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anonymous
 4 years ago
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?
anonymous
 4 years ago
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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1351143411523:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0tan(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
 4 years ago
Best ResponseYou've already chosen the best response.0can you explain the diferrentiatie both sides with respect to time part?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.0derivative of the 'outside' is...?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so it's x = 50/sec^2(theta)?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yep, so sec^2 (f(t) ) * f '(t)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0f(t) is theta yeah, I wrote it that way because theta depends on time and I wanted you to see the chain rule

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so LHS is sec^2(theta) * d(theta) /dt RHS still needs to be differentiated...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0d/dt ( 1/(g(t)) ) =...?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what is g(t) in this case?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well, just because I used f(t) for theta... now we're talking about x, which is a different function of time...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so:\[\frac{ d \theta }{dt } = \frac{ 1 }{x^2 \sec^2\theta }\frac{ d x}{ dt }\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dx/dt is given.. x and theta are easy to find...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0d(theta)/dt = 8/x^2*sec(theta)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0x= sqrt(300^2 50^2) theta = arctan (50/x)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wouldn't x = sqrt(300^2 + 50^2)?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so x is longer than the hypotenuse?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so d(theta)/dt = 8/((295.803)^2*(sec^2(.16744 rads)) which is incorrect.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well you said d(theta)/dt = 1/(x^2*sec^2(theta)) * dx/dt

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you didn't get 8.87E5?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and it says it's wrong...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dunno let me look it all over. I don't see any glaring mistakes...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.08.89E5 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
 4 years ago
Best ResponseYou've already chosen the best response.0May I have the answer to the question?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@RolyPoly id like the answer too!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@callisto is checking it over...

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.1I... 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
 4 years ago
Best ResponseYou've already chosen the best response.1My bad, I meant with respect to t

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you're saying the differentiation is wrong?

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.1\[tan \theta = \frac{50}{x}\] 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
 4 years ago
Best ResponseYou've already chosen the best response.050*8/((295.803^2)*sec^2(.16744)) = correct answer

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thank you @Callisto and @Algebraic!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes thanks @Callisto !

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.1\[sec^2\theta=\frac{x^2+50}{x^2}\]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
 4 years ago
Best ResponseYou've already chosen the best response.0only took me 4hrs but i got my question worth 1pt right now to get some sleep

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you guys have a good night thanks again

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0A revised Mathematica solution, posted on 25 October 2012, 22:15 California time, is attached.
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