JamesJ
ah, in which case, the graph was right.
Mimi_x3
  • Mimi_x3
Can I differentiate it keeping h^2 ?
phi
  • phi
@JamesJ True, but the question asks for h(t)
JamesJ
  • JamesJ
Let f(t) = h^2(t). Differentiate this function.
Mimi_x3
  • Mimi_x3
f(t) ?
anonymous
  • anonymous
won't make any difference differentiate with h^2 so it is easy
JamesJ
  • 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.
JamesJ
  • JamesJ
But if this causes anyone confusion, differentiate the function h(t)
Mimi_x3
  • Mimi_x3
h^2 = 5000t-1400 h = squareroot ( 5000t-1400) Is that right >
JamesJ
  • JamesJ
No. As you write above: H^2 = 2500t^2 - 14000t + 20000
Mimi_x3
  • Mimi_x3
Don't I differentiate it >
anonymous
  • anonymous
Yes, but you made a few mistakes and your notation is wrong. (h^2) ' = 5000t - 14000
Mimi_x3
  • Mimi_x3
Okay, the its h' = squareroot (5000t-1400)
Mimi_x3
  • Mimi_x3
then*
JamesJ
  • JamesJ
No, no, definitely not. (h^2) ' does not equal (h')^2
phi
  • phi
you are dropping a zero on 14000
JamesJ
  • JamesJ
Just set (h^2) ' = 0 and solve for t
JamesJ
  • JamesJ
i.e., (h^2) ' = 5000t - 14000 = 0 means t = .... Now for that t, h(t) = ....
Mimi_x3
  • Mimi_x3
Oh I got it 2.8 thank you (:

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