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