- #26

lanedance

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Imagine one half of the equilateral triangle, by Pythagorean

[tex] b^2 = (b/2)^2 + h^2 [/tex]

[tex] 3b^2/4 = h^2 [/tex]

then

[tex] b^2 = 4.h^2/3 [/tex]

so

[tex] b(h) = \frac{2h}{\sqrt{3}} [/tex]

so the area is

[tex] A(h) = \frac{h.b(h)}{2} = \frac{2h^2}{2 \sqrt{3}}= \frac{h^2}{ \sqrt{3}}[/tex]

differentiating

[tex] \frac{d}{dt} A(h) = \frac{d}{dt} \frac{h^2}{ \sqrt{3}} = \frac{2h}{ \sqrt{3}}\frac{dh}{dt}[/tex]