Question

The sides of an equilateral triangle are increasing at the rate of 2 ft/min.
Find the rate of change of the Perimeter and Area of the triangle when the side is 3 ft.

please help me with this one....

Answer 1

optimization?

Answer 2

i dont know.. :(

Answer 3

what class are you in?

Answer 4

it's a lesson in my calculus subject...

Answer 5

yeah it is optimization, I would recommend watching videos from patrickjmt on youtube.

http://patrickjmt.com/optimization-problem-1/

Answer 6

So they gave us this information.
Let's think about the perimeter as a function of the side length.\[\Large\rm P(x)=3x\]Right?

Answer 7

I added up the 3 sides, x+x+x, that step ok? :o

Answer 8

then...?

Answer 9

Ummm we want to take the derivative of our function `with respect to time`.
So what will happen with our x when we do that?

Answer 10

Remember, we're not differentiating with respect to x.

Answer 11

\[\Large\rm P'(x)=?\]

Answer 12

Where you at jesssss >.< You can do itttttt

Answer 13

If this is confusing, we can switch to Leibniz notation.\[\Large\rm P=3x\]And we want to find:\[\Large\rm \frac{dP}{dt}\]

Answer 14

3 dx?

Answer 15

Ok good!\[\Large\rm P'(x)=3(1)x'\]

Answer 16

P' <--- is this the rate of change of the perimeter?

Answer 17

Yes,
\[\Large\rm P'=\frac{dP}{dt}\]the rate of change of perimeter over time

Answer 18

so i'll just substitute 2 in P'(x) = 3(1)x' ...?

Answer 19

x'=2
dx/dt=2

This is how the length is changing over time.
So plug that in first of all :O

Answer 20

And then figure out P'(3).
The rate of change of perimeter over time when the side length is 3.

Answer 21

I hope I don't trick you up the way I worded that >.<
since there are no x's to plug in for...

Answer 22

P'(3) is 6...?

Answer 23

Ok great! :)
So we've found the rate of change of the perimeter.
It turns out it's independent of the side lengths.
it only depends on the rate of change of the side.

Answer 24

So for area, ummmmm area of an equilateral triangle... i better look that one up >.<

Answer 25

Google is telling me that it's,\[\Large\rm A(x)=\frac{\sqrt3}{4}x^2\](We don't want another variable, so we needed a formula that doesn't involve height).

Answer 26

I suppose we could have used Heron's Formula to derive that, but Google saves time >.< heh

Answer 27

So what do you think we'll do here? c:

Answer 28

differentiate the Area, then plug in x' ... then solve for A'(3)...? :)

Answer 29

Yah good c:
Since we have a square on x, it looks like our area function will end up depending on both x and x'.

So we'll have more stuff to plug in this time.

Answer 30

Do eet! >.< tell me what you get!

Answer 31

\[3\sqrt{3}\] ...?

Answer 32

Mmm good that's what I'm coming up with also \c:/ yay team!
Hopefully we didn't make any boo boo's along the way :U
Seems pretty straight forward.

Answer 33

what is the unit for A'...? ft^2/min. ..?

Answer 34

ft/min for P' ...?

Answer 35

Ummm yah I guess that would make sense.
I'm trying to remember how the units pop out when we do it the legit way...
Oh I guess plugging in the values is where we get the dimensions from.\[\Large\rm A'(3)=\frac{\sqrt3}{2}\left(3ft\right)\left(2\frac{ft}{\sec}\right)\]

Answer 36

/min* soz

Answer 37

thanks for your help! :)

Answer 38

\c:/

Answer 39

I missed the show! :(