Question

Can someone help me with this rekated rate problem: A child is standing still and flying a kite. The kite remains at an altitude of 30 ft above the childs hand while traveling parrallel to the ground at a rate of 10 ft per second. When the kite is 50 feet away from the child, how fast is the kite string leaving the child's hand?

Answer 1

y = 30 ; dy/dt = 0
x = 10t ; dx/dt = 10

h = hypotenuse of a rt triangle in an ideal environment

dh/dt = (dh/dx) (dx/dt) if anyting; but lets use the pythag thrm and see if we can get an answer

Answer 2

ok

Answer 3

h^2 = x^2 + y^2

2h = 2x + 2y

dh/dt = x(dx/dt) + y(dy/dt) ; plug in the values and see what we get

Answer 4

where did you get 10t from?

Answer 5

x is moving at a rate of 10 feet per unit of time .... its right there in the problem

y is moving at a rate of 0 feet per unit of time as well

Answer 6

i gues im just not visulaizing the problem

Answer 7

do you know what it would look like if i tried to draw a picture and lable it with the info i have?

Answer 8

and to fix an error it shoud be this
\[\frac{dh}{dt} = \frac{x(\frac{dx}{dt}) + y(\frac{dy}{dt})}{h}\]

Answer 9

basically i am trying to find out, in relation to the pic, the rate of change of the hypotenuse?

Answer 10

if I interpret it correctly, yes, what is the rate of change of the hypotenuse; the speed at which the string is leaving the childs hand

at x= 50; what does h = ?

h^2 = 50^2 + 30^2

Answer 11

dh/dt = 50(10)/h

Answer 12

8.575 ft/sec or so is what I get ...

Answer 13

I get something like this: xx'+yy'=hh'. Now we know that x should be 50, also we know that dx/dt=10. y'=0, so that yy' goes to 0. The h, after using the pyth. throem, is around 58.3. So after substituting i have, 500=58.30h'. Then dividing i get h=8.574

Answer 14

very good :)

Answer 15

I have 8 for the answer, but i think there no other way to get 8 exactly, might be a type o, huh?

Answer 16

my just be a rounding or truncating; either way, its colse

Answer 17

or close too lol