This problem is impossible.
 

The area of an equilateral triangle is increasing at the rate of 4 in²/min. At what rate is one side increasing when the area is 16in²?

Been working on it for like 5 hours now. Can't figure it out. About to throw someone through a wall.

I assume you're in Calc and working with derivatives and related rates?

< EDIT >BLAH, BLAH, BLAH. THERE WAS A TYPO IN MY FIRST ANSWER. GIMME A MINUTE TO REWRITE IT.< /EDIT >

A great online resource for math, especially college (Algebra, Calc I/II/III and DE) is "Paul's online math notes" (apparently a professor at Lamar Univ in Texas)

Here's a link to his Calc I notes.
http://tutorial.math.lamar.edu/Classes/CalcI/CalcI.aspx

Here's his main site.
http://tutorial.math.lamar.edu/

< edit > I screwed up again, thanks dad911 < /edit >

This is more difficult, but it still works out the same way. Getting the measure of a side when the area is 16 is messier with the correct formula (assuming you can't use a calculator and give the answer in decimals).

16 = √3/4•s²

s = 8/fourth root of 3
s' = looking for
A = 16
A' = 4

A = √3/4•s²
A' = √3/2•ss'

s' = 2A'/(s•√3)

s' = 8/(8/(fourth root of 3)•√3)

Assuming I didn't screw up again....

s' = 1/fourth root of 3

Sheesh, I hope that's right, but having to use the fourth root would be unusual for most math classes

Nerd. :)

what masraum said
steve

Damn, that was my line ;)

Hahaha, yeah.

This is home

http://forums.pelicanparts.com/uploa...1240269257.jpg

If I can't get the company to spring for a couple of better monitors (1280x1024 left and 1440x900 right) I'm going to have to buy my own.

Notice the titanium 911 con rod, G50.50 pinion (From Hank Watt's Green Monster), and 2.7L P&C (pen holder).

I need a Mg intake valve cover and standard steel rod. I also want to bring in my old fan and shroud. You can never be surrounded by enough greasy old car parts.

Equilateral is all 3 sides equal. Height is not equal to base......

area is represented as:

http://forums.pelicanparts.com/uploa...1240269702.gif

http://www.mathwords.com/a/area_equilateral_triangle.htm

how do you type that square root symbol?

Crap!

Hahahah, damn!

hold down the <alt> key and type 251 on the number pad.

Nah, I cheated. Uploaded the graphic from the mathworlds site.

A few beers less, and I would try to figure out the answer, maybe tomorrow.... but then again, a few beers more, and I would have agreed with masraum(steve)

Actually the answer is 1(3^1/4) in/minute.

• Yes, the area of the equilateral triangle is ½s².
• s=4√2
• dA/dt = 4 in²/min
• So if A=½s² , dA/dt = 2(½)s(ds/dt)
• simplify: dA/dt=s(ds/dt), 4=4√2(ds/dt),
• ds/dt = 1/√2 which is WRONG and why I can't figure out how to do this problem and want to stab someone in the throat.

Hmm that Area equation is probably where I was going wrong too.... WHY WAS GEOMETRY 17 YEARS AGO?!

This forum/thread may shed some light:
http://www.physicsforums.com/showthread.php?t=297625

Dude that was your 911th post. :)

I first attempted using Pythagoreus but got stuck a hundred and twenty one different ways.

Ok, Use Heron's formula, (not Pythagorean) to get the area, and you will get the same area formula that I showed above. http://en.wikipedia.org/wiki/Heron%27s_formula

1. rate of increase of side length is a function of rate of increase of area

2. area (per se) = 16

3. all sides equal ====> only need to know about any one side

you need the formula to derive the area from the length of a side

then rearrange to focus on side length

differentiate

if non-linear, evaluate the rate at #2 above

& you are done

go for a walk when frustrated

good luck and you can tell your prof. that I did NOT give you the answer.

I KNOW the process. I can't do the math or put the picture together.

Well, after doing it multiple times (obviously making stupid errors along the way), I think I got it.

A = (s²√3)/4

s= 8/(3^¼)

A'=(√3/4)(2s)(s')

4=(√3/4)(2)[8/(3^¼)](s')
2=(√3/4)(8/(3^¼)](s')
(8/√3)=[8/(3^¼)](s')
(8/(3^¼)) / (8√3) = s'
s' = 3^¼ / 3^ ½ = 3^(-¼)

s' comes out to be 1/(3^¼).

Wow am I dumb.

Ok, it is late and no time to check this, but what do you think of this:

We know equilateral triangle's area A = (3^0.5)/4 * S^2 where S is length of one side.
Write side of triangle as a function of time, S = S(t). We don't know what the S(t) formula is, but no matter.
So can write area as function of time, A(t) = (3^0.5)/4 * [S(t)]^2

Now, differentiate dA(t)/dt. Using the chain rule, we know
dA(t)/dt = dA(S)/dS * dS(t)/dt
The power rule gives us dA(S)/dS = (3^0.5)/4 * 2 * S(t) = (3^0.5)/2 * S(t)
Since we don't know what the S(t) formula is, we do not know exactly what the dS(t)/dt formula is, but no matter.
The chain rule gives us dA(t)/dt = (3^0.5)/2 * S(t) * dS(t)/dt

We are looking at the instant when A = 16. So
16 = (3^0.5)/4 * [S(t)]^2
[S(t)]^2 = 16 *4 / (3^0.5) = 64 / (3^0.5)
S(t) = 8 / (3^0.25)

And at this instant, dA(t)/dt = 4. So
4 = (3^0.5)/2 * S(t) * dS(t)/dt
Substitute in S(t) = 8 / (3^0.25)
And get 4 = [ (3^0.5)/2 ] * [8 / (3^0.25)] * dS(t)/dt
4 = 4 *(3^0.25) * dS(t)/dt
dS(t)/dt = 1/(3^0.25)

Sorry, I just read your initial post then posted an answer, I didn't see that you'd already solved it. Good job.

What was interesting to me is that we have no idea if the side is growing linearly, logrithmically, etc - the solution is a general one.

Oh, I see others beat me to it also. Never mind . . .

Thanks for the answers though. I have no idea what was going wrong when I was doing it. I might have just been the fractions in the exponents, or forgetting the chain rule. I dunno. But you figured it out and it didn't take you 8 hours I'm guessing.

The fractional exponents are confusing.

Thanks for reminding me why I decided to become an engineering MANAGER.:D

Hah :)

HA AWESOME! Few more to go until the 924th, then 928th, then 944th! Woohoo! Porsche can't keep up with me!

and THAT is why I studied chemistry

What?!? It's interesting, but awful. Entropy and Enthalpy, etc, etc.... Blech!

Something about using numbers as in this problem gives me awful brain cramps. Chemistry, even P-Chem was not as painful for me.

I never learned much chemistry. My vague recollection is that it involved way too much memorization for me. For math and physics, you only have to memorize a few formulas and constants (maybe today's students are permitted to have them all programmed into their calculators for all I know).

The awful thing about physics and engineering, IMO, is that you have to carry over numbers and decimals and units for pages and pages, and at the end you've dropped some decimal place 30 lines ago, or have ended up with an area in cubic meters, and you have to go back and check every all over again.

Much easier to do math. It is usually clear when you have the answer. The theorem is proved, or it isn't.

In my line of work the airplane either flies or it doesn't. Simple.:D

Yeah but you don't find out by trial and error - or do you . . .

if it doesn't fly, I bet there will be a trial!