Finding how fast the area is changing of an equilateral triangle

  • #26
lanedance
Homework Helper
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I get a slightly different value for b, though your derivative is looking good now

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]
 
  • #27
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Your b value may look different from mine, but as far as I can tell by checking h=5, both our b values are exactly the same.
 
  • #28
lanedance
Homework Helper
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yep - you're correct, i wasn't 100% how to read ur post, got it now

but as
[tex] \frac{2.\sqrt{3}}{3} = \frac{2}{ \sqrt{3}} [/tex]
 
Last edited:
  • #29
HallsofIvy
Science Advisor
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I, on the other hand would have used the Pythagorean theorm, mainly because that is how I remember the cosine of 60 degrees! The perpendicular from one vertex to the opposite side divides an equilateral triangle into two right triangles haveing hypotenuse of length b and two legs of lengths h and b/2. [itex]b^2= h^2+ b^2/4[/itex] or [itex]h^2= b^2- b^2/4= (3/4)b^2[/itex] so [itex]h= (\sqrt{3}/2) b[/itex] or [itex]b= (2\sqrt{3}/3)h[/itex].
 

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