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

- chestercat

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- anonymous

|dw:1351143411523:dw|

- 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

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

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## More answers

- 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

derivative of the 'outside' is...?

- anonymous

sec^2

- anonymous

:)

- anonymous

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

- anonymous

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

- 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

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

- anonymous

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

- anonymous

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

- anonymous

what is g(t) in this case?

- anonymous

well, just because I used f(t) for theta... now we're talking about x, which is a different function of time...

- anonymous

ah okay

- anonymous

-1/x^2*dx/dt

- anonymous

great:)

- anonymous

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

- anonymous

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

- anonymous

dx/dt = 8

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

x = 304.138

- anonymous

theta = .1629 rads

- anonymous

- anonymous

so x is longer than the hypotenuse?

- anonymous

x = 295.803

- anonymous

k:)

- anonymous

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

- anonymous

-8.87E-5?

- anonymous

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

- anonymous

yes

- anonymous

you didn't get -8.87E-5?

- anonymous

yes I did

- anonymous

and it says it's wrong...

- anonymous

yep

- anonymous

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

- anonymous

alright

- 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

May I have the answer to the question?

- anonymous

go for it.

- anonymous

- anonymous

@RolyPoly id like the answer too!

- anonymous

@callisto is checking it over...

- anonymous

ty @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

My bad, I meant with respect to t

- anonymous

you're saying the differentiation is wrong?

- 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

arg

- anonymous

forgot the 50 rfl

- anonymous

you so smart cally

- anonymous

:)

- anonymous

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

- anonymous

thank you @Callisto and @Algebraic!

- anonymous

yes thanks @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

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

- anonymous

heh

- anonymous

you guys have a good night thanks again

- Callisto

Good night!!~

- anonymous

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

##### 1 Attachment

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