Problem

A kite flies with a constant height of 60 meters above ground. The speed is 2ms2ms2 \frac ms. How fast is the line released the moment 100100100 meters of it is released?

My attempt

I visualized this as a right triangle, where the hypotenuse LL is the line, the horizontal edge is xx with x′=2msx’ = 2\frac ms and the vertical edge is 60m60\mathrm m.

I tried creating a function, x(L)=√L2−602x(L) = \sqrt{L^2 – 60^2}

which gives x′(L)=L√L2−602x'(L) = \frac{L}{\sqrt{L^2 – 60^2}}

From here I’m unsure what to do. What I want to know is L′L’ when L=100L = 100 right?

This is where I’m stuck. Any help appreciated!

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1 Answer

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Note that the length of the horizontal edge as a function of time is x(t)=2tx(t)=2t. Hence by the Pythagorean theorem: L(t)=√602+x(t)2=√602+4t2.L(t)=\sqrt{60^{2}+x(t)^{2}}=\sqrt{60^{2}+4t^{2}}.

Let’s see when L(t)=100L(t)=100: 100=√602+4t2⇔t=40.100=\sqrt{60^{2}+4t^{2}} \Leftrightarrow t=40.

And at t=40t=40 the line is released with speed L′(40)=8⋅402√602+4⋅402=1.6 ms.L'(40)=\frac{8\cdot 40}{2\sqrt{60^{2}+4\cdot40^{2}}}=1.6\ \frac{m}{s}.

Ah, see that’s where I mislead myself. I should have treated each length as a function of time, rather than just have them as functions of each other. Thanks!

– Alec

2 days ago