Question

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?

Answer 1

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

Answer 2

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

Answer 3

@Algebraic!

Answer 4

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) ) =

Answer 5

derivative of the 'outside' is...?

Answer 6

sec^2

Answer 7

:)

Answer 8

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

Answer 9

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

Answer 10

f(t) is theta yeah, I wrote it that way because theta depends on time and I wanted you to see the chain rule

Answer 11

so LHS is sec^2(theta) * d(theta) /dt

RHS still needs to be differentiated...

Answer 12

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

Answer 13

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

Answer 14

what is g(t) in this case?

Answer 15

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

Answer 16

ah okay

Answer 17

-1/x^2*dx/dt

Answer 18

great:)

Answer 19

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

Answer 20

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

Answer 21

dx/dt = 8

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

x = 304.138

Answer 26

theta = .1629 rads

Answer 27

@Algebraic!

Answer 28

so x is longer than the hypotenuse?

Answer 29

x = 295.803

Answer 30

k:)

Answer 31

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

Answer 32

-8.87E-5?

Answer 33

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

Answer 34

yes

Answer 35

you didn't get -8.87E-5?

Answer 36

yes I did

Answer 37

and it says it's wrong...

Answer 38

yep

Answer 39

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

Answer 40

alright

Answer 41

-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...

Answer 42

May I have the answer to the question?

Answer 43

go for it.

Answer 44

@eSpeX

Answer 45

@RolyPoly id like the answer too!

Answer 46

@callisto is checking it over...

Answer 47

ty @Callisto

Answer 48

I... am no good at maths...
\[tan \theta = \frac{50}{x}\]
Differentiate both sides with respect to x. Probably you won't get:
\[\frac{ d \theta }{dt } = \frac{ -1 }{x^2 \sec^2\theta }\frac{ d x}{ dt }\]

Answer 49

My bad, I meant with respect to t

Answer 50

you're saying the differentiation is wrong?

Answer 51

\[tan \theta = \frac{50}{x}\]
Diff. both sides w.r.t. t
\[sec^2\theta \frac{d\theta}{dt} = \frac{-50}{x^2} \frac{dx}{dt}\]\[\frac{d\theta}{dt} = \frac{-50}{x^2sec^2\theta } \frac{dx}{dt}\]

Answer 52

arg

Answer 53

forgot the 50 rfl

Answer 54

you so smart cally

Answer 55

:)

Answer 56

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

Answer 57

thank you @Callisto and @Algebraic!

Answer 58

yes thanks @Callisto !

Answer 59

\[sec^2\theta=\frac{x^2+50}{x^2}\]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

Answer 60

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

Answer 61

heh

Answer 62

you guys have a good night thanks again

Answer 63

Good night!!~

Answer 64

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