Realted Rates Problem help!

A girl flies a kite at a height of 200 ft, the wind carrying the kite horizontally away from her at a rate of 4 ft/sec. How fast must she let out the string when the kite is 428 ft away from her?

Question

Realted Rates Problem help!

A girl flies a kite at a height of 200 ft, the wind carrying the kite horizontally away from her at a rate of 4 ft/sec. How fast must she let out the string when the kite is 428 ft away from her?

anonymous · Answer

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anonymous · Answer

another pythagoras

$$200^2+x^2=y^2$$
$$2xx'=2yy'$$
$$x'=4$$ so 
$$4x=yy'$$ and you want $y'$ when $y=428$

anonymous · Answer

Nice work satellite.  @payal0912: also bear in mind that when you are taking the derivative of x and y, they are both functions of time.  Related rates always are implicitly defined as functions of time.

anonymous · Answer

so, y'=dy/dt

anonymous · Answer

take care

anonymous · Answer

I'm still confused on where to get the x value from

anonymous · Answer

hi paya are you in calculus class or in algebra or trig ?

anonymous · Answer

calculus

anonymous · Answer

there are two unknowns in this equation. i tried using the pythagorean to get x but my answer is incorrect. 

I got $$\sqrt{143184}$$ as my x value using y=428 and z=200 for z^2 + x^2 = y^2. but then using that x value in the differentiation, I got an incorrect answer