Check my logic...and derivatives

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saxything43
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Check my logic...and derivatives

Postby saxything43 » Thu Oct 04, 2007 1:26 am UTC

The problem is, I have to use derivatives to find the rate of change of a triangle's area in respect to its side length(the triangle is equilateral). Because the triangle is equilateral, I know that as the length of one leg changes, the others change with it to match it exactly. This would mean that the formula I COULD use to find the area of an equilateral triangle would be

s^2/2

which would be read as "s squared over two."

The order of the squaring and dividing would not matter even if one ignored the order of operations...so i'm pretty sure I have that much right.

Let's take the derivative to find the instantaneous rate of change!

According to my logic, I can turn my equation into 1/2 times s^2 and it's the same thing.

Applying the power rule, the derivative of the above equation is simply s. (read "s".)

BUT the back of the book (where the answers are hidden) has the first formula as sqrt3/4 times s^2 and the derivative of that as sqrt3/2 times s.

Am I really that far off? What am I doing wrong? Help!

OH and I forgot to read the rules for this section of the forums, so if I'm breaking some rule by posting for help with homework, feel free to delete this thread, mods.

Thanks
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CodeLabMaster
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Re: Check my logic...and derivatives

Postby CodeLabMaster » Thu Oct 04, 2007 1:40 am UTC

The general policy on homework is to give hints, not answers, so I'll follow suit.

I'm guessing you derived s2/2 from a*b/2, right? That formula only works in certain cases. You can't just take any two sides of a triangle, mutliply them, then divide by 2. In order to use the a*b/2 thing to find the area, you have to have a triangle with one of its angle at 90 degrees. Try using this to derive that formula for an equilateral triangle (all angle at 60), then take the derivative.
Last edited by CodeLabMaster on Thu Oct 04, 2007 1:42 am UTC, edited 1 time in total.

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Vaniver
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Re: Check my logic...and derivatives

Postby Vaniver » Thu Oct 04, 2007 1:40 am UTC

Your formula for the area of a triangle is wrong. It's bh/2- b is s, but h is not s- it's sqrt(3)/2 s. Using that, you get the area formula that the book has, and then using your steps, you get the answer that the book has.

Whenever you get an answer that's different from the book but similar, divide them and you'll get the scale that you're off by. If it's something easily recognizable like 2 or sqrt(3)/2, that's a hint as to where you may have messed up.
Last edited by Vaniver on Thu Oct 04, 2007 1:42 am UTC, edited 1 time in total.
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Token
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Re: Check my logic...and derivatives

Postby Token » Thu Oct 04, 2007 1:41 am UTC

Easy mistake to make. The formula for the area of a triangle that you're using is 1/2 x base x height, right? The base of the equilateral triangle is s, but the height is not. Call the height h. To find it, we just use the Theorem of Pythagoras.

(s/2)2 + h2 = s2

Solve that for h, and you get the correct value to plug in to the equation.
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Re: Check my logic...and derivatives

Postby saxything43 » Thu Oct 04, 2007 1:50 am UTC

OH! 1/2 base times height only works for right triangles.....ah. That's where my problem lies....thanks a ton guys. I must've reworked that thing ten times until I gave up. And now, off to recalculate.
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Re: Check my logic...and derivatives

Postby Pathway » Thu Oct 04, 2007 3:48 am UTC

saxything43 wrote:OH! 1/2 base times height only works for right triangles.....ah. That's where my problem lies....thanks a ton guys. I must've reworked that thing ten times until I gave up. And now, off to recalculate.


Nah, 1/2 base times height definitely works for all triangles. The problem was that s was your base, but the height was not s.

Draw the triangle out. Then draw a line from the top corner to the middle of the opposite sides.

That divides your triangle into two right triangles. Pick either of them. Since you know the hypotenuse is equal to s, and one side is equal to s/2, you can use the Pythagorean Theorem to calculate the third side. That third side is the height of your equilateral triangle.

s2 = (s/2)2 + (height)2

That gives you sqrt( s2 - (s/2)2 ) = height

So height = sqrt( s2 - s2/4) = sqrt(3s2/4) = sqrt(3)/sqrt(4) * s = s * sqrt(3)/2

Then 1/2 * base * height = 1/2 * s * s * sqrt(3)/2
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Re: Check my logic...and derivatives

Postby Nimz » Thu Oct 04, 2007 4:01 am UTC

There is, of course, another way to get the formula for the area (there's always another way to do just about anything in maths!). Heron's Formula doesn't require you to know the height. Just the three sides, which are all s.
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