Asked by Naz
A girl is flying a kite at a height of 320 ft. If the wind blows the kite horizontally at a rate of 15 feet per second away from the girl, at what rate is the girl releasing the string when the length of the string is 400 ft?
[kite] <---15 ft/sec
. .
. .
. .
. .
. .
. .
320 ft .
. .
. .
. .
._____________.[girl]
So I listed my given data:
a=320 ft
b=240 ft
c=400 ft
And I know I have to find dc/dt when c=400 ft - using the Pythagorean theorem. I differentiated it but now I realize that I don't have da/dt and db/dt...what do I do?
[kite] <---15 ft/sec
. .
. .
. .
. .
. .
. .
320 ft .
. .
. .
. .
._____________.[girl]
So I listed my given data:
a=320 ft
b=240 ft
c=400 ft
And I know I have to find dc/dt when c=400 ft - using the Pythagorean theorem. I differentiated it but now I realize that I don't have da/dt and db/dt...what do I do?
Answers
Answered by
Reiny
I can see from your data that you labeled your triangle with a, b, and c
a being the vertical (fixed at 320) , b the horizontal and c the hypotenuse (the string)
so c^2 = b^2 + 320^2
you also found correctly that if c= 400, b = 240
differentiate ...
2c dc/dt = 2b db/dt + 0
dc/dt = b db/dt/c
= 240(15)/400 = 9
check my arithmetic
a being the vertical (fixed at 320) , b the horizontal and c the hypotenuse (the string)
so c^2 = b^2 + 320^2
you also found correctly that if c= 400, b = 240
differentiate ...
2c dc/dt = 2b db/dt + 0
dc/dt = b db/dt/c
= 240(15)/400 = 9
check my arithmetic
Answered by
Reiny
I was going to mention, that since a is a constant, da/dt = 0 as seen above.
Answered by
Naz
Ohh thank you-that is what I had overlooked: the fact that the derivative of a constant is = to 0 and that is why I did not need da/dt. Thank You I understand where I went wrong now.
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