anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1351143411523:dw|
anonymous
  • anonymous
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
anonymous
  • anonymous
can you explain the diferrentiatie both sides with respect to time part?

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anonymous
  • anonymous
@Algebraic!
anonymous
  • anonymous
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) ) =
anonymous
  • anonymous
derivative of the 'outside' is...?
anonymous
  • anonymous
sec^2
anonymous
  • anonymous
:)
anonymous
  • anonymous
so it's x = 50/sec^2(theta)?
anonymous
  • anonymous
yep, so sec^2 (f(t) ) * f '(t)
anonymous
  • anonymous
f(t) is theta yeah, I wrote it that way because theta depends on time and I wanted you to see the chain rule
anonymous
  • anonymous
so LHS is sec^2(theta) * d(theta) /dt RHS still needs to be differentiated...
anonymous
  • anonymous
d/dt ( 1/(g(t)) ) =...?
anonymous
  • anonymous
-1/ ..........
anonymous
  • anonymous
what is g(t) in this case?
anonymous
  • anonymous
well, just because I used f(t) for theta... now we're talking about x, which is a different function of time...
anonymous
  • anonymous
ah okay
anonymous
  • anonymous
-1/x^2*dx/dt
anonymous
  • anonymous
great:)
anonymous
  • anonymous
so:\[\frac{ d \theta }{dt } = \frac{ -1 }{x^2 \sec^2\theta }\frac{ d x}{ dt }\]
anonymous
  • anonymous
dx/dt is given.. x and theta are easy to find...
anonymous
  • anonymous
dx/dt = 8
anonymous
  • anonymous
d(theta)/dt = -8/x^2*sec(theta)
anonymous
  • anonymous
x= sqrt(300^2 -50^2) theta = arctan (50/x)
anonymous
  • anonymous
wouldn't x = sqrt(300^2 + 50^2)?
anonymous
  • anonymous
x = 304.138
anonymous
  • anonymous
theta = .1629 rads
anonymous
  • anonymous
@Algebraic!
anonymous
  • anonymous
so x is longer than the hypotenuse?
anonymous
  • anonymous
x = 295.803
anonymous
  • anonymous
k:)
anonymous
  • anonymous
so d(theta)/dt = -8/((295.803)^2*(sec^2(.16744 rads)) which is incorrect.
anonymous
  • anonymous
-8.87E-5?
anonymous
  • anonymous
well you said d(theta)/dt = -1/(x^2*sec^2(theta)) * dx/dt
anonymous
  • anonymous
yes
anonymous
  • anonymous
you didn't get -8.87E-5?
anonymous
  • anonymous
yes I did
anonymous
  • anonymous
and it says it's wrong...
anonymous
  • anonymous
yep
anonymous
  • anonymous
dunno let me look it all over. I don't see any glaring mistakes...
anonymous
  • anonymous
alright
anonymous
  • anonymous
-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...
anonymous
  • anonymous
May I have the answer to the question?
anonymous
  • anonymous
go for it.
anonymous
  • anonymous
@eSpeX
anonymous
  • anonymous
@RolyPoly id like the answer too!
anonymous
  • anonymous
@callisto is checking it over...
anonymous
  • anonymous
ty @Callisto
Callisto
  • Callisto
I... am no good at maths... \[tan \theta = \frac{50}{x}\] Differentiate both sides with respect to x. Probably you won't get: \[\frac{ d \theta }{dt } = \frac{ -1 }{x^2 \sec^2\theta }\frac{ d x}{ dt }\]
Callisto
  • Callisto
My bad, I meant with respect to t
anonymous
  • anonymous
you're saying the differentiation is wrong?
Callisto
  • Callisto
\[tan \theta = \frac{50}{x}\] Diff. both sides w.r.t. t \[sec^2\theta \frac{d\theta}{dt} = \frac{-50}{x^2} \frac{dx}{dt}\]\[\frac{d\theta}{dt} = \frac{-50}{x^2sec^2\theta } \frac{dx}{dt}\]
anonymous
  • anonymous
arg
anonymous
  • anonymous
forgot the 50 rfl
anonymous
  • anonymous
you so smart cally
anonymous
  • anonymous
:)
anonymous
  • anonymous
-50*-8/((295.803^2)*sec^2(.16744)) = correct answer
anonymous
  • anonymous
thank you @Callisto and @Algebraic!
anonymous
  • anonymous
yes thanks @Callisto !
Callisto
  • 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
anonymous
  • anonymous
only took me 4hrs but i got my question worth 1pt right now to get some sleep
anonymous
  • anonymous
heh
anonymous
  • anonymous
you guys have a good night thanks again
Callisto
  • Callisto
Good night!!~
anonymous
  • anonymous
A revised Mathematica solution, posted on 25 October 2012, 22:15 California time, is attached.
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