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A kite 50ft above the ground moves horizontally at a speed of 8ft/s. At what rate is the angle between the string and the horizontal decreasing when 300ft of string has been let out?

Mathematics
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|dw:1351143411523:dw|
tan(theta) = 50/x differentiate both sides with respect to time find x when sqrt(x^2+50^2) =300 (*spoiler*) find theta when x= sqrt(300^2 -50^2) plug x, theta and dx/dt into the result of the differentiation to find d(theta)/dt
can you explain the diferrentiatie both sides with respect to time part?

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Other answers:

I guess, it's just implicit differentiation really...or the chain rule... or whatever you are comfortable thinking of it as... d/dt ( tan ( f(t) ) =
derivative of the 'outside' is...?
sec^2
:)
so it's x = 50/sec^2(theta)?
yep, so sec^2 (f(t) ) * f '(t)
f(t) is theta yeah, I wrote it that way because theta depends on time and I wanted you to see the chain rule
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?
well, just because I used f(t) for theta... now we're talking about x, which is a different function of time...
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
-8.89E-5 rad.s/sec .. best I can do... let's see who else is on, might be able to see if I made a mistake...
May I have the answer to the question?
go for it.
@RolyPoly id like the answer too!
@callisto is checking it over...
I... am no good at maths... \[tan \theta = \frac{50}{x}\] Differentiate both sides with respect to x. Probably you won't get: \[\frac{ d \theta }{dt } = \frac{ -1 }{x^2 \sec^2\theta }\frac{ d x}{ dt }\]
My bad, I meant with respect to t
you're saying the differentiation is wrong?
\[tan \theta = \frac{50}{x}\] Diff. both sides w.r.t. t \[sec^2\theta \frac{d\theta}{dt} = \frac{-50}{x^2} \frac{dx}{dt}\]\[\frac{d\theta}{dt} = \frac{-50}{x^2sec^2\theta } \frac{dx}{dt}\]
arg
forgot the 50 rfl
you so smart cally
:)
-50*-8/((295.803^2)*sec^2(.16744)) = correct answer
thank you @Callisto and @Algebraic!
yes thanks @Callisto !
\[sec^2\theta=\frac{x^2+50}{x^2}\]You can cancel the x^2 since \[\frac{d\theta}{dt} = \frac{-50}{x^2sec^2\theta } \frac{dx}{dt}=\frac{-50}{x^2(\frac{x^2+50}{x^2}) } \frac{dx}{dt}\]\[=\frac{-50}{(x^2+50) } \frac{dx}{dt}\]And x^2 is easy to find
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.
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