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|dw:1440103563755:dw|

\(\frac{dT}{dx} = 0 \) is the measure of what you are trying to answer
go ahead and solve

so i dont differentiate the v ???

no!
v is a constant. it could be 10, or 6 or anything

put the v on the LHS!

the answer reads:
\[\frac{ dt }{ dx }=\frac{ x }{v \sqrt{x^2+900} }-\frac{ 1 }{ 2v}\]

which looks good

no, you are after t, first and foremost.

you are minimising t. wrt x
so you set dt/dx = 0

v is just a number, it could be 9 or 774
who cares what it is . i don't !! i'm caling it v.

sorry I meant to say
\[\Large \frac{ dt }{ dx }=\frac{ 2x- \sqrt{x^2+900} }{2v \sqrt{x^2+900} }\]

@IrishBoy123 i might have a question for u . give me a min

okay i think i know where im tripping up,,
Differentiating (200-x)(2v)

i guess what you did was treated the V variable as a constant - but didnt eliminate it ...

you mean apply the quotient rule to (200-x)/(2v) ?

omggg i was doing silly things all this time!

that's ok

it's not a waste if you learned something though, so no worries