Question

*Calculus*
Car B is 30 miles directly east of Car A and begins moving west at 90 mph. At the same moment car A begins moving north at 60 mph. What will be the minimum distance between the cars and at what time t does the minimum distance occur ?

Answer 1

Grr, hate these kind of problems, haha. Okay, related rates x_X

Answer 2

The answer is B. the minimum distances.

Answer 3

There are no answer choices, this is free response, please refrain from trolling my questions, tima :)

Answer 4

@awkwardpanda I LOVE YOU <3 LMAO MADE LAUGH SO HARD!

Answer 5

*me

Answer 6

Shamil I'm sorry, my duty is to help others in here if you don't appreciate my help, then others will.

Answer 7

Im sure you know @zepdrix lol

Answer 8

thinkinggg :c

Answer 9

I dont know what to do with y, lol. I can get the set-up but with the stranded y variable there I dont know xD

Answer 10

It's 69.

Answer 11

Still trying to figure this out, but side note:
Why is car B going 90 miles per hour? that seems rather dangerous +_+

Answer 12

Best response ever.

Answer 13

I don't know man, because simon said to D:

Answer 14

\[x^{2} + y^{2} = d^{2}\]
\[2x \frac{ dx }{ dt }+ 2y\frac{dy}{dt}= 2s\frac{ds}{dt}\]
x = 30, dx/dt = -90, dy/dt = 60
\[2(30)(-90) + 2y(60) = 2s\frac{ds}{dt}\]

And thats all I got, haha.

Answer 15

I was thinking of approaching it like so
variable x is the distance car A travels in t hours, and variable y is the distance car B travels in t hours. and L is e the distance between cars A and B after t hours..

Answer 16

@Psymon

. Don't spam! Common forms of spam include: •Posting a reply that is not relevant to the topic of discussion within that question.

http://openstudy.com/code-of-conduct

Answer 17

How is that not relevant? :/

Answer 18

@Directrix I don't think that's spam..

Answer 19

My post was more relevant than much of theothercrap above -_-

Answer 20

@awkwardpanda

Don't be offensive, inappropriate, or creepy! So avoid:
Sexual innuendo

http://openstudy.com/code-of-conduct

Answer 21

-sigh- sorry.. -_- ;-;

Answer 22

and to find the distance of L
\[L = \sqrt{x^2 +(30-y)^2}\]
and then because the travel is at a constant
using this (distance traveled) = (rate of travel) (time elapsed)

Answer 23

I still dont see how I did any spam, im actually posting an idea to the problem, whether its anywhere near the correct answer or not. Otherwise you might as well report this entire question and everyone in it -_-

Answer 24

And how is that 30-y?

Answer 25

That drawing is so confusing :( why is the `vertical component` x?

Answer 26

and 60mph is car a so it would be x = 60t
and car B went 90mph is y = 90t

Answer 27

I've worked out this much so far..

Answer 28

Start out with \[
z^2=x^2+y^2
\]So \[
2zz'=2xx'+2yy'
\]

Answer 29

Okay, at least I got that far, haha.

Answer 30

Then substitutuing those values for the equation of L, right?..

Answer 31

\[L =\sqrt{(60t)^2 +(30-90y)^2}\]
\[=\sqrt{3600t^2 + (30-90y)^2}\]

Answer 32

So you want\[
z'=0=\frac{xx'+yy'}{z}=\frac{90x+60y}{\sqrt{x^2+y^2}}
\]

Answer 33

Or\[
z' = 90x+60y
\]

Answer 34

Wouldn't i just differentiate with the chain rule?..

Answer 35

can one of you guys help me when you're done helping him?
http://openstudy.com/study#/updates/5258f3d5e4b002bdb08f5d62

Answer 36

Tima got dis.

Answer 37

Wio and Pysmon, That is so sweet, and kind of you two to take your time and help this kid. i hope sham apreciates your help at least.

Answer 38

Shamil you still confused ? need more help ?

Answer 39

Except I cant help. I only have an idea xD I can get stuff like wio did, but I dont know what to do with it.

Answer 40

@iambatman

Answer 41

hollaaa

Answer 42

I'll continue then ._.
\[L' =(1/2)(3600t^2+(30-90t)^2)^-1/2 \]

Answer 43

You still tried Pysmon which is reeally great of you <3

Answer 44

But yes, I thought sham was somewhat on the right track.
\[d = \sqrt{(30-90t)^{2}+ (60t)^{2}}\]
we want ds/dt to be a minimum, so we differentiate and set the derivative = 0 to try and find critical points.

Answer 45

\[\frac{ ds }{ dt }= \frac{ 1 }{ 2 }((30-90t)^{2}+ (60t)^{2})^{-1/2}((2)(30-90t)(-90)+120t)\]

Answer 46

\[(7200t^2 + 2(30-90t)(-90))\]

Answer 47

@TakeA.I.M.

Answer 48

here is what i get..
L = [ (30-90t)^2 + (60t)^2 ] ^(1/2)

the derivative with respect to time is:
(30 (13t-3) ) / sqrt( 13t^2 - 6t + 1)

(30 (13t-3) ) / sqrt( 13t^2 - 6t + 1) = 0, solve for t
t = 0.23 hr

i'll hand check the times 0.24 and 0.22 to make sure they merge at 0.23 :)

Answer 49

\[= \frac{ 23,400t - 5400}{ 2\sqrt{3600t +(30-90t)^2} }\]

Answer 50

= 0
if a/b= 0
then
23,400t - 5400 =0
23,400t = 5400
t \[\approx 23\]

Answer 51

t = 0.23 = 13.8 mins
x = 13.85 mi, then y = 20.77 mi
so the shortest possible distance would be
\[L \approx24.96 mi\]

Answer 52

I just need someone to check my work now ._.

Answer 53

yeah its right

Answer 54

made me erase all my work ty takeaim

Answer 55

Yeah, id have never gotten it.

Answer 56

Thanks, everyone, excluding the irrelevant posters tima and panda..
:)
My brain is exhausted a bit now though aha xD
again
Thank you everyone ! :)

Answer 57

Glad to help :D @timaashorty

Answer 58

You are very welcome Shamil98 & Panda, I believe you did a brilliant job in helping him, and correcting my answer. Thank you.

Answer 59

Thank you, Tima :D now time for cake.

Answer 60

Let's go eat that cake hun (:

Answer 61

wasn't this around the time where I got you two banned? lel

Answer 62

oh my, READING THIS MAKES ME LOL!

Answer 63

Yes it is !WOW 23 days ? ;o

Answer 64

ikr o.o

Answer 65

Sham was so mean ;-; btw I can't get chat ):

Answer 66

Why can't you get chat ?

Answer 67

I'm on my iPod xD

Answer 68

awwh ;c