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|dw:1351143411523:dw|

can you explain the diferrentiatie both sides with respect to time part?

derivative of the 'outside' is...?

sec^2

:)

so it's x = 50/sec^2(theta)?

yep, so sec^2 (f(t) ) * f '(t)

so LHS is sec^2(theta) * d(theta) /dt
RHS still needs to be differentiated...

d/dt ( 1/(g(t)) ) =...?

-1/ ..........

what is g(t) in this case?

ah okay

-1/x^2*dx/dt

great:)

so:\[\frac{ d \theta }{dt } = \frac{ -1 }{x^2 \sec^2\theta }\frac{ d x}{ dt }\]

dx/dt is given..
x and theta are easy to find...

dx/dt = 8

d(theta)/dt = -8/x^2*sec(theta)

x= sqrt(300^2 -50^2)
theta = arctan (50/x)

wouldn't x = sqrt(300^2 + 50^2)?

x = 304.138

theta = .1629 rads

so x is longer than the hypotenuse?

x = 295.803

k:)

so d(theta)/dt = -8/((295.803)^2*(sec^2(.16744 rads)) which is incorrect.

-8.87E-5?

well you said d(theta)/dt = -1/(x^2*sec^2(theta)) * dx/dt

yes

you didn't get -8.87E-5?

yes I did

and it says it's wrong...

yep

dunno let me look it all over. I don't see any glaring mistakes...

alright

May I have the answer to the question?

go for it.

My bad, I meant with respect to t

you're saying the differentiation is wrong?

arg

forgot the 50 rfl

you so smart cally

:)

-50*-8/((295.803^2)*sec^2(.16744)) = correct answer

thank you @Callisto and @Algebraic!

only took me 4hrs but i got my question worth 1pt right now to get some sleep

heh

you guys have a good night thanks again

Good night!!~

A revised Mathematica solution, posted on 25 October 2012, 22:15 California time, is attached.