A kite 100 ft above the ground moves horizontally at a speed of 8 ft/s. At what rate is the angle (in radians) between the string and the horizontal decreasing when 200 ft of string have been let out? i got -8/400 but it was wrong

- anonymous

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- freckles

|dw:1434325170769:dw|
picture looks something like this

- freckles

|dw:1434325227788:dw|

- freckles

"A kite 100 ft above the ground moves horizontally at a speed of 8 ft/s. At what rate is the angle (in radians) between the string and the horizontal decreasing when 200 ft of string have been let out? i got -8/400 but it was wrong
"
---
Moves horizontally at a speed of 8ft/s
this means x'=8ft/s
we are looking for theta (in rad) when y=200ft

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## More answers

- freckles

well what is theta when y=200ft

- freckles

what trig function can you use here

- freckles

|dw:1434325371019:dw|

- freckles

hint you have opp and hyp measurements given

- anonymous

ok i did sin of 100/200 and got .479

- freckles

arcsin(1/2)?
which is pi/6

- anonymous

why arcsin?

- freckles

to solve sin(theta)=1/2 you normally take arcsin( ) of both sides since arcsin(sin(x))=x where x is between -pi/2 and pi/2
|dw:1434325633246:dw|
anyways we are going to use x (because we are given x's rate) and theta for sure
and I will go ahead and use that constant side
\[\tan(\theta)=\frac{100}{x}\]
differentiate both side

- anonymous

sec^2x=0?

- freckles

how did you get that equation?

- freckles

sec is never 0

- anonymous

im confused...

- freckles

Like can you tell me what is confusing you please? Because I don't know how to help if I don't know where to start to help.

- anonymous

the derivative of tan=100/x thats where i got confused

- freckles

well the equation is actually
\[\tan(\theta)=\frac{100}{x} \text{ or you can write it as } \\ \tan(\theta)=100x^{-1}\]
so you don't have to use quotient rule if you prefer

- freckles

you will need to use chain rule on both sides though

- freckles

do you know how to evluate either one of these:
\[\frac{d \tan(t)}{dt}=? \\ \frac{d(t^{-1})}{dt}=?\]

- freckles

for the first it is easy just to remember derivative of tan(t) w.r.t t is sec^2(t)
and the second one I asked you about can be done by power rule
have you every used power rule before? I hope so because this question is about application of chain rule

- freckles

\[\tan(\theta)=100x^{-1} \\ \text{ differentiate both sides } \\ \sec^2(\theta) \cdot \frac{d \theta}{d t}=100 \cdot \frac{d(x^{-1})}{dt}\]
you still to differentiate the right hand side

- anonymous

would it be 100/-x^2?

- freckles

don't forget you need to write times dx/dt

- freckles

\[\sec^2(\theta) \cdot \frac{d \theta}{ d t}=-\frac{100}{x^2} \cdot \frac{dx }{dt}\]

- freckles

plug in all the values you found and are given and solve for dtheta/dt

- anonymous

would it be sec^2(pi/6)*dtheta/dt=- 100/1ooroot3* 200?

- freckles

well the x has a square on it and I thought dx/dt was given as 8ft/sec
why you put 200 for dx/dt?
dx/dt represents the rate and which the kite moves horizontally

- freckles

\[\sec^2(\frac{\pi}{6}) \cdot \frac{ d \theta}{ d t}=\frac{-100}{(100 \sqrt{3})^2} \cdot 8\]

- anonymous

i solved the right hand side but then how do i divide by the sec^2 im havinf trouble calculating that?

- freckles

1/sec(k)=cos(k)

- freckles

\[\frac{d \theta}{dt} =\frac{1}{\sec^2(\frac{\pi}{6})} \frac{-100}{(100 \sqrt{3})^2} \cdot 8 \\ \frac{d \theta}{ dt}=\cos^2(\frac{\pi}{6}) \frac{-100}{(100\sqrt{3})^2} \cdot 8\]

- anonymous

okay i got -.o2?

- freckles

ok and i have -1/50
those should be equal

- freckles

don't forget to put your units

- freckles

have fun

- anonymous

wait

- anonymous

my answer was -8/400 and i got it wrong?

- anonymous

or is it not supposed to be negative?

- freckles

you can try without
did you put the units though ?

- anonymous

the units are already tehre rad/s so i just had to type my answer in

- freckles

k

- freckles

have you got it yet?

- anonymous

i only have one more try so im scared to type it and get it wrong

- freckles

ok
don't know what to tell you we have the answer above says it is decreasing at a rate of 1/50 rad/sec

- anonymous

i know this thing is weird -_- but thank you for your help though!!

- anonymous

it was supposed to be positive got it right!!!

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