NickR
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?
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|dw:1351143411523:dw|
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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
NickR
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can you explain the diferrentiatie both sides with respect to time part?
NickR
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@Algebraic!
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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) ) =
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derivative of the 'outside' is...?
NickR
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sec^2
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:)
NickR
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so it's x = 50/sec^2(theta)?
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yep, so sec^2 (f(t) ) * f '(t)
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f(t) is theta yeah, I wrote it that way because theta depends on time and I wanted you to see the chain rule
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so LHS is sec^2(theta) * d(theta) /dt
RHS still needs to be differentiated...
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d/dt ( 1/(g(t)) ) =...?
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-1/ ..........
NickR
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what is g(t) in this case?
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well, just because I used f(t) for theta... now we're talking about x, which is a different function of time...
NickR
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ah okay
NickR
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-1/x^2*dx/dt
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great:)
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so:\[\frac{ d \theta }{dt } = \frac{ -1 }{x^2 \sec^2\theta }\frac{ d x}{ dt }\]
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dx/dt is given..
x and theta are easy to find...
NickR
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dx/dt = 8
NickR
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d(theta)/dt = -8/x^2*sec(theta)
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x= sqrt(300^2 -50^2)
theta = arctan (50/x)
NickR
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wouldn't x = sqrt(300^2 + 50^2)?
NickR
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x = 304.138
NickR
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theta = .1629 rads
NickR
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@Algebraic!
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so x is longer than the hypotenuse?
NickR
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x = 295.803
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k:)
NickR
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so d(theta)/dt = -8/((295.803)^2*(sec^2(.16744 rads)) which is incorrect.
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-8.87E-5?
NickR
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well you said d(theta)/dt = -1/(x^2*sec^2(theta)) * dx/dt
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yes
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you didn't get -8.87E-5?
NickR
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yes I did
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and it says it's wrong...
NickR
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yep
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dunno let me look it all over. I don't see any glaring mistakes...
NickR
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alright
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-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...
RolyPoly
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May I have the answer to the question?
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go for it.
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@eSpeX
NickR
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@RolyPoly id like the answer too!
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@callisto is checking it over...
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ty @Callisto
Callisto
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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
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My bad, I meant with respect to t
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you're saying the differentiation is wrong?
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\[tan \theta = \frac{50}{x}\]
Diff. both sides w.r.t. t
\[sec^2\theta \frac{d\theta}{dt} = \frac{-50}{x^2} \frac{dx}{dt}\]\[\frac{d\theta}{dt} = \frac{-50}{x^2sec^2\theta } \frac{dx}{dt}\]
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arg
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forgot the 50 rfl
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you so smart cally
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:)
NickR
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-50*-8/((295.803^2)*sec^2(.16744)) = correct answer
NickR
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thank you @Callisto and @Algebraic!
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yes thanks @Callisto !
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\[sec^2\theta=\frac{x^2+50}{x^2}\]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
NickR
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only took me 4hrs but i got my question worth 1pt right now to get some sleep
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heh
NickR
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you guys have a good night thanks again
Callisto
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Good night!!~
robtobey
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A revised Mathematica solution, posted on 25 October 2012, 22:15 California time, is attached.