# Thread: Another Related Rate Problem

1. ## Another Related Rate Problem

Here is the text of my problem:

A boy is flying a kite at a height of 150ft. If the kite moves gorizontally away from the boy at 20ft/s, how fast is the string being paid out when the kite is 250ft from him?

Given:
• y=150ft
• x=250ft
• $\frac{dx}{dt} = 20ft/s$
Find:
• $\frac{dz}{dt} when x = 250$
Work:
• by the theorem of pythagoras:
• $z = 50\sqrt{34}$
• $\displaystyle\frac{dz}{dt} = \frac{x*\frac{dx}{dt} + y*\frac{dy}{dt}}{z}$

• $\displaystyle\frac{dz}{dt} = \frac{(250*20)+ (150*0)}{50\sqrt{34}}$

• $\displaystyle\frac{dz}{dt}=\frac{5000}{50\sqrt{34} }= \frac{100}{\sqrt{34}}}$
The book goves an answer of 16ft/s. Can someone give me a hint of where I might have gone astray? Thanks!

2. Originally Posted by dbakeg00
Here is the text of my problem:

A boy is flying a kite at a height of 150ft. If the kite moves gorizontally away from the boy at 20ft/s, how fast is the string being paid out when the kite is 250ft from him?

Given:
• y=150ft
• x=250ft
• $\frac{dx}{dt} = 20ft/s$
Find:
• $\frac{dz}{dt} when x = 250$
Work:
• by the theorem of pythagoras:
• $z = 50\sqrt{34}$
• $\displaystyle\frac{dz}{dt} = \frac{x*\frac{dx}{dt} + y*\frac{dy}{dt}}{z}$

• $\displaystyle\frac{dz}{dt} = \frac{(250*20)+ (150*0)}{50\sqrt{34}}$

• $\displaystyle\frac{dz}{dt}=\frac{5000}{50\sqrt{34} }= \frac{100}{\sqrt{34}}}$
The book goves an answer of 16ft/s. Can someone give me a hint of where I might have gone astray? Thanks!
The distance from the person is the hypotenuse not the adjacent side.

3. ok, I see what you mean.

That gives me this triangle:
x=200
y=150
z=250

which now gives me:
$\displaystyle\frac{dz}{dt}=\frac{(x*\frac{dx}{dt}) +(y*\frac{dy}{dt})}{z}$

$\displaystyle\frac{dz}{dt}=\frac{(200*20)+(150*0)} {250}$

$\displaystyle\frac{dz}{dt}=\frac{4000}{250}=16ft/s$

Thanks for the help, appreciate it very much. These word problems are sometimes difficult for me to visualize if I'm not careful.

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### a boy flying a kite at a height of 150 ft

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